Download Chapters 6-10 POLYNOMIALS

Chapters 6-10 POLYNOMIALS
EXPRESSIONS
Examples: 3+2(1-4)2, -
EQUATIONS
Examples: 3x + 2 = 5, 5x
polynomial – is a term or sum of terms in which all variables have
whole number exponents. Example: 3x, or x2 + 1, or -3x2 + 3x + 1
monomial – a number, a variable, or a product of numbers and
variables.
Example: 3, 2x, -4x2 are all monomials.
binomial – the sum of two monomials that are unlike terms.
trinomial – the sum of three monomials that are unlike terms.
like terms – terms of a variable expression that have the same
variable and the same exponent.
Example: 3x and 3x2 are unlike terms, but 3x and 2x are like terms.
factor – (in multiplication) a number being multiplied.
Example: What are the factors of 121? 1, 11, and 121.
121 = 11 X 11, 121 = 1 X 121
to factor a polynomial – to write a polynomial as a product of other
polynomials
to factor a trinomial of the form ax2 + bx + c - to express the
trinomial as the product of two binomials.
Example: x2 + 5x + 6 = (x+2)(x+3)
to factor by grouping – to group and factor terms in a polynomial in
such a way that a common binomial factor is found.
Example: 2x(x+1) – 3(x+1) = (x + 1)(2x – 3)
factor completely – to write a polynomial as a product of factors that
are nonfactorable over the integers.
FOIL method – A method of finding the product of two binomials in
which the sum of the products of the First terms, of the Outer
terms, of the Inner terms, and of the Last terms is found.
Example: (x+2)(x+3) =x2+ 3x + 2x + 2*3 = x2 + 5x + 6
common factor – a factor that is common to two or more numbers.
Example: What are the common factors of 12 and 16x2?
The factors of 12x are 1,2,3,4,6,12, and x
The factors of 16x2are 1,2,4,8,16, x, x
The common factors of 12x and 16x2 are 1x, 2x, and 4x
The Greatest Common Factor of 12x and 16x2 is 4x.
3x+x, (3x+2)2
+ 5y=10, x(x-5) = -6
Can be simplified
Ex1: Don’t forget order of
operations!
3+2(1-4)2 can be simplified to
3+2(-3)2 which becomes
3+2(9) which becomes
3+18 which becomes 21.
Ex2:
-3x + x can be simplified to
-2x
Can be solved:
(a+b) = a + 2ab + b
5 y  5 x 10


5
5
5
y  x  2
Solution is all (x, y)
that make this linear
equation true.
When graphed
this is a line.
Ex3 :
x( x  5)  6
x 2  5 x  6
x 2  5 x 6  0
( x  3)( x  2)  0
x  3, x  2
Two solutions
Can be evaluated: (3x+2)2 can be
evaluated at x= -1. (3(-1)+2)2 is (3+2)2 which is (-1)2 which is 1.
2
Examples:
Difference of squares
a 2 - b 2 = (a+b)(a-b)
Rules for Variable Expressions:
Only like terms can be added, and when adding like terms, do not change the
exponent of the variable.
5x2 + 3x2 = 8x2
When multiplying variable expressions, add exponents of like variables
(5xy3)(2y2)=10xy3+2 = 10xy5
When taking powers of variable expression that is a monomial (one term),
multiply exponents of EVERY term inside the parentheses.
(2x3y4)3 = 23x3*3y4*3 = 8x9y12
When taking powers of a variable expression that is a binomial, trinomial or
some other polynomial, use the rules of polynomial multiplication.
For example: (x+2)2 ≠ x2 + 22
(x+2)2 = (x+2)(x+2) = x2+4x + 4 (FOIL METHOD)
Example 2:
(x2+3x+5)2 = (x2+3x+5)(x2+3x+5)
=(x2+3x+5)x2 + (x2+3x+5)3x + (x2+3x+5)5 (DISTRIBUTIVE
PROPERTY)
One solution
5 y  5 x  10
Can be reduced:
2
3x 3

3 3
x 1
Ex2 :
Perfect-Square Trinomials
2
Ex1 :
3x  5  2  3
20 4

25 5
25 x 2 y 3 5 x

3y
15 xy 4
Can be evaluated to see if a solution
is true.
Is (3,4) a solution of y=-x+2?
4 = -3+2=-1 NO
A General Strategy for Factoring a Polynomial
1.
Do all the terms in the polynomial have a common factor? If so,
factor out the
Greatest Common Factor. Make sure that you don’t forget it in
your final answer.
Example: 24x4 - 6x2 = 6x2(4x2 - 1). Also look to see if the other
polynomial factor and be factored more. (4x2-1)=(2x-1)(2x+1),
so the final answer is
24x4 - 6x2 =6x2(2x-1)(2x+1),
2.
After factoring out the GCF (if there is one) count the number of
terms in the remaining polynomial.
Two terms: Is it a difference of squares? Factor by using: a2-b2
= (a+b)(a-b)
Example: 36x2 – 49 = (6x)2 – 72 = (6x-7)(6x+7)
If the polynomial can’t be factored, it is PRIME.
SOLVING POLYNOMIAL EQUATIONS (2 SOLUTIONS):
Put Equation in standard form, ax2 + bx + c = 0.
3 methods:
1) SQUARE ROOT METHOD: If there is no bx term or if the equation
is in the form (x+ k)2 + c = 0, then just get the constant c on one side, take
± the square root of both sides and get x by itself. x 2  5  x   5
2) FACTORING METHOD: If product ac has two factors that add up
to b, then it is factorable. Factor it and use the Zero Product Product to find
the solutions. This says that if A*B=0, then A=0 or B=0. Example: x2 -2x = 3
Standard form: x2 – 2x + 3 = 0.
Factored: (x+1)(x-3) = 0
So x+1 = 0, which gives x = -1, or another possible solution is x-3 = 0,
which gives x = 3.
Three terms: Is it a perfect square trinomial?
If it is it would be in the form a2x2 + 2abx + b2 , which
is factored as (a+b)2
or a2x2 + 2abx + b2 which is factored as (a-b)2
Example: 4x2 + 12x + 9 = (2x)2 + 2(2)(3)x + 32 = (2x
+ 3)2
Is it of the form x2 + bx + c?
Factor by finding two numbers that multiply to c and add
to b.
Example: x2 -3x - 4 = (x+1)(x-4) because
1*-4 = -4 and 1 + -4 = -3
Can’t find the numbers? Maybe the polynomial is
PRIME.
Is it of the form ax2 + bx + c?
Try factoring by the Grouping Method (or ac Method) or
Trial and Error.
Example: 2x2 + 13x + 15 (the a*c method means
multiply 2*15 which is 30.
Find factors of 30 that add up to the middle term’s
coefficient, which in this case is 13. 3*10=30 and
3+10 = 13. Split the middle term into two parts:
2x2 + 10x + 3x + 15 and then factor by grouping.
2x(x+5)+3(x+5) = (2x+3)(x+5)
3) USE QUADRATIC FORMULA: If the equation ax2+ bx+ c =0 isnot
factorable, then use quadratic formula.
Quadratic Formula
if ax 2 + bx + c = 0
(standard form of a quadratic equation) then
x = ( -b
(b 2 - 4ac) ) / 2a
RATIONAL EXPRESSIONS
A rational expression is a fraction in which the numerator or denominator is
a variable expression (such as a polynomial). A rational expression is
undefined if the denominator has a value of 0.
A rational expression is in SIMPLEST form when the numerator and
denominator have no common factors other than 1.
6x
is not in simplest form.
9x2
2
is in simplest form.
3x
Reducing to simplest form – factor the numerator and denominator, then
cancel out any common factors in the numerator and denominator (not
common factors that are both in the numerator or both in the denominator, e.g.
side by side).
Those methods don’t work? Maybe the polynomial is
PRIME.
Four terms: Try Factoring by Grouping. Group the 1st two
terms and the last two terms. Factor out the Greatest Common
Factor from each grouping. Then factor out the common binomial
term.
3.
4.
Always factor completely. Double check that each of your factors
can not be factored more.
Multiplying Rational Expressions – factor the numerators and
denominators then cancel out common factors as above, then multiply the
numerators and multiply the denominators.
Dividing Rational Expressions – change to a multiplication problem by
changing the DIVISOR into it’s RECIPROCAL.
Check your work by multiplying the factors together. Does it
result in the original polynomial?
2
21 1
2  
5
52 5
Adding and Subtracting Rational Expressions –
x
4x  1
2

1
2x  x
2
Step 1: Factor the denominators, then find the LCM. The LCM of two
polynomials is the simplest polynomial that contains the factors of each
polynomial. To find the LCM of two or more polynomials, first factor each
polynomial completely. The LCM is the product of each factor the greater
number of times it occurs in any one factorization.
x
2 x  12 x  1

1
x(2 x  1)
Similar Triangles
Triangles are similar if at least 2
corresponding angles are the same
in each triangle.
1st denominato r : 2 x  12 x  1
2 nd denominato r : x(2 x  1)
LCM  x(2 x  1)2 x  1
Step 2: Change each rational expression so that the new denominator will be
the LCM. You will multiply the numerator and denominator of each
expression by whatever it takes to get the LCM as the new denominator.
x
1
(2 x  1)
x2
2x 1
 



2 x  12 x  1 x x(2 x  1) (2 x  1) x(2 x  1)(2 x  1) x(2 x  1)(2 x  1)
x
Step 3: Add the two new fractions by adding the numerators and keeping the
denominator (the LCM) the same.
x2  2x 1
x(2 x  1)(2 x  1)
Step 4: Now factor the resulting expression and cancel out any common
factors in the numerator and denominator.
x2  2x 1
( x  1) 2

x(2 x  1)(2 x  1) x(2 x  1)(2 x  1)
Simplify Complex Fractions – Complex fractions are just rational expressions
with fractions within fractions. To simplify, find the LCM of all the
denominators of every fraction in the expression, then multiply the main
numerator and denominator by that LCM. Then simplify as usual.
1
1
1
1
1( x 3 )  ( x 3 )
3
x3  x 2
x 2 ( x  1)
x 
x x 
x


 x2
1
1
1
1 x3
1
1
x 1
x 1
3
3


(
x
)

(
x
)
x 2 x3
x 2 x3
x2
x3
1
WORK
Rate of Work * Time Worked = Part of Tasked Completed
If someone can do a job in 60min, their rate of work is 1/60min.
If someone else can do the same job in 40minutes, their rate of work is
1/40min.
The TIME to get the same job done TOGETHER can be found by
Adding their parts together to make 1 whole job.
LCD= x3
Solving Equations with Fractions – multiply BOTH SIDES of
the equation by the LCM of all denominators in the equation.
Then solve as usual. 3  x  1
LCD  x( x  1)
x x 1
3
x( x  1)   x x( x  1)   x( x  1)
x
x 1
3( x  1)  x 2  x( x  1)
3x  3  x 2  x 2  x
2 x  3
x  3
2
CHECK :
3
3
3
2  1  2 
2  2  3  1

3
 3 1
1
2
2
2
If the equation is one fraction set equal to another, this is
called a PROPORTION. Solve by CROSS-MULTIPLYING,
then isolating the variable.
x 12

3 18
18 x  3(12)  36
18 x 36

18 18
x2
Find LCD of 60 and 40. LCD = 120.
Multiply both sides of the equation by 120 to remove fractions.
t
t
(120)  (120)  1(120)
60
40
2t 3t  120
5t  120
t  24 minutes to complete the job working together.
RADICAL EXPRESSIONS
A square of a positive number x is a number whose
square is x. The square root of a negative number is
not a real number.
 x
2
 25 
2
x
 5  25
2
The Product Property of Square Roots says you
can split a radical expression into its factors.
ab  a  b
225  25  9  25  9  5  3  15
18
Simplify:
18  9  2  9  2  3 2
Taking a number or variable expression to the
power ½ is the same as taking the square root.
x2  (x2 )
1
x  (x )
1
6
6
2
x
2
x
Simplify b15
3
 b14  b
 b14  b
 b7 b
Property of Squaring Both Sides of an Equation
If a and b are real numbers and a=b, then a2=b2