Upload
others
View
50
Download
0
Embed Size (px)
Citation preview
1
Unit 5 Notes FACTORING POLYNOMIALS
Factorization
How many pairs of monomials can you find whose product is
416x ?
The process they went through, finding pairs of monomials with a
product of
416x is called FACTORING.
Each pair of monomials is called a FACTORIZATION.
FACTOR MONOMIAL
Example 1
Find the factorization of 15π₯3
Factors are 1,-1,3,-3,5,-5,15,-15,x, π₯2, π₯3.
Some factorizations are
(15x)( π₯2) (5x)( 3π₯2
) (3)( 5π₯3)
(1)( 15π₯3) (3x)( 5π₯2
) (5)( 3π₯3)
Since (-1)(-1) = 1, we could have
(-15π₯3)(-1) (-3)( β5π₯3
) (βπ₯3)(-15)
There are still other ways to factor 15π₯3.
Try This. Find 3 factorizations of each monomial.
a. 8π₯4
b. 6π5
c. 12π3π2
2
FACTORING TERMS WITH A COMMON FACTOR
Factor out is explained as the opposite of the Distributive Property.
Always make sure to use the greatest possible coefficient and the variable
to the greatest power.
DISTRIBUTE (Multiply) FACTOR (Divide)
2 2
2
3 ( 2) 3 ( ) 3 (2)
................. 3 6
a b a b a
ab a
25 15 5 5 3
................ 5 ( 3)
x x x x x
x x
Examples
1. 3π₯2 + 3 = 3(π₯2) + 3(1)
= 3(π₯2 + 1)
2. 16π2π2 + 20π2 = 4π2(4π2) + 4π2(5)
= 4π2(4π2 + 5)
3. 15π₯5 β 12π₯4 + 27π₯3 β 3π₯2
= 3π₯2(5π₯3) β 3π₯2(4π₯2) + 3π₯2(9π₯) β 3π₯2(1)
= 3π₯2(5π₯3 β 4π₯2 + 9π₯ β 1)
Try This
d. π₯2 + 3π₯
e. 3π₯6 β 5π₯3 + 2π₯2
f. 9π₯4 β 15π₯3 + 3π₯2
g. 12π4π4 + 3π3π2 + 6π2π β
3
Factoring Difference of Two Squares
DIFFERENCE OF TWO SQUARES
For a binomial to be the difference of 2 squares; two conditions must be
met:
There must be two terms, both perfect squares (
2 4 816, ,9 ,25y x x )
There must be a minus sign between the two terms (difference)
Examples
1. Is 9π₯2 β 36 a difference of two squares?
The first term is a square. 9π₯2 = (3π₯)2
The second term is a square. 36 = 62
There is a minus sign between them. Thus we have a difference of
two squares.
2. Is β4π₯2 + 16 a difference of two squares?
β4π₯2 + 16 = 16 β 4π₯2
16 = 42 πππ β4π₯2 = (2π₯)2
Since there is a minus sign between 16 and β4π₯2 , we have a
difference of two squares.
Try This. State whether each expression is a difference of two squares.
a. π₯2 β 25
b. π₯2 β 24
c. β36 β π₯2
d. 4π₯2 β 15
e. β49 + 16π2
4
FACTORING THE DIFFERENCE OF TWO SQUARES
2 2( )( )a b a b a b β¦if we look carefully, the product IS the
difference of two squares; so writing the equation in reverse is factoring
the difference of two squaresβ¦
2 2( ) ( )( )a b a b a b
Examples
3. 4π₯2 β 25 = (2π₯)2 β (5)2 = (2π₯ + 5)(2π₯ β 5)
=(π΄)2 β (π΅)2 = (π΄ + π΅)(π΄ β π΅)
4. π6 β 16π2 = (π3)2 β (4π)2 = (π3 + 4π)(π3 β 4π)
5. 36π₯2 β 25π¦6 = (6π₯)2 β (5π¦3)2 = (6π₯ β 5π¦3)(6π₯ + 5π¦3)
6. 9π8π4 β 49 = (3π4π2 + 7)(3π4π2 β 7)
Try This. Factor
f. 4π¦2 β 49
g. 16π₯2 β 25π¦2
h. π8π4 β 4
i. 25π10 β 36π8
If the terms of the binomial have a common factor, first factor out the
common factor. Then continue factoring.
4 6 4 2
4 2 2
4
49 9 (49 9 )
.................. (7) (3 )
................. (7 3 )(7 3 )
x x x x
x x
x x x
5
Example
9. 18π2π2 β 50π6 = 2π2(9π2 β 25π4)
= 2π2[(3π)2 β (5π )2]
= 2π2(3π β 5π2)(3π + 5π2)
Try This. Factor
j. 32π¦6 β 8π¦2
k. 5 β 20π¦6
l. π3π β 4ππ3
m. 64π₯6π¦4 β 25π₯4π¦8
FACTORING COMPLETELY
After you factor the difference of 2 squares, you can sometimes continue
to factor. Factoring Completely means factor until factoring is no longer
possible (other than for a common factor of 1)
Example. Factor
10. 1 β 16π₯12 = (1)2 β (4π₯6)2
= (1 β 4π₯6)(1 + 4π₯6)
= [12 β (2π₯3)2](1 + 4π₯6)
6
= (1 + 2π₯3)(1 β 2π₯3)(1 + 4π₯6)
Reminder, that one factoring does NOT mean they are finished. Assume
there will always be another factoring available and that way never stop
after the first step.
Try This. Factor
h. 81π₯4 β 1
i. 16π4 β π8
TRINOMIAL SQUARES.
RECOGNIZE A TRINOMIAL SQUARE
2( 3)x 2( 3)x
2 6 9x x 2 6 9x x
Use the following to help identify a trinomial square:
Two of the terms must be squares (
2 2A andB )
There must be NO minus sign before either of the two squares
If we multiply A and B and double the result we get the third term,
either 2AB or the additive inverse, -2AB
Examples
1. Is π₯2 + 6π₯ + 9 a trinomial square?
A. π₯2 = (π₯)2 πππ 9 = (3)2
B. There is no minus sign before π₯2 and 9.
C. If we multiply x and 3 and double the results, we get the third
term, 2 β 3 β π₯, ππ 6π₯
7
Thus π₯2 + 6π₯ + 9 is the square of the binomial (x + 3).
2. Is π₯2 + 6π₯ + 11 a trinomial square?
The answer is no because only one term is a square.
3. Is 16π2 β 56ππ + 49π2 a trinomial square?
A. 16π2 = (4π)2 πππ 49π2 = (7π)2
B. There is no minus sign before 16π2and 49π2
.
C. If you multiply 4a and 7b and double the result, we get the
additive inverse of the third term, 2 β 4π β 7π = 56ππ
Thus 16π2 β 56ππ + 49π2 is the square of (4a - 7b)
Try This. Which of the following are trinomial squares?
a. π₯2 + 8π₯ + 16
b. π₯2 β 12π₯ + 4
c. 4π₯2 + 20π₯ + 25
d. 9π₯2 β 14π₯ + 16
e. 16π₯2 + 40π₯π¦ + 25π¦2
FACTORING TRINOMIAL SQUARES
2 2 22 ( )A AB B A B Notice the addition sign between the first 2
terms gives an addition sign in the factorization.
2 2 22 ( )A AB B A B Again notice, the subtraction sign between the
first two terms, gives a subtraction sign in the factorization.
***Remember to ALWAYS factor out a common factor first if possible***
8
Examples. Factor
4. π₯2 + 6π₯ + 9 = π₯2 + 2 β π₯ β 3 + 32 = (π₯ + 3)2
5. π₯2 β 14π₯ + 49 = π₯2 β 2 β π₯ β 7 + 72 = (π₯ β 7)2
6. 16π2 β 40ππ + 25π2 = (4π β 5π)2
7. 27π2 + 72ππ + 48π2 = 3(9π2 + 24ππ + 16π2)
=3(3π + 4π)2
Try This. Factor
f. π₯2 + 2π₯ + 1
g. π₯2 β 2π₯ + 1
h. 25π₯2 β 70π₯ + 49
i. 48π2 + 120ππ + 75π2
FACTORING
2x bx c , where c > 0 (Constant term is
Positive)
Explain to students that in the polynomial
2x bx c , c is called
the constant term or just the constant.
If the constant of the polynomial is NOT a perfect square, the trinomial
cannot be factored into the square of a binomial like
2 8 16x x 2( 4)x ,is for example.
It may be possible to be factored as the product of two DIFFERENT
binomials.
Example 1
9
Factor π₯2 + 5π₯ + 6.
The first term of each factor is x.
(x+_)(x+_)
Look for the two numbers whose product is 6 and whose sum is 5.
Product of 6 Sum
1,6 7
2,3 5
π₯2 + 5π₯ + 6 = (π₯ + 2)(π₯ + 3)
Try This. Factor
a. π₯2 + 7π₯ + 12
b. π₯2 + 13π₯ + 36
In Example 2 on we now have to worry about the second term being
negative. Using FOIL in reverse tells us that out factors will be negative
(negative times a negative is a positive) and the sum will be a negative
(negative plus negative is negative)
Example 2
Factor π₯2 β 8π₯ + 12
Since the coefficient of the middle term is negative, we need two negative
numbers whose products are 12 and whose sum is -8.
Product of 12 Sum
-1,-12 -13
-2,-6 -8
-3,-4 -7
10
π₯2 β 8π₯ + 12 = (π₯ β 2)(π₯ β 6)
Try This. Factor
c. π₯2 β 8π₯ + 15
d. π₯2 β 9π₯ + 20
e. π₯2 β 7π₯ + 12
Example 3 takes what we have done and added the twist of having two
variables.
Factor π2 + 7ππ + 10π2
Since π2 is the product of a times a, and π2
is the product of b times b, we
are looking for binomials of the form.
(a + _b)(a + _b)
Find the two numbers whose sum is 7 and whose product is 10.
Product of 10 Sum
1,10 11
2,5 7
π2 + 7ππ + 10π2 = (π + 2π)(π + 5π)
Try This . Factor
f. π2 + 8ππ + 15π2
g. π2 + 5ππ + 6π2
11
h. π2 + 6ππ + 8π2
CONSTANT TERM IS NEGATIVE
2x bx c , where c < 0
Because the constant term is negative, or less than zero, the middle term
MAY be positive or negative.
2
2
( 5)( 2) 2 5 10
.......................... 3 10
x x x x x
x x
2
2
( 5)( 2) 5 2 10
............................ 3 10
x x x x x
x x
In both cases the constant term is less than zero, or negative, so the
second term will determine how we factor.
Example 4
Factor π₯2 β 8π₯ β 20.
Find two numbers whose sum is -8 and whose product is -20.
Product of -20 Sum
-1,20 19
1,-20 -19
-2,10 8
2,-10 -8
4,-5 -1
-4,5 1
π₯2 β 8π₯ β 20 = (π₯ + 2)(π₯ β 10)
12
Example 5 demonstrates 2 variables.
Factor π2 β ππ β 6π2
We are looking for binomials of the form (a_b)(a_b). Find two numbers
whose sum is 1 and whose product is -6.
Product of -6 Sum of 1
1,-6 -5
-1,6 5
2,-3 -1
-2,3 1
π2 β ππ β 6π2 = (π + 2π)(π β 3π)
Try This. Factor.
i. π₯2 + 4π₯ β 12
j. π₯2 β 13π₯ + 12
k. π2 + 5ππ β 14π2
l. π₯2 β π₯π¦ β 30π¦2
FACTORING
2ax bx c
What is (2x+5)(3x+4) ?
13
Students should work this out using FOIL or the Box method;
26 23 20x x
Factoring Trinomials
To factor ππ₯2 + ππ₯ + π, we look for binomials
(_x + _)(_x + _) where products of numbers in the blanks are as follows.
1. The numbers in the first blanks of each binomial have a product of
a.
2. The numbers in the last blanks of each binomial have a product of
c.
3. The outside product and the inside product have a sum of b.
Examples
1. Factor 3π₯2 + 5π₯ + 2.
First look for a factor common to all terms. There is none. Next look for
two numbers whose product is 3.
1,3 -1,-3
Now look for numbers whose product is 2.
1,2 -1,-2
Since the last term of the trinomial is positive, the signs of the second
terms must be the same. Here are some possible factorizations.
(x + 1)(3x + 2) (x + 2)(3x + 1)
(x β 1)(3x β 2) (x - 2)(3x β 1)
When we multiply, the first term will be 3π₯2 and the last term will be 2 in
each case. Only the first multiplication gives a middle term of 5x.
3π₯2 + 5π₯ + 2 = (π₯ + 1)(3π₯ + 2)
2. Factor 2π₯2 + 5π₯ β 12.
14
First term: Find two numbers whose product is 2.
Last term: Find two numbers with product of -12.
(2x + 3)(x β 4) (2x β 2)(x + 6) (2x β 1)(x + 12)
(2x β 3)(x + 4) (2x + 2)(x β 6) (2x - 12)(x + 1)
The outside product plus the inside product must equal 5x.
2π₯2 + 5π₯ β 12 = (2π₯ β 3)(π₯ + 4)
Try This
a. 6π₯2 + 7π₯ + 2
b. 8π₯2 + 10π₯ β 3
c. 6π₯2 + 41π₯ β 7
3. Factor 8π2 + 8π β 6.
8π2 + 8π β 6 = 2(4π2 + 4π β 3)
1st
term: Find two numbers whose product is 4.
2nd
term: Find two numbers whose product is -3.
(4m + 3)(m - 1) (4m β 3)(m + 1) (2m + 3)(2m β 1)
(4m β 1)(m + 3) (4m + 1)(m - 3) (2m β 3)(2m + 1)
The outside product plus the inside product must be equal to 4m.
8π2 + 8π β 6 = 2(4π2 + 4π β 3) = 2(2π + 3)(2π β 1)
15
Try This
4. 3π₯2 β 21π₯ + 36
5. 9π2 β 15π β 6
6. 4π2 + 2π β 6
7. 6π2 + 15ππ β 9π2
FACTORING BY GROUPING
Can I use the distributive property to factor this polynomial? If so, show
me. 4 3 23 12 6 9x x x x
3 23 ( 4 2 3)x x x x The common factor 3x was removed from each
term within the polynomial.
Can I use the distributive property to factor this polynomial?
3 2 2 2x x x β¦..Unfortunately there is no one common factor
for all the terms, but maybe students may see the following:
16
3 2x x has a common factor (
2x )(x+1)
2x + 2 has a common factor (2)(x + 1)
Now, if we put this all together we would get
2
2
( 1) 2( 1)
( 2)( 1)
x x x
x x
The (x+1) can be factored from both.
Because we factored in sections or groups set apart using parentheses,
this is called factoring by grouping.
Examples
1. 6π₯3 β 9π₯2 + 4π₯ β 6 = (6π₯3 β 9π₯2) + (4π₯ β 6)
= 3π₯2(2π₯ β 3) + 2(2π₯ β 3)
= (2π₯ β 3)(3π₯2 + 2)
2. π₯3 + π₯2 + π₯ + 1 = (π₯3 + π₯2) + (π₯ + 1)
= π₯2(π₯ + 1) + 1(π₯ + 1)
= (π₯ + 1)(π₯2 + 1)
3. π₯3 + 2π₯2 β π₯ β 2 = (π₯3 + 2π₯2) + (βπ₯ β 2)
= π₯2(π₯ + 2) + 1(βπ₯ β 2)
= π₯2(π₯ + 2) β 1(π₯ + 2)
= (π₯ + 2)(π₯2 β 1)
= (π₯ + 2)(π₯ β 1)(π₯ + 1)
4. π₯2π¦2 + ππ¦2 + ππ + ππ₯2 = π¦2(π₯2 + π) + π(π₯2 + π)
= (π₯2 + π)(π¦2 + π)
5. π₯3 + π₯2 + 2π₯ β 2 = π₯2(π₯ + 1) + 2(π₯ β 1)
Cannot be factored.
Try This. Factor
a. 8π₯3 + 2π₯2 + 12π₯ + 3
b. 4π₯3 β 6π₯2 β 6π₯ + 9
17
c. π₯3 + π₯2 β π₯ β 1
d. 3π β 6π + 5π2 β 10ππ
FACTORING POLYNOMIALS
A. ALWAYS look first for a common factor
B. Then look at the number of terms
2 terms: Is it the difference of 2 squares
3 terms: Is the trinomial a square of a binomial? If not, test the
Factors of the terms
4 terms: try factoring by grouping
C. ALWAYS factor completely
SOLVING EQUATIONS BY FACTORING
The Principle of Zero Products
The product of two or more factors is zero if any of the factors are
equal to zero.
If a product is zero, then one or more of the factors must be zero
For any rational number a and b, if ab=0, then a=0 or b=0 and if
a=0 or b=0, then ab=0
If we have an equation with zero on one side and a factorization on the
other, we can solve the equation by finding the values that make the
factors zero. This means there can be more than ONE right answer.
18
(5 1)( 7) 0
5 1 0......... ........ 7 0
5 1.......................... 7
1
5
x x
x or x
x x
x
Check to see if these solutions
work by substituting, one at a time, both values for x and calculate
1 1(5( ) 1)( 7)
5 5
1( 1 1)( 7 )
5
10( 7 )
5
0
(5 7 1)(7 7)
(35 1)(0)
(36)(0)
0
FACTORING and SOLVING
Use the following steps to solve equations using the principle of zero
Get zero on one side of the equation by using the addition property
Factor the expression on the other side of the equation
Set each factor equal to zero
Solve each equation.
Check your solutions.
Example 1. Solve.
π₯2 β 6π₯ = 16
π₯2 β 8π₯ + 16 = 0
(π₯ β 8)(π₯ + 2) = 0
π₯ β 8 = 0 ππ π₯ + 2 = 0
π₯ = 8 ππ π₯ = β2
Check:
19
π₯2 β 6π₯ = 16
(8)2 β 6(8) = 16
64 β 48 = 16
π₯2 β 6π₯ = 16
(β2)2 β 6(β2) = 16
4 + 12 = 16
Example 2
π₯2 + 5π₯ + 6 = 0
(π₯ + 2)(π₯ + 3) = 0
π₯ + 2 = 0 ππ π₯ + 3 = 0
π₯ = β2 ππ π₯ = β3
Check:
π₯2 + 5π₯ + 6 = 0
(β2)2 + 5(β2) + 6 = 0
4 β 10 + 6 = 0
π₯2 + 5π₯ + 6 = 0
(β3)2 + 5(β3) + 6 = 0
9 β 15 + 6 = 0
The solutions are -2 and -3.
Try This
a. π₯2 β π₯ β 6 = 0
20
b. π₯2 β π₯ = 56
c. π₯2 β 3π₯ = 28
The answers found to solving the equations are called the root of the
polynomial; the value of the variable that makes the polynomial equal to
zero.
Example 3 shows how the word βrootβ will be used.
Find the roots of 4π₯2 β 25
4π₯2 β 25 = 0
(2π₯ β 5)(2π₯ + 5) = 0
2π₯ β 5 = 0 ππ 2π₯ + 5 = 0
2π₯ = 5 ππ 2π₯ = β5
π₯ =5
2 ππ π₯ =
β5
2
Check:
4π₯2 β 25 = 0
4(5
2)2 β 25 = 0
4(25
4) β 25 = 0
0 = 0
21
4π₯2 β 25 = 0
4(β5
2)2 β 25 = 0
4(25
4)2 β 25 = 0
0 = 0
Try This. Find the roots.
d. π₯2 + 6π₯ + 9
e. 25π₯2 β 16
SOLVE PROBLEMS BY WRITING & SOLVING EQUATIONS
Examples Translate to an equation and solve.
1. The product of one more than a number and one less than the
number is 8. Find the number.
Let x = the number.
One more than a number times one less than the number is 8.
(x + 1)(x β 1) = 8
π₯2 β 1 = 8
22
π₯2 β 1 β 8 = 0
π₯2 β 9 = 0
(x β 3)(x + 3) = 0
x β 3 = 0 or x + 3 = 0
x = 3 or x= -3
Check: Both 3 and -3 are the solutions.
2. The square of a number minus twice the number is 48. Find the
number.
Let x = the number.
π₯2 β 2π₯ = 48
π₯2 β 2π₯ β 48 = 0
(π₯ β 8)(π₯ + 6) = 0
π₯ β 8 = 0 ππ π₯ + 6 = 0
π₯ = 8 ππ π₯ = β6
Both 8 and -6 check. They are both solutions.
Try This. Translate to an equation and solve.
a. The product of seven less than a number and eight less than the
number is 0.
.
b. The product of one more than a number and one less than the
number is 24.
23
c. The square of a number minus the number is 20.
d. One more than twice the square of a number is 73.
3. The area of the foresail on a 12-meter racing yacht is 93.75 square
meters. The sailβs height is 8.75 meters greater than its base. Find
its base and height.
Area = 1
2β πππ π β βπππβπ‘
Let h = the sailβs height and h β 8.75 = the length of the sailβs base.
1
2(β β 8.75)β = 93.75
(β β 8.75)β = 187.5
(β2 β 8.75β) = 187.5
(β2 β 8.75β β 187.5) = 0
(β β 18.75)(β + 10) = 0
β β 18.75 = 0 ππ β + 10 = 0
β = 18.75 ππ β = β10
The solution of the equation are 18.75 and -10. The height of the sail
cannot have a negative value, so the height must be 18.75 meters. The
height of the base is then 8.75 meters shorter, or 10 meters.
24
4. The product of two consecutive integers is 156. Find the integers.
Let x represent the first integer, Then x+1 represent the second
integer.
π₯ β (π₯ + 1) = 156
π₯(π₯ + 1) = 156
π₯2 + π₯ = 156
π₯2 + π₯ β 156 = 0
(π₯ β 12)(π₯ + 13) = 0
π₯ β 12 = 0 ππ π₯ + 13 = 0
π₯ = 12 ππ π₯ = β13
When x = 12, x+1=13 and 12(13)=156
When x = -13, x+1=-12 and -13(-12)=156
We have two pairs of solutions, 12 and 13 and -12 and -13. Both are
pairs of consecutive integers whose product is 156.
Try This
e. The width of a rectangular card is 2cm less than the length. The
area is 15 cm squared. Find the length and width?
f. The product of two consecutive integers is 462. Find the integers?