Download §5.1 Exponents and Scientific Notation Definition of an exponent ar

§5.1 Exponents and Scientific Notation
Definition of an exponent
ar =
Example:
a)
34
Expand and simplify
b)
(1/4)2
c)
(0.05)3
d)
(-3)2
Difference between (-a)r and –ar
(-a)r =
–ar =
Note: The 1st says use –a as the base, the second says the opposite of the answer of ar. The reason that this
is true is because the exponent only applies to the number to which is written to the right of and –a is –1  a
and therefore the –1 isn't being raised to the "r" power.
Example:
a)
(-5)2
Example:
a)
(-3)2
Name the base, exponent, rewrite using repeated multiplication, and
simplify to a single number.
b)
-52
Simplify
-32
b)
c)
- 0.252
Definition of zero exponent if a  0
a0 =
Anything to the zero power is 1. Be careful because we still have to be certain what the
base is before we rush into the answer!
Example:
a)
20
d)
Simplify each by removing the negative exponent.
b)
(1/2) 0
c)
-2x0
-10  10
Y. Butterworth
e)
2x0  y0
Ch. 5 – Martin-Gay/Greene
d)
(7 + x)0
1
Product Rule for Exponents
ar as =
Example:
a)
x2 x 3
Example:
Use Product Rule of Exponents to simplify each of the following. Write
the answer in exponential form.
b) (-7a2 b)(5ab)
c) 32  37
d)
y2  y3  y7  y8
Simplify using the product rule
y 2m + 5 y m  2
Definition of a negative exponent (Shorthand for take the reciprocal of the base)
a-r =
Anytime you have a negative exponent you are just seeing short hand for take the
reciprocal. When a negative exponent is used the negative in the exponent reciprocates
the base and the numeric portion tells you how many times to use the base as a factor. A
negative exponent has nothing to do with the sign of the answer.
Example:
a)
2 –1
Simplify each by removing the negative exponent.
b)
(1/2) –1
When a negative exponent has a numeric portion that isn't one, start by taking the
reciprocal of the base and then doing the exponent (using it as a base the number of times
indicated by the exponent).
Example:
a)
2 –2
Simplify each by removing the negative exponent.
b)
(1/2) –3
Don't let negative exponents in these bother you. Copy the like bases, subtract (numerator
minus denominator exponents) the exponents and then deal with any negative exponents. If
you end up with a negative exponent it just says that the base isn't where it belongs – if
it's a whole number take the reciprocal and if it is in the denominator of a fraction taking
the reciprocal moves it to the numerator. Let's just practice that for a moment.
Example:
a)
a –2
Y. Butterworth
Simplify each by removing the negative exponent.
b)
1
b–3
Ch. 5 – Martin-Gay/Greene
2
The Quotient Rule of Exponents
ar =
as
Example:
a)
a 12
a7
d)
2b
16 b3
Simplify each. Don't leave any negative exponents.
b)
2b5
c)
x8
3
b
x2
e)
(a + b) 15
(a + b)7
f)
2a3b2
4ab3
Now, let's practice the quotient rule with negative exponents. You have 2 choices in
doing these problems: 1) Use the quotient rule on integers (can be tricky if your integer
subtraction is not what it should be)
or 2) Remove the negative exponents and use
product/quotient rules as needed.
Example:
a)
a8
a10
d)
2 b–3
4
Example:
Simplify each. Don't leave any negative exponents.
b)
2
c)
x -8
–3
b
x -2
e)
(a + b) 5
(a + b)-7
Simplify using the quotient rule
f)
2a3b-2
ab–3
15x 7n  5
3x 2n  3
Our next topic is an application of exponents that makes writing very large and very
small numbers much easier, especially in application. Think of scientific notation as
what you see every day in written expression of large numbers – like in a newspaper
article. You might see 5,000, 000 written a 5 million. This is the same thing that
scientific notation does, but it uses multiplication by factors of ten, which are decimal
point movers.
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
3
Scientific Notation
When we use 10 as a factor 2 times, the product is 100.
102 = 10 x 10 = 100
second power of 10
When we use 10 as a factor 3 times, the product is 1000.
103 = 10 x 10 x 10 = 1000
third power of 10.
When we use 10 as a factor 4 times, the product is 10,000.
104 = 10 x 10 x 10 x 10 = 10,000
fourth power of 10.
From this, we can see that the number of zeros in each product equals the number of
times 10 is used as a factor. The number is called a power of 10. Thus, the number
100,000,000
has eight 0's and must be the eighth power of 10. This is the product we get if 10 is used
as a factor eight times!
Recall earlier that we learned that when multiplying any number by powers of ten that
we move the decimal to the right the same number of times as the number of zeros
in the power of ten!
Example :
1.45 x 1000 = 1,450
Recall also that we learned that when dividing any number by powers of ten that we
move the decimal to the left the same number of times as the number of zeros in the
power of ten!
Example :
5.4792  100 = 0.054792
Because we now have a special way to write powers of 10 we can write the above two
examples in a special way – it is called scientific notation .
Example :
1.45 x 103 = 1,450 ( since 103 = 1000 )
Example:
5.4792 x 10-2 = 0.054792
( since 102 = 100 and
[ 102 ]-1 = 1 which
100
means divided by 100)
Writing a Number in Correct Scientific Notation:
Step 1: Write the number so that it is a number  1 but < 10 (decimals can and will be used)
Step 2: Multiply this number by 10x ( x is a whole number ) to tell your reader where the
decimal point is really located. The x tells your reader how many zeros you took
away! (If the number was 1 or greater, then the x will be positive, telling your reader that you
moved the decimal to the right to get back to the original number, otherwise the x will be
negative telling the reader to move the decimal left to get back to the original number.)
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
4
Example : Change 17,400 to scientific notation.
1) Decimal
1 7 4 0 0
2) Multiply
x 10
Example : Write 0.00007200 in scientific notation
1) Decimal
0 0 0 0 7 2 0 0
2) Multiply
x 10
Example :
Change each of the following to scientific notation
a)
8,450
b)
104,050,001
d)
0.00902
e)
0.00007200
f)
c)
34
0.92728
Note: When a number is written correctly in scientific notation, there is only one number to the left of the
decimal. Scientific notation is always written as follows: a x 10x, where a is ≥1 and <10 and x is an
integer.
We also need to know how to change a number from scientific notation to standard form.
This means that we write the number without exponents. This is very simple, we just
use the definition of scientific notation to change it back – in other words, multiply the
number by the factor of 10 indicated. Since multiplying a number by a factor of 10
simply moves the decimal to the right the number of times indicated by the # of zeros,
that’s what we do! If the exponent is negative, this indicates division by that factor of 10
so we would move the decimal to the left the number of times indicated by the
exponent.
Example :
Change 7.193 x 105 to standard form
1) Move Decimal to the Right ________ times.
2) Giving us the number …
Example : Change 6.259 x 10-3 to standard form.
1) Move Decimal Left ____ times
2) Giving us the number …
Example:
Write each of the following to standard form.
a)
-7.9301 x 10-3
b) 8.00001 x 105
c) 2.9050 x 10-5
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
d) -9.999 x 106
5
§5.2 More Work with Exponents and Scientific Notation
This section just takes what is left of exponent rules and scientific notation and discusses
those topics. We have power rules and their combinations with other rules to discuss
about exponents. In scientific notation, we need to discuss how to multiply and divide
and use this in applications.
Power Rule of Exponents
(ar)s =
or
(ab)s =
or
(a/b)r =
Example:
a)
(a3)2
Use Power Rule of Exponents to simplify each of the following. Write the
answer in exponential form.
b)
(10xy4)2
c)
(-75)2
d)
(-3a/5b)3
Note: A negatvie number to an even power is positive and a negative number to an odd power is negative.
Now that we have discussed all the rules for exponents, all we have left is to put them
together. Let's practice with some examples that use the power rule, the product rule and
also the quotient rule. Use the power rule 1st, then the product rule and finally the
quotient rule. Deal with the negative exponents last.
Example:
a)
Use the properties of exponents and definitions to simplify each of the
following. Write in exponential form.
2
2
(x y) (xy2)4
b)
(-8)3 (-8)5
c)
(3/2)3  34
2
2
2 5
d)
[ ]
g)
[ ]
9x y
15xy8
9x2y5
15xy8
Y. Butterworth
e)
-2
h)
2 3
(2xy )
4x2y
[ ]
f)
[ ]
i)
(2xy3)-2
xy2
2 3
-5x y
4x-1
-3
-5x2y3
4x-1
Ch. 5 – Martin-Gay/Greene
6
Your Turn
Example:
Simplify each. Don't leave any negative exponents.
a)
(ab) 8
b)
(-2xy)3
c)
x -8
10
–3
a
3x y
x -2
-3
[ ]
d)
(5xy2)2 (4x -1y -3) -3
5x –3 (2 x3 y -5) -2
f)
2a3b-2
ab–3
e)
x –3  x –5  x7
Now, for the next step with scientific notation – it can used to multiply and divide
large/small numbers. This is really quite easy. Here is some explanation and how we can
do it!
What happens if we wish to do the following problem,
7 x 102 x 103 = (7 x 102)(1 x 103)
We can think of 102 and 103 as "decimal point movers." The 102 moves the decimal
two places to the right and then the 103 moves the decimal three more places to the right.
When we are finished we have moved the decimal five places to the right. This is an
application of the product rule.
Steps for Multiplying with Scientific Notation:
Step 1: Multiply the whole numbers
Step 2: Add the exponents of the "decimal point movers", the factors of 10.
Step 3: Rewrite in scientific notation where the number multiplied by the factor of 10 is
 1 but < 10.
Before we begin practicing this concept, I want to practice a skill. I want to learn to write
a number in correct scientific notation.
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
7
Steps for Writing in correct scientific notation
Step 1: Write the number in correct scientific notation
Step 2: Add the exponent of the new “number’s factor of 10 and the one at the start.
Example:
Write in correct scientific notation.
a)
14.4 x 105
b) 105.4 x 10 -3
c) 0.0005 x 1015
d) 0.098 x 10 -4
Example : Multiply and write the final answer in correct scientific notation.
a)
(3 x 102 ) ( 2 x 104)
b)
(2 x 10-2 ) (3 x 106)
c)
(1.2 x 10-3 ) (12 x 105)
d)
(9 x 107 ) (8 x 10-3)
Note: In part c) & d) once you multiply the numbers you have a number that is greater than 10 so it must
be rewritten into correct scientific notation by thinking about the number that 14.4 x 10 10 actually
represents and changing that to scientific notation.
Steps for Dividing with Scientific Notation:
Step 1: Divide the whole numbers
Step 2: Subtract the exponents of the "decimal point movers" (numerator minus
Denominator exponents)
Step 3: Rewrite in scientific notation where the number multiplied by the factor of 10 is
 1 but < 10.
Example: Divide using scientific notation. Be sure final answer is in correct sci. note.
a)
( 9 x 105 )
b)
( 2.5 x 107 )
c)
( 2 x 10 -2 )
2
5
( 3 x 10 )
( 2.5 x 10 )
( 1.5 x 105 )
Example:
Perform the indicated operation using scientific notation.
a)
3000 x 0.000012
b)
0.0007 x 11,000
400
0.001 x 0.0001
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
8
§5.3 Polynomials and Polynomial Functions
First, we need to have some definitions. You should already know several, and several
may be new, make an effort to learn vocabulary as math is like a foreign language!
Constant – A number
Numeric Coefficient – A number multiplied by a variable
Term – A part of a sum (constant, variable, variable times a constant) separated by
addition/subtraction symbol
Like Terms – Terms which have the same variable component (must be to same powers)
Polynomial – A term or the sum of two or more terms of the form axn (an algebraic
expression in only one variable)
Monomial – A polynomial with a single term.
Binomial – A two termed polynomial.
Trinomial – A three termed polynomial.
Degree of a Term – The sum of the powers of the variables in a term. The degree of a
constant is zero.
Degree of a Polynomial – The degree of the highest degreed term within the polynomial.
Example:
Each of the following form algebraic expressions (some polynomials). I want
to discuss each in terms of the above vocabulary: constant, variable,
numeric coefficient, and type of polynomial.
b)
2x2 + 2x
c)
2(x  5) + 3x
a)
3x
d)
x2 + 5x + 5
/2
Example:
Are the following like terms? Why or why not?
3
a)
-5x & 25x
b)
/4xy & 3/8 x2y
d)
54x
/3 &
Example:
a)
b2
-5 & 5x
d)
1
-54xy
/3
What is the degree of the term?
b)
2 b3
c)
x2 y2z
Example:
What is the degree of the polynomial?
2
a)
a + 3a + 5
b) 3a + 4 a3  2a2  6
Y. Butterworth
c)
c) ab2 + 3a2b3 + 2a2  4
Ch. 5 – Martin-Gay/Greene
9
Method 1: Adding/Subtracting Polynomials (Horizontal – Combining Like Term)
Step 1: Remove grouping symbols (This is really distribute the subtraction!!)
Step 2: Group Like Terms
Step 3: Combine Like Terms
Example:
Add/Subtract the following polynomials/algebraic expressions.
2
a)
½ x + 5y  1/3 x2 + 3y
b)
2(x  5) + 3x
c)
[(8xy2 + 2x) + (-7xy + 3)] – (xy2 + 3x)
e)
(7xy2  2x + 3)  (5x2y  4x  9)
f)
½ (x  y)  ¼ (x + 3)  ½ y
d)
x2 + 5x + 5
Note: You can not clear an expression of fractions as you can an equation. You must deal with them as
they are.
There is another way to think about adding and subtracting polynomials. This is
columnar addition and subtraction. We must really focus on ordering the polynomial to
do this. Ordering a polynomial means putting the terms in order of descending degree.
Example:
Order the following polynomials/algebraic expressions
2
a)
7x + 9x3  x
b)
xy  9 + xy2
Method 2: Adding/Subtracting Polynomials (Algebraic Expressions) in Columns
Step 1: Order polynomials being added or subtracted (leave blanks for missing degrees)
Step 2: Remove subtraction
Step 3: Stack in columns (like terms over one another, leave blanks where there are no like terms)
Step 4: Add
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
10
Example:
a)
Add/Subtract the following polynomials/algebraic expressions
(7x  2x2 + 3)  (5 + x2  2x)
b)
(9x2  9) + (x2 + x + 7)
c)
(2/3x2  9)  (1/5x2 + 2 x  1/3)
d)
Subtract 3x + 3 from 5x2 + 4x – 7
A polynomial function is a function that contains a polynomial.
P(x) = axn + axn–1 + axn–2 + … + axn–(n–1) + a
Finding the Value of a Polynomial Function(Evaluating an polynomial/alg. exp.)
Step 1: Leave blanks for variables
Step 2: Fill the blanks with the value of the variable, the number inside the P(?)
Step 3: Simplify the resulting numeric expression using order of operations
Example:
For P(x) = (x2  3x + 3) find P(-1)
Example:
For P(x) = (x2  5)2 find P(4)
Example:
For P(x) = 2(x + 2) – 3 find P(x + h)
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
11
We can even use this in terms of application problems:
Example:
The average cost of a college education per year for a full-time student in
two year public colleges can be approximated by the polynomial function
D(x) = 6.4x2 + 37.9 x + 2856.8 for the years 1984 through 2006.
Predict how much it will cost in 2010: D(26). Round to the nearest
dollar. (Martin-Gay, Beginning Algebra, 5th Edition, p. 323 #111)
Oh, and life would not be complete without the use of function notation in adding and
subtracting polynomials. Just substitute and simplify!
Example:
If P(x) = 4x2 – 6x + 3 & Q(x) = 5x2 – 7 find the following
a)
2[P(x)] + 7[Q(x)]
b)
Q(x) – P(x)
Also in review for us this section is the Quadratic and Cubic functions that we saw in
§2.1. The reason they are in review in this section is for the interest in their degree. A
quadratic function, P(x) = ax2 + bx + c, is of degree 2 and all polynomial functions of
degree 2 will be quadratic and will have a graph that looks like a parabola. A cubic
function, P(x) = ax3 + bx2 + cx + d, is of degree 3 and all polynomial functions of degree
3 will be cubic and will have a graph that looks like a lazy s. Also under review is the bit
of information that we learned in §2.3, about the y-intercept of functions being the
constant value. You will need to review these two concepts from these two sections, to
complete problems like the following and problems like those found in exercises 89-92:
Example:
a)
Give the coordinates of the y-intercept of the graph of each equation and
describe the shape of the graph.
f(x) = x3 – 6x + 4
b)
g(x) = -2x2 + 11x – 3
c)
h(x) = 5x2 – 2x + 1
Y. Butterworth
d)
Ch. 5 – Martin-Gay/Greene
m(x) = -x3 – 3x2 + 2x – 5
12
§5.4 Multiplying Polynomials
Multiplying a monomial by a monomial is really an application of the properties of
exponents combined with the associative property. I will review slightly, but this is not
something that should be difficult.
Monomial x Monomial
Multiply the coefficients
Add the exponents of like bases
Example:
Find each product.
a)
(5a2)(-6ab2)
b)
(-r2s2)(3ars)
Multiplying polynomials is an application of the distributive property. This is also called
expanding.
Distributive Property Review
Monomial x Polynomial
Example:
Multiply.
2
a)
2x (x + 2x + 3)
a ( b + c ) = ab + ac
b)
-4x2 y(x2y  2xy + 3y)
Binomial x Binomial
Now we'll extend the distributive property further and to help us remember how we will
have an acronym called the FOIL method.
F
=
( a + b ) ( d + c ) = ad + ac + bd + bc
F
O
I
L
First
O
=
Outside
I
=
Inside
L
=
Last
}
These add & are like terms if the binomials are alike.
Example:
a)
Multiply. Think about only sum & product of the constants to get the
middle & last terms of your polynomial.
(x + 2) (x + 3)
b)
(x − 5)(x + 9)
c)
(x − 4)(x − 3)
Y. Butterworth
d)
Ch. 5 – Martin-Gay/Greene
(x + 7)(x − 9)
13
Note: A binomial of the form (x + c) multiplied by a binomial of the same form will always yield a
trinomial. The first terms yield the highest degree term(of degree 2), the inside and outside add to give the
middle term and the last will yield the constant.
Example:
a)
Multiply. Remember that it is the outside & inside products that
will add & the last & first terms multiply to give first and last terms of the
trinomial.
(x  5) (2x + 3)
b)
(3x + 4)(x + 5)
c)
(2x − 5)(x − 4)
d)
(2x − 5)(x + 3)
e)
(2x + 3)(5x + 4)
f)
(2x − 4)(3x + 5)
g)
(2x + 4)(3x − 5)
h)
(3x + 5) (4x + 1)
Example:
Multiply.
(x2 + y) (x  y)
Note: This didn’t yield a trinomial because the binomials are not of the same form.
Example:
Multiply.
a)
(5x + 1)(3x + 1)
b)
(4x − 1)(3x − 1)
(5x − 1)(6x + 1)
d)
(8x + 1)(9x − 1)
c)
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
14
Note: These are just like multiplying when the numeric coefficient of the x is one. The only differences are
that the leading coefficient is the product, not the constants, and you must watch the sign on the middle
term a little more carefully. Product of x coefficients gives leading coefficient & sum/difference of the x’s
coefficients gives the middle term (just watch if positive or neg. since that comes from outside & inside products).
Example:
Find the product.
a)
5x(x + 2)(x  2)
b)
-3r(5r + 1)(r + 3)
Note: The 1st example is possibly more easily done by doing the product of the binomials 1st and then
multiplying by the monomial. Always remember that multiplication is commutative and associative so you
can choose which product you find 1st.
Special Products
Squaring a Binomial (Returns a Perfect Square Trinomial)
(a + b)2 = a2 + 2ab + b2
Example:
Expand.
a)
(2x + 3)2
d)
(a – b)2 = a2 – 2ab + b2
or
b)
(2x – 3)2
c)
(2x + 3y)2
[(3x + 2) + y]2
Product of Conjugates (Returns the Difference of Two Perfect Squares)
(a + b)(a – b) = a2 – b2
Example:
Expand.
a)
(3x + 5)(3x – 5)
b)
(2x – y)(2x + y)
c)
[(x + 1) + 3][(x + 1) – 3]
d)
(4/3 x + 5)(4/3 x – 5)
Polynomial x Polynomial
Multiplying two polynomials is an extension of the distributive property. The first term
of the first polynomial gets multiplied by each term of the second, and then the next term
of the first gets multiplied by each term of the second and so on and the end result is the
sum of all the products.
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
15
Of course there is a second way to do the distributive property and it takes care of
keeping all the degrees of variables in order. This method looks like long multiplication.
All polynomials must be ordered for this method to be successful. I will do one of the
examples below using both methods, and all others using just the new method.
Example:
Multiply using long multiplication.
a)
(2x + 3) (x2 + 4x + 5)
b)
c)
(x2 + 2x  7) (x2  2x + 1)
(2x2 + 4x  8)(1/2 x + 3)
In this last example we have to simplify by multiplying the first two polynomials out and
then multiplying that result by the third polynomial. We’ll only do this example one
way, but I will use one method on the first 2 and then the other method on the result times
the last. It is interesting to point out that sometimes it is useful to use the commutative
property to rearrange the order thus deciding the two you will multiply first, as some
binomial products are much simpler than others!
Example:
Simplify
a)
(x + 1)(2x  3)(x + 1)
Y. Butterworth
b)
Ch. 5 – Martin-Gay/Greene
(2x + 1)(3x + 2)(2x – 1)
16
Now for some application of our polynomial functions.
Example: For P(x) = x2 + 5 and Q(x) = x2 – 2 find
a)
[P(x)]2
b)
P(x)•Q(x)
Example:
For the following figure compute the area of the shaded region.
x–7
2x
Y. Butterworth
x
Ch. 5 – Martin-Gay/Greene
17
§5.5 The Greatest Common Factor and Factoring by Grouping
Recall that the greatest common factor(GCF) is the largest number that two or more
numbers are divisible by. A GCF isn’t a big number at all, it is a little piece of a number
and it is the biggest, little piece that all the numbers have in common. The largest a GCF
can ever be is the smallest number.
Finding Numeric GCF
Step 1: Factor the numbers.
a) Write the factors in pairs so that you get all of them starting with 1  ? = #
Step 2: Find the largest that both have in common.
Example:
Find the GCF of the following.
a)
18 & 36
b)
12, 10 & 24
We are going to be extending this idea with algebraic terms. The steps are:
1) Find the numeric GCF (negatives aren’t part of GCF)
2) Pick the variable(s) raised to the smallest power that all have in common (this
btw also applies to the prime factors of the numbers)
3) Multiply number and variable and you get the GCF
Example:
For each of the following find the GCF
2 5
a)
x ,x ,x
b)
x2y, x3y2, x2yz
c)
8 x3, 10x2, -16x2
d)
15 x2y3, 20 x5y2z2, -10 x3y2
Now we'll use this concept to factor a polynomial. Factor in this sense means change
from an addition problem to a multiplication problem! This is the opposite of what we
did in multiplying a monomial by a polynomial.
Factoring by GCF
Step 1: Find the GCF of all terms
Step 2: Rewrite as GCF times the sum of the quotients of the original terms divided by
the GCF. Be sure that your polynomial is the same length as the polynomial that
you started with.
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
18
Let's start by practicing the second step of this process. Dividing a polynomial that is
being factored by its GCF.
Example:
For each of the following, factor using the GCF
3
a)
8 x  4x2 + 12x
b)
27 a2b + 3ab  9ab2
c)
18a  9b
d)
108x2y2  12x2y+ 36xy2 + 96xy
e)
1
f)
-x2 + x – 1
/7 x3 + 4/7 x
Note: When the GCF involves fractions with like denominators the GCF of the numerators is what matters
and the denominator tags along. Remember when dividing by a fraction it is multiplying by the reciprocal.
Note2: It is considered unpalatable for the leading coefficient of a polynomial to be negative, and it is
therefore desirable to factor a negative one from such a polynomial.
Sometimes a greatest common factor can itself be a polynomial. These problems help it
make the transition to factoring by grouping a lot easier.
Example:
a)
In the following problems locate the 2 terms (the addition that is not in
parentheses separate terms), and then notice the binomial GCF, and
factor it out just as you did in the previous problems
t2(t + 2) + 5(t + 2)
b)
5(a + b) + 25a(a + b)
Our next method of factoring will be Factoring by Grouping. In this method you rewrite
the polynomial so that terms with similar variable(s) are grouped together. This type of
factoring will take some practice, because the idea is to get a polynomial which will have
a binomial in each term that we will then be able to factor out as in the last two examples.
Factoring By Grouping
Step 1: Group similar terms and factor out a GCF from each grouping (keep in mind the aim
is to get a binomial that is the same out of each grouping(term) – look for a GCF)
Step 2: Factor out the like binomial and write as a product (product of 2 binomials)
Hint: Trinomials are prime for factoring with the GCF and a polynomial with 4 terms is prime for this
method
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
19
Example:
a)
In the following problems factor out a GCF from binomials in such a way
that you achieve a binomial in each of the resulting terms that can be
factored out.
8x + 2 + 3y2 + 12xy2
b)
2zx + 2zy  x  y
Note: In b), to get the binomial term to be the same you must factor out a negative one. This is the case in
many instances. The way that you can tell if this is the case, look at your binomials if they are exact
opposites then you can factor out a negative one and make them the same.
c)
b2 + 2a + ab + 2b
d)
xy  2 + 2y  x
Note for c): Terms must be rearranged to factor a GCF from a binomial. There are several different
possibilities, so don't let it worry you if you would have chosen a different arrangement. [b 2 + 2b + 2a
+ ab is one and b2 + ab + 2a + 2b is the other]
Note for d): Terms must be rearranged to factor a GCF from a binomial. There are several different
possibilities, so don't let it worry you if you would have chosen a different arrangement. [xy  x + 2y 
2 is one and xy + 2y  x  2 is the other]
Note 2 for d): In addition to rearranging the grouping the order of the terms can also be rearranged in
each grouping resulting in the necessity to see that terms are commutative, when they are added to one
another. [xy  x  2 + 2y, results in x(y  1) + 2(-1 + y) or if you factored out a -1, then it looked
really different x(y  1)  2(1  y), but you can recognize that (1  y) is the opposite of (y  1) and get
the situation turned in your favor!]
e)
6x2y + 15x2  6xy  15x
f)
5x2y + 10xy − 15xy − 6y
Note for e)&f): These one is a bit trickier still! It has a GCF 1 st and then factoring by grouping.
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
20
Your Turn
1.
a)
Find the GCF of the following
28, 70, 56
b)
x2, x, x3
d)
x2y3, 3xy2, 2x
2.
a)
Factor the following by factoring the GCF
28x3  56x2 + 70x
b)
9x3y2  12xy  9
c)
15a2  60b2
d)
a2(a + 1)  b2(a + 1)
3.
a)
Factor the following by grouping
xy2 + y3  x  y
b)
5x2 + x  y  5xy
c)
12y3 + 9y2 + 16y + 12
Y. Butterworth
e)
c)
3x3, 15x2, 27x4
12xy2, 20x2y3, 24x3y2
Ch. 5 – Martin-Gay/Greene
21
§5.6 Factoring Trinomials
It is important to point out a pattern that we see in the factors of a trinomial such as this:
(x + 2)(x + 1)
= x2

xx

Product of 1st 's
+ 3x
+ 2


2+1
21


nd
Sum of 2 's
Product of 2nd 's
Because this pattern exists we will use it to factor trinomials of this form.
Factoring Trinomials of the Form x2 + bx + c
Step 1: Start by looking at the constant term. Think of all it's possible factors
Step 2: Find two factors that add/subtract to give middle term's coefficient
a)
They will ADD if the constant’s sign is positive
b)
They will SUBTRACT if the constant’s sign is negative
Step 3: Write as
(x ± 1st factor)(x ± 2nd factor) ;where x is the variable in question &
signs depend upon last & middle terms’
signs (c is positive both will be the same
as middle term, c is negative larger factor
gets middle terms’ sign)
Step 4: Check by multiplying
Example:
x2 + 5x + 6
1) Factors of 6?
2) Which add to 5?
3) Write as a product of 2 binomials.
Example:
x2  x  12
1) Factors of -12?
2) Which add to -1?
3) Write as a product of 2 binomials.
Example:
x2 + x  12
1) Factors of -12?
2) Which add to 1?
3) Write as a product of 2 binomials.
Example:
x2 + xy  2y2
2
1) Factors of 2y ?
2) Which add to 1y?
3) Write as a product of 2 binomials.
Example:
x2  5x + 6
1) Factors of 6?
2) Which add to -5 ?
3) Write as a product of 2 binomials.
Note: If 2nd term and 3rd term are both
positive then factors are both positive.
If 2nd term and 3rd are both negative or
2 term is positive and 3rd term is negative then
one factor is negative and one is positive.
nd
If the 2nd term is negative and 3rd is
positive then both factors are negative.
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
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Example:
a2 + 8a + 15
1) Factors of 15?
2) Which add to 8?
3) Write as a product of 2 binomials.
Example:
x2 + x  6
1) Factors of -6?
2) Which add 1?
3) Write as a product of 2 binomials.
Example:
z2  2z  15
1) Factors of -15?
2) Which add to -2 ?
3) Write as a product of 2 binomials.
Example:
x2  17x + 72
1) Factors of 72?
2) Which add to -17?
3) Write as a product of 2 binomials.
Example:
x2  3xy  4y2
1) Factors of -4 y2?
2) Which add to -3y?
3) Write as a product of 2 binomial
Sometimes it is just not possible to factor a polynomial. In such a case the polynomial is
called prime. This happens when none of the factors of the third term (constant usually) can
add to be the 2nd numeric coefficient.
Example:
x2  7x + 5
If the leading coefficient (the first term in an ordered polynomial) is not one, try to factor out a
constant first, then factor as usual. In this section, any time the leading coefficient is not
1, there will be a GCF, but that is not always true in “the real world.”
If there is a variable common factor in all terms try to factor out that first.
Example:
Factor completely.
2
a)
2x + 10x + 12
b)
5x2 + 10x  15
7x2  21x + 14
d)
x3  5x2 + 6x
c)
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
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Sometimes the common factor is a more than a number and a variable, sometimes it is the
product of several variable and sometimes it is even a binomial. Here are some
examples.
Example:
Factor each completely. (Warning: Sometimes after you
factor out the GCF you will be able to factor the remaining
trinomial, and sometimes you won't.)
a)
(2c  d)c2  (2c  d)c + 4(2c  d)
b)
x3z  x2z2  6z2
c)
(a + b)a2 + 4(a + b)a + 3(a + b)
Another new idea in intermediate algebra is the concept of factoring using substitution.
If we see a polynomial in “quadratic form” even though it may not look like a second
degree polynomial, it can still be factored. There are two types of substitutions:
1)
Higher degree than 2.
a)
Middle term’s exponent, times 2 must give exponent of
highest degreed term & there still must be a constant.
i) Use middle term’s variable as “x” in binomial
Example:
2)
x4 – 10x2 + 24
Binomial of descending degree
a)
2nd then 1st then zero degree
i) Use “u” to block out the binomial so you see
u2 ± u ± c and factor
ii) Re-substitute binomial for u and simplify
Example:
(x + 2)2 + 8(x + 2) + 15
Note: Don’t confuse this example with a binomial GCF! The binomial GCF is same degree in every term.
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
24
Next we tackle the trinomials where the leading coefficient is not 1. These are the
buggers! We are going to learn a trick first, and then we will come back to doing it the
“old –fashioned” way!
Factoring a Trinomial by Grouping
Step 1: Find the product of the 1st and last numeric coefficients
Step 2: Factor the product in one so that the sum of the factors is the 2nd coefficient
Step 3: Rewrite the trinomial as a four termed polynomial where the 2nd term is now 2
terms that are the factors in step 2
Step 3: Factor by grouping
Step 4: Rewrite as a product
1)
2)
3)
4)
Example:
12x2  11x + 2
Multiply the numeric coefficients of 1st and last terms
Factors of number from step 1? (Hint: Use the prime factors of a &c to help you!)
Rewrite as the four termed polynomial where the middle terms are factor from step 2
that sum to 11 (note the middle term is negative so both must be negative)
Factor by grouping
Example:
12x2 + 7x  12
1) Multiply the numeric coefficients of 1st and last terms
2) Factors of number from step 1?
3) Rewrite as the four termed polynomial where the middle terms are factor from step 2
that yield a difference of 7 (note the middle term is positive so the larger factor is positive and the
other is negative)
4) Factor by grouping
1)
2)
3)
4)
Example:
4x2 − 9x  9
Multiply the numeric coefficients of 1st and last terms
Factors of number from step 1?
Rewrite as the four termed polynomial where the middle terms are factor from step 2
that yield a difference of 9 (note the middle term is negative so the larger factor must be negative)
Factor by grouping
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
25
1)
2)
3)
4)
1)
2)
3)
4)
1)
2)
3)
4)
Example:
24x2 − 58x + 9
Multiply the numeric coefficients of 1st and last terms
Factors of number from step 1?
Rewrite as the four termed polynomial where the middle terms are factor from step 2
that yield a sum of 58 (note the middle term is negative so both must be negatives)
Factor by grouping
Example:
12x2 + 17x + 5
Multiply the numeric coefficients of 1st and last terms
Factors of number from step 1?
Rewrite as the four termed polynomial where the middle terms are factor from step 2
that yield a sum of 17
Factor by grouping
Example:
2x2 + 17x + 10
Multiply the numeric coefficients of 1st and last terms
Factors of number from step 1?
Rewrite as the four termed polynomial where the middle terms are factor from step 2
that yield a difference of 17
Factor by grouping
Note: Sometimes these will be prime too!
1)
2)
3)
4)
5)
Example:
24x2y2 – 42xy2 + 9y2
Factor out the GCF first
Multiply the numeric coefficients of 1st and last terms
Factors of number from step 2?
Rewrite as the four termed polynomial where the middle terms are factor from step 3
that yield a sum of 14
Factor by grouping, don’t forget to write the GCF
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
26
1)
2)
3)
4)
5)
6)
1)
2)
3)
4)
5)
6)
Example:
3(x + 2)2 – 19(x + 2) + 6
Make your substitution first to create 3u2 – 19u + 6
Multiply the numeric coefficients of 1st and last terms
Factors of number from step 2?
Rewrite as the four termed polynomial where the middle terms are factor from step 3
that yield a sum of 19
Factor by grouping
Re-substitute (x + 2) and simplify
Example:
14x6 – x3 – 3
Make your substitution first to create 14u2 – u – 3
Multiply the numeric coefficients of 1st and last terms
Factors of number from step 2?
Rewrite as the four termed polynomial where the middle terms are factor from step 3
that yield a difference of 1
Factor by grouping
Re-substitute x3
Now let’s return to the traditional method. Don’t be surprised if I skip this altogether and
return to it in a few days.
We will be using the same pattern as with x2 + bx + c, but now we have an additional
factor to look at, the first factor.
Factoring Trinomials of Form – ax2 + bx + c
Step 1: Find the factors of a
Step 2: Find the factors of c
Step 3: Find all products of factors of a & c (a1x + c1)(a2x + c2) where a1x  c2 and
c1  a2x are the products that must add to make b! (This is the hard part!!!) The
other choice is (a1x + c2)(a2x + c1) where a1x  c1 and c2  a2x must add to
make b. And then of course there is the complication of the sign. Pay attention
to the sign of b & c still to get your cues and then change your signs accordingly.
(But you have to do this for every set of factors. You can narrow down your possibilities by
thinking about your middle number and the products of the factors of a & c. If “b” is small, then
the sum of the products must be small or the difference must be small and therefore the products
will be close together. If “b” is large then the products that sum will be large, etc.)
Step 4: Rewrite as a product.
Step 5: Check by multiplying. (Especially important!)
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
27
Example:
2x2 + 5x + 2
1) Look at the factors of the 1st term
2) Look at the factors of the last term
3) Sum of product of 1st and last factors that equal the middle term
Ask yourself – What plus what equals my 2nd term?
Lucky here that the 1st and last terms are both prime that makes life very easy.
Example:
10x2 + 9x + 2
1) Factors of 10?
2) Factors of 2?
3) Product of factors that sum to 9?
9 is relatively small so we probably won’t be multiplying 10 and 2! This eliminates at least one
combination!
Since 2 is prime and we know that 10  2 won’t work that narrows our possibilities a lot!
Example:
15x2  4x  4
1) Factors of 15?
2) Factors of -4?
3) Product of factors that sum to -4?
The difference is relatively small, so I won’t be using 15  4 and 1  1 or 15  1 and 4  1, which
actually eliminates quite a bit, since the only other factors of –4 are –2 and 2, which means that
we just have to manipulate the sign.
Sometimes when factoring the leading coefficient will be negative. It is easier to deal
with problems that involve a negative coefficient if the negative is factored first and the
focus can return to the numbers and not deal with unfamiliar signs.
Example:
Y. Butterworth
Factor the following using techniques from this section
and by factoring out a –1 first.
-2a2  5a  2
Ch. 5 – Martin-Gay/Greene
28
Sometimes there will be a common factor in a trinomial, just like those found in
binomials in section 3. This does not change what we must do, but after factoring out the
binomial we must continue to look for factorization.
Your Turn
1.
Factor the following trinomials.
a)
x2 + 3x + 2
b)
y2 + 2y  15
c)
z2  12z  28
d)
r2  7r + 12
2.
a)
Factor each trinomial completely.
9x2  18x  27
b)
2x2y + 6xy  4y
c)
(2a + 1)a2  5(2a + 1)a  6(2a + 1)
d)
(2z + 1)z2  4(2z + 1)z  6(2z + 1)
3.
a)
Factor each of the following trinomials completely.
2x2 + 5x  3
b)
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
12x2 + 7x + 1
29
4.
a)
Factor the following completely.
15x2(r + 3)  34x(r + 3)  16(r + 3)
c)
6x2y + 34xy − 84y
5.
a)
Factor each of the following using factoring by grouping.
10x2  13xy  3y2
b)
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
b)
21x2 − 48x − 45
72x2  127x + 56
30
§5.7 Special Products
Sometimes we will see some of the special patterns that we talked about in §5.3, such as:
a2 + 2ab + b2 = (a + b)2
or
a2  2ab + b2 = (a + b)2
These are perfect square trinomial. They can be factored in the same way that we've been
discussing or they can be factored quite easily by recognizing their pattern.
Factoring a Perfect Square Trinomial
Step 1: The numeric coefficient of the 1st term is a perfect square
i.e. 1,4,9,16,25,36,49,64,81,100,121,169,225, 256 etc.
Step 2: The last term is a perfect square
Step 3: The numeric coefficient of the 2nd term is twice the product of the 1st and last
terms' coefficients’ square roots
Step 4: Rewrite as:
(1st term + last term )2 or (1st term - last term )2
Note: If the middle term is negative then it's the difference of two perfect squares and if it is positive then
it is the sum.
Note2: that whenever we see the perfect square trinomial, the last term is always positive, so if the last
term is negative don't even try to look for this pattern!!
1)
2)
3)
4)
Example:
x2 + 6x + 9
st
Square root of 1 term?
Square root of last term?
Twice numbers in two and three?
Factor, writing as a square of a binomial
1)
2)
3)
4)
Example:
4x2  12x + 9
st
Square root of 1 term?
Square root of last term?
Twice numbers in two and three?
Factor, writing as a square of a binomial
1)
2)
3)
4)
5)
Example:
32x2 + 80x + 50
GCF 1st
Square root of 1st term?
Square root of last term?
Twice numbers in two and three?
Factor, writing as a square of a binomial
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
31
Difference of Two Perfect Squares
Remember the pattern:
(a + b)(a  b) = a2  b2
Example:
(x  3)(x + 3) = x2  9
Now we are going to be "undoing" this pattern.
Factoring the Difference of Two Perfect Squares
Step 1: Look for a difference binomial and check
a) Is there a GCF? If so, factor it out and proceed with b) & c)
b) Is 1st term coefficient is a perfect square? (If no, then stop, problem is complete)
c) Is 2nd term is a perfect square? (If no, then stop, problem is complete)
Step 2: Yes to both b) and c) then factor the difference binomial in the following way
(1st term + 2nd term) (1st term  2nd term)
Step 3: If there was a GCF don’t forget to multiply by that GCF.
Example:
Factor completely.
a)
x2  y2
b)
d)
4x2  81
27z2  3y2
c)
z2  1/16
e)
12x2  18y2
Note: Sometimes there is a common factor that must be factored 1 st & sometimes after factoring a GCF,
the remaining binomial can’t be factored.
Any time an exponent is evenly divisible by 2 it is a perfect square. If it is a perfect
cube it is evenly divisible by 3, and so forth. So, in order to factor a perfect square
binomial that doesn’t have a variable term that is square you need to divide the exponent
by 2 and you have taken its square root.
Example:
Factor completely.
a)
16x4  81
b)
9x6  y2
Note: Sometimes the terms can be rewritten in such a way to see them as perfect squares. In order to see
x4 as a perfect square, think of (x2)2.
Sometimes we will need to use our factoring by grouping skills to factor special
binomials as well as just a GCF. Here is an example.
Example:
Y. Butterworth
Factor the following completely.
3x2  3y2 + 5x  5y
Ch. 5 – Martin-Gay/Greene
32
The SUM of Two Perfect Squares
a2 + b2
is
PRIME
unless there is a GCF
Example:
Factor completely.
a)
25x2 + 49y2
b)
25x2 + 100
Note: Part a is prime, and part b only factors because it has a GCF. Remember that even when the
polynomial you create is prime, if you’ve created a product, you have factored and you do not write prime.
Sum and Difference of Two Perfect Cubes
This is the third pattern in this section. The pattern is much like the pattern of the
difference of two perfect squares but this time it is either the sum or the difference of two
perfect cubes. At this time it might be appropriate to review the concept of a perfect cube
and or finding the cube root of a number. It may also be appropriate to review some
perfect cubes: 13=1, 23=8, 33=27, 43=64, 53=125, 103=1000
If you have the difference of two perfect cubes then they factor as follows:
a3  b3 = (a  b)(a2 + ab + b2)
If you have the sum of two perfect cubes then they factor as follows:
a3 + b3 = (a + b)(a2  ab + b2)
Factoring the Sum/Difference of Two Perfect Cubes
Step 1: Look for a sum or difference binomial and check
a) Is there a GCF? If so, factor it out and proceed with b) & c)
b) Is 1st term coefficient is a perfect cube? (If no, then stop, problem is complete)
c) Is 2nd term is a perfect cube? (If no, then stop, problem is complete)
Step 2: Yes to both b) and c) then factor the difference binomial in the following way
Where
a = cube root of the 1st term and
b = the cube root of the 2nd term
If the binomial is the difference
(a  b)(a2 + ab + b2)
If the binomial is the sum
(a + b)(a2  ab + b2)
Step 3: If there was a GCF don’t forget to multiply by that GCF.
Note: In the factored form, the binomial follows the binomial’s sign, and the middle term of the trinomial
is the opposite of the binomials sign, and the sign of the last term is always positive.
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
33
Example:
Factor each of the following perfect cube binomials.
3
a)
125x + 27
b)
27b3  a3
c)
24z3 + 81
d)
48x3  54y3
e)
x6 + 125
f)
x3y6  64
g)
(x + 1)3 – 8
h)
(2x – 1)3 + 27
One last type of special factoring for intermediate algebra: A difference of a perfect
square trinomial and a perfect square. This one is tricky because it is again a 4 termed
polynomial, and it is the first time we will have seen this pattern so we won’t be
expecting it.
Here’s what to look for:
1)
4 terms
2)
3 perfect squares, two of the same sign, and one of the opposite
a)
Usually 2 positive & one negative
Here’s what to do:
1)
Separate perfect square trinomial & perfect square
2)
Factor perfect square trinomial
3)
Factor difference of 2 perfect squares using substitution
4)
Re-substitute and simplify
Example:
Factor
2
a)
a + 2a + 1 – b2
Warning:
Y. Butterworth
b)
x2 – 18x + 81 – y2
Integrated Review follows this section. Homework due next class!
Ch. 5 – Martin-Gay/Greene
34
§5.8 Solving Equations by Factoring and Problem Solving
A quadratic equation is any polynomial which is a 2nd degree polynomial and is set equal
to zero. A quadratic is said to have the form:
ax2 + bx + c = 0
where a, b, c are real numbers and a  0.
A quadratic equation can be written other than in the above form, which is called
standard form, but it can always be put into standard form. To get a quadratic in standard
form, simply use the addition property of equality to move all terms to one side of the
equal sign. It is best to keep the leading coefficient positive. Let's practice.
Example:
Put the following into standard form.
2
a)
x  2x = 5
b)
c)
2x + 5 = x2  2
15  2x = x2  5x + 3
BTW: A polynomial function where P(x) = 0 is a quadratic in standard form. This fact
will come into play in a while.
Our next task is solving a quadratic equation. Just as with any algebraic equation such as
x + 5 = 0, we will be able to say that x = something. This time however, x will not
have just one solution, it will have up to two solutions!! In order to solve quadratics we
must factor them! This is one reason why we learned to factor. There is a property
called the zero factor property that allows us to factor a quadratic, set those factors
equal to zero and find the solutions to a quadratic equation. The zero factor property is
based upon the multiplication property of zero – anything times zero is zero. Thus if one
of the factors of a quadratic is zero then the whole thing is zero and that is the setup!
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
35
Solving Quadratic Equations
Step 1: Put the equation in standard form
Step 2: Factor the polynomial
Step 3: Set each term that contains a variable equal to zero and solve for the variable
Step 4: Write the solution as: variable = or variable =
Step 5: Check
Sometimes book exercises give you equations where step 1 or steps 1 and 2 have already
been done. Don’t let this fool you, the steps from there on are the same.
Example:
Solve the following:
a)
x  4x = 0
b)
x2  6x = 16
c)
x2 = 4x + 3
d)
-2 = -27x2  3x
2
Sometimes it is necessary to multiply a factor out in order to arrive at the problem in standard form. You
will realize that this is necessary when you see an equation that has one side that is factored but those
factors are equal to some number or when there are sums of squared binomials or squared binomials that
equal numbers.
e)
(2x  5)(x + 2) = 9x + 2
f)
y(2y  10) = 12
f)
a2 + (a + 1)2 = -a
There is also the case where we have a greatest common factor or which can be solved by factoring by
grouping. These all use the same principles.
Example:
a)
Solve each of the following by applying the zero factor
property to give the solution(s).
3x3 + 5x2  2x = 0
Y. Butterworth
b)
Ch. 5 – Martin-Gay/Greene
4y3 = 4y2 + 3y
36
c)
(2x + 1)(6x2  5x  4) = 0
d)
125x2 = 245
X-Intercepts of a Parabola
We already know that
y = ax2 + bx + c
can be graphed and that the graph of
a quadratic equation in two variables is a parabola. We have also discussed that this
equation will be increasing or decreasing according to the sign of a. We know that it has
a vertex and is symmetric [vertex is at (-b/2a, f(-b/2a))] about the line of symmetry which goes
through this vertex [and is therefore the line x = -b/2a]. What we have not discussed is that just
as with linear equations, parabolas also have intercepts. Recall that an x-intercept is a
place where the graph crosses the x-axis. Lines only do this at one point, but because of
the nature of a parabola, it is possible for this to happen twice (vertex is below x-axis and
parabola opens up, or vertex is above the x-axis and the parabola opens down), once [vertex is on x-axis]
or even never (vertex is above the x-axis). Just as with lines, the x-intercept is found by
letting y = 0 and solving for x. This gives us our relationship to quadratic equations,
when we look at a quadratic as a polynomial function, and set the function (the y-value)
equal to zero – P(x) = 0.
Finding X-Intercept(s) of a Parabola
Step 1: Let y = 0, if no y is apparent, set the quadratic equal to zero or let P(x) = 0
Step 2: Use skills for solving a quadratic to find x-intercept(s)
Step 3: Write them as ordered pairs (remember there could be one or two or even none)
Example:
Find the x-intercept(s) for the following parabolas
a)
y = (2x + 1)(x  1)
c)
y = x2  4
b)
y = x2 + 2x + 1
Pythagorean Theorem
The Pythagorean Theorem deals with the length of the sides of a right triangle. The two
sides that form the right angle are called the legs and are referred to as a and b. The side
opposite the right angle is called the hypotenuse and is referred to as c. The Pythagorean
Theorem gives us the capability of finding the length of one of the sides when the other
two lengths are known. Solving the Pythagorean theorem for the missing side can do
this. One of the legs of a right triangle can be found if you know the equation:
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
37
Pythagorean Theorem
a2 + b2 = c2
Solving the Pythagorean Theorem
Step 1: Substitute the values for the known sides into the equation
Note: a is a leg, b is a leg and c is the hypotenuse
Step 2: Square the values for the sides
Step 3: Solve using methods for solving quadratics or using principles of square roots
Example:
One leg of a right triangle is 7 ft. shorter than the other. The
length of the hypotenuse is 13 ft. Find the lengths of the legs.
Example:
Two cyclists, Tim and Joe, start at the same point. Tim rides
west and Joe rides north. At some time they are 13 miles apart.
If Tim traveled 7 miles farther than Joe, how far did each
travel?
Geometry Problems
Geometry problems are problems that deal with dimensions, so always remember that
negative answers are not valid. As with number problems (as encountered in Algebra) it is
possible to get more than one set of answers. Geometry problems that we will encounter
will deal with the area of figures and the Pythagorean Theorem. We've just discussed the
Pythagorean Theorem.
Example:
Y. Butterworth
Find the dimensions of a rectangle whose length is twice
its width plus 8. Its area is 10 square inches.
Ch. 5 – Martin-Gay/Greene
38
Projectile Problems
Projectiles follow a parabolic course. Therefore anything that is thrown or launched can
be expected to have a quadratic equation to describe its course. As a result, when the
object reaches the ground (keep in mind if there is a constant the ground has been adjusted, for
instance in the following example the ball is being thrown from a building or a hill or something that is 960
above ground level) the value of the function will be the same as if the function were equal
to zero (think about the symmetry of a parabola). It is also known that at the time that an object
reaches its highest point it has reached its vertex.
Example:
A softball thrown into the air travels in a parabola. Its height is a function
of the time from which it was thrown and is described by
h(t) = -16x2 + 64x + 960. Find the time it takes for the ball to reach the
the ground.
Note: The leading coefficient in a parabolic motion problem represents the acceleration
due to gravitational pull, the middle term’s coefficient represents the speed at which the
object is “launched” and the constant represents the height with respect to the ground
from which the object was launched.
WARNING: We just finished the chapter. The Chapter 5 test is due next class period!
Y. Butterworth
Ch. 5 – Martin-Gay/Greene
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