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SS 5.1 The Greatest Common Factor and Factoring by Grouping
Recall the Greatest Common Factor – The largest number that two or more
numbers are divisible by
Example:
18
36
Example:
12
10
24
We are going to be extending this idea with algebraic terms. The trick is to:
1) Find the numeric GCF
2) Pick the variable raised to the smallest power
3) Multiply number and variable and you get the GCF
Example:
x2, x5, x
Example:
x2y, x3y2, x2yz
Example:
8 x3, 10x2, -16x2
Example:
15 x2y3, 20 x5y2, -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!
Steps in Factoring by GCF
1) Find the GCF of all terms
2) Rewrite as GCF times the sum of the quotients of the original
terms divided by the GCF
Example:
8 x2  4x2 + 12x
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Example:
18a  9b
Example:
27 a2b + 3ab  9ab2
Note: Sometimes a greatest common factor can itself be a polynomial.
Example:
t2(t + 2) + 5(t + 2)
Example:
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 polynomial in each term
that we will then be able to factor out as in the last two examples.
Steps in Factoring By Grouping
1) Group similar terms and factor out a GCF from each grouping
(keep in mind the aim is to get a polynomial that is the same out of
each grouping – look for a GCF)
2) Factor out the like polynomial and write as a product
Hint: Trinomials are prime for factoring with the GCF and a polynomial
with 4 terms is prime for this method
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Example: 2zx + 2zy  x  y
Example:
xy  2 + 2y  x
Example:
b2 + 2a + ab + 2b
Your Turn
Example: xy2 + y3  x  y
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SS 5.2 Factoring Trinomials of the Form x2 + bx + c
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
+ 3x
+ 2



xx
2+1
21



st
nd
Product of 1 's
Sum of 2 's
Product of 2nd 's
Because this pattern exists we will use it to attempt factoring trinomials of
this form
Steps to Factoring Trinomials of the Form x2 + bx + c
1) Start by looking at the constant term (including its sign). Think of
all it's possible factors
2) Find two factors that add to give middle term's coefficient
3) Write as
(x + 1st factor)(x + 2nd factor)
4) Check by multiplying
Example:
x2 + 5x + 6
1) Factors of 6?
2) Which add to 5?
3) (x +
)(x +
)
Example: x2 + x  12
1) Factors of -12?
2) Which add to 1?
3) (x +
)(x +
)
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Example: x2  5x + 6
1) Factors of 6?
2) Which add to -5 ?
3) (x +
)(x +
)
Example: x2  x  12
1) Factors of -12?
2) Which add to -1?
3) (x +
)(x +
)
Example: x2 + xy  2y2
1) Factors of 2y2?
2) Which add to 1y?
3) (x +
)(x +
)
Note: If 2nd term and 3rd term are both positive then factors are both
positive.
If 2nd term and 3rd are both negative or 2nd term is positive and 3rd
term is negative then one factor is negative and one is positive.
If the 2nd term is negative and 3rd is positive then both factors are
negative.
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Example: x2 + 8x + 15
1) Factors of 15?
2) Which add to 8?
3) (x +
)(x +
)
Example: x2  2x  15
1) Factors of -15?
2) Which add to -2 ?
3) (x +
)(x +
)
Example: x2 + x  6
1) Factors of -6?
2) Which add 1?
3) (x +
)(x +
)
Example: x2  17x + 72
1) Factors of 72?
2) Which add to -17?
3) (x +
)(x +
)
Example: x2  3xy  4y2
1) Factors of -4 y2?
2) Which add to -3y?
3) (x +
)(x +
)
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 number can add to be the 2nd numeric coefficient.
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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.
If there is a variable common factor in all terms try to factor out that first.
Example:
2x2 + 10x + 12
Example:
5x2 + 10x  15
Example:
7x2  21x + 14
Example:
x3  5x2 + 6x
Your Turn
Example: 9x2  18x  27
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Example:
2x2y + 6xy  4y
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SS 5.3 Factoring Trinomials of the Form
ax2 + bx + c
We will be using the same pattern as with x2 + bx + c, but now we have
an additional factor to look at.
Example: 2x2 + 4x + 2
1) Look at the factors of the 1st term
2) Look for 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?
Example: 10x2 + 9x + 2
1) Factors of 10?
2) Factors of 2?
3) Product of factors that sum to 9?
Example: 15x2  4x  4
1) Factors of 15?
2) Factors of -4?
3) Product of factors that sum to -4?
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Your turn
Example: 2x2 + 5x  3
1) Factors of 2?
2) Factors of -3?
3) Product of factors that sum to 5?
Example: 12x2 + 7x + 1
1) Factors of 12?
2) Factors of 1?
3) Product of factors that sum to 7?
Sometimes we will see some of the special patterns that we talked about in
chapter 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
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, etc.
2) The last term is a perfect square
3) The numeric coefficient of the 2nd term is twice the product of the
1st and last terms coefficients
4) Rewrite as:
(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.
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Example: x2 + 6x + 9
1) Square root of 1st term?
2) Square root of last term?
3) Twice numbers in two and three?
4) ( x +
)2
Example: x2  4x + 1
1) Square root of 1st term?
2) Square root of last term?
3) Twice numbers in two and three?
4) ( x +
)2
Your Turn
Example: x2 + 10x + 25
1) Square root of 1st term?
2) Square root of last term?
3) Twice numbers in two and three?
4) ( x +
)2
Notice 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!!
Factoring a Trinomial by Grouping
1) Find the product of the 1st and last numeric coefficients
2) Factor the product in one so that the sum of the factors is the 2nd
coefficient
3) Rewrite the trinomial as a four termed polynomial where the 2nd
term is now 2 terms that are the factors in step 2
4) Factor by grouping
5) Rewrite as a product
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Example: x2 + 12x + 36
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 2 that sum to 12
4) Factor by grouping
5) ( x +
)(
x +
)
Example: 2x2 + 13x  7
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 2 that sum to 13
4) Factor by grouping
5) ( x +
)(
x +
)
Your Turn
Example: 10x2  13xy  3y2
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 2 that sum to 12
4) Factor by grouping
5) ( x +
)(
x +
)
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Example: 2x2 + 4x + 2
1a) There is a common factor – remove it first
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 2 that sum to 2
4) Factor by grouping
5)
( x +
)( x +
)
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SS5.4 Factoring Binomials
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.
Steps for Factoring the Difference of Two Perfect Squares
1) If I have a difference binomial check if the 1st term coefficient is a
perfect square
2) Check if 2nd term is a perfect square
3) Yes to both 1 and 2 then
(1st term + 2nd term) (1st term  2nd term)
Example: x2  16y2
1) Square root of 1st term
2) Square root of 2nd term
3) ( x +
)( x 
)
Example: 4x2  81
1) Square root of 1st term
2) Square root of 2nd term
3) ( x +
)( x 
)
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Your Turn
Example: 25x2  4
1) Square root of 1st term
2) Square root of 2nd term
3) ( x +
)( x 
)
Example: 16x2  4y2
1) Square root of 1st term
2) Square root of 2nd term
3) ( x +
)( x 
)
Sometimes there may be a factor that needs to be factored out 1st, so we
factor it out and then follow the same steps!
Example: 2x2  32
1a) Factor out common term
1) Square root of 1st term
2) Square root of 2nd term
3)
( x +
)( x 
)
Example: 32x2  72
1a) Factor out common term
1) Square root of 1st term
2) Square root of 2nd term
3) ( x +
)( x 
)
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What if our binomial has a cubed term?
Then we have the sum or difference of two perfect cubes
Pattern
(a + b)(a2  ab + b2) = a3 + b3
(a  b)(a2 + ab + b2) = a3  b3
If the cube is the sum of two cubes then the binomial is the sum and the
second term of the trinomial is negative.
If the cube is the difference of two cubes then the binomial is the difference
and the trinomial is the sum of all terms.
Recall:
13 =1, 23 =8, 33 =27, 43 =64, 53 =125, 63 =216, etc.
The reverse of a cube is called the cube root, just as the reverse of a square is
called the square root. A cube root is written 3x3 = x
Factoring a Sum or Difference of Perfect Cubes
1) What is the cube root of the 1st term
2) What is the cube root of the 2nd term?
3) Put in the following form
(31st term 
2nd term)[( 31st term)2 + (31st term )(32nd term) + (31st term)2]
or
3
st
3
nd
( 1 term + 2 term)[( 31st term)2  (31st term )(32nd term) + (31st term)2]
3
Example: x3  125
1) What is cube root of the 1st coefficient?
2) What is the cube root of the 2nd coefficient?
3) ( x 
)(
x2 +
x +
)
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Example: 8x3 + 64
1a) Factor out common factor
1) What is cube root of the 1st coefficient?
2) What is the cube root of the 2nd coefficient?
3)
(
x 
)(
x2 
x +
)
Example: 8x3  27y3
1) What is cube root of the 1st coefficient?
2) What is the cube root of the 2nd coefficient?
3) (
x 
)(
x2 +
x
+
)
Your Turn
Example: 2x3  16
1a) Factor out common factor
1) What is cube root of the 1st coefficient?
2) What is the cube root of the 2nd coefficient?
3) (
x 
)(
x2 +
x +
)
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SS 5.5 Choosing a Factoring Strategy
1) GCF – This removes a monomial from an entire polynomial and
sometimes will accomplish factoring by grouping if the GCF is a
binomial
Example: 4t2 + 36
2) Factoring by Grouping – This is a great way to factor a polynomial with
4 terms
2
Example: 4x  8xy  3x + 6y
3) Factoring Trinomials of the Form x2 + bx + c or ax2 + bx + c
Example: x2 + x  12
Example: 3x2 + 10xy  8y2
4) Factoring a Perfect Square Trinomial -- If the 1st and last term are perfect
squares and the middle term is twice the product of the 1st and last terms
then it is a perfect square trinomial
Example: 16a2  56ab + 49b2
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5) Factoring Trinomial by Grouping -- If you can't factor a trinomial
easily, this is a method that will allow you to find the factors more easily.
Example: 42a2  43a + 6
6) Factoring a Binomial -- If the binomial is:
1) The sum of 2 perfect squares it is prime
2) The difference of 2 perfect squares check
(a + b)(a  b)
3) The sum of 2 perfect cubes
4) The difference of 2 perfect cubes
Example: 4a2 + b2
Example: (x  y)2  z2
Example: x3y3 + 8z3
Example: 125  8y3
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SS5.6 Solving Quadratic Equations
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. Let's
practice.
Example:
Put
x2  2x = 5
into standard form.
Hint: The trick is to get zero on the right side.
Example: 2x + 5 = x2  2
Put all terms on the left and leave zero on the right.
Your Turn
Example: 15  2x = x2  5x + 3
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Our next task is solving a quadratic equation. Just as with an 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 only, it will have up to
two solutions!! In order to solve quadratics we must factor them! This is
why we learned to factor!
Steps to Solving Quadratic Equations
1) Put the equation in standard form
2) Factor the polynomial
3) Set each term that contains a variable equal to zero and solve for the
variable
4) Write the solution as: variable = or variable =
5) Check
Example:
x2  4x = 0
Example:
x2  6x = 16
Example:
x2 = 4x + 3
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Your Turn
Example: 8x2  8x = 6
Example:
8x + 3 = - 4x2
Just as with a linear equation, a quadratic equation in 2 variables, such as
y = ax2 + bx + c
can be graphed. The graph of a quadratic equation
in two variables is a parabola. A parabola looks like a right-side up or upside down u in this section.
We will explore the graph of a parabola slightly by exploring it's xintercepts. A quadratic in two variables will have zero, one or two xintercepts. The intercepts are the solutions to the quadratic equation where
the y-variable is set equal to zero and the x-variable is solved for.
Example: y = x2 + 2x + 1
1) Set y equal to zero
2) Solve the quadratic equation
3) x-intercepts are the solutions
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Example:
y = (2x + 1)(x  1)
Your Turn
Example: y = x2  2x  8
124
SS5.7 Quadratic Equations and Problem Solving
The whole reason that we've learned to solve quadratic equations is because
many things in our world can be described by a quadratic equation. If you
are going on into physics or chemistry or any field which requires these
studies you will need to solve quadratic equations.
Example: Find the dimensions of a rectangle whose length is twice its
width plus 4 if its area is 10 square inches.
One of the legs of a right triangle can be found if you know the quadratic
equation:
a2 + b2 = c2 where a and b are the legs and c is the
hypotenuse. (This is called the Pythagorean Theorem)
Example: Find the missing length in the following triangle
3
4
Example: One number is 3 more than another number. If their
product is 54, find the numbers.
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Example: The product of two consecutive even numbers is 48. Find
the numbers.
Example: A company can manufacture x hundred items per month at
a cost of C = 500 + 600x  100x2 Find the number of items it can
produce for a total cost of $1300.
Your Turn
Example: One leg of a right triangle is 2 more than twice the other. The
hypotenuse is 13 meters. Find the lengths of the two legs.
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Example: The product of two consecutive odd integers is seven more than
their sum. Find the integers.
Example: A manufacturer knows that the number of items he can sell each
week, x, is related to the price of each item, p, by the equation
x = 900  100p. What price should he charge for each item in order to have
a weekly revenue of $1,800? (Revenue is xp)
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