Transcript No Slide Title

Digital Lessons on Factoring
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Factoring using the Greatest Common Factor and Factoring by Grouping
1. List all possible factors for a given number.
2. Find the greatest common factor of a set of numbers
or monomials.
3. Write a polynomial as a product of a monomial GCF
and a polynomial.
4. Factor by grouping.
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Factored form: We say that a number or expression is in
its factored form if it is written as a product of factors.
Following are some examples of factored form:
An integer written in factored form with integer factors:
28 = 2 • 14
A monomial written in factored form with monomial
factors: 8x5 = 4x2 • 2x3
A polynomial written in factored form with a monomial
factor and a polynomial factor: 2x + 8 = 2(x + 4)
A polynomial written in factored form with two
polynomial factors: x2 + 5x + 6 = (x + 2)(x + 3)
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Slide 13- 3
Listing all possible factors for a given number.
Example:
List all natural number factors of 24.
Solution:
To list all the natural number factors, we can divide 24 by 1, 2, 3, and so on,
writing each divisor and quotient pair as a product until we have all possible
combinations.
1 • 24, 2 • 12, 3 • 8, 4 • 6
The natural number factors of 24 are 1, 2, 3, 4, 6, 8, 12
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Slide 13- 4
Find the greatest common factor of a set of numbers or
monomials.
Greatest common factor (GCF) is the largest natural number that
divides all given numbers perfectly or without remainder.
Steps for finding the GCF of a set of numbers by listing:
1. List all possible factors for each given number.
2. Search the lists for the largest factor common to all
lists.
Example: Find the GCF of 24 and 36.
Solution:
Factors of 24: 1, 2, 3, 4, 6, 8, 12
Factors of 36: 1, 2, 3, 4,6, 9, 12,18
The GCF of 24 and 36 is 12.
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Slide 13- 5
Prime Factorization Method for Finding GCF
To find the greatest common factor of a given set of numbers:
1. Write the prime factorization of each given number in exponential form.
2. Create a factorization for the GCF that includes only those prime factors
common to all the factorizations, each raised to its smallest exponent in
the factorization.
3. Multiply the factors in the factorization created in Step 2.
Note: If there are no common prime factors, then the GCF is 1.
Example: Find the GCF of 25a3b and 40a2.
Solution:
Write the prime factorization of each monomial, treating the
variables like prime factors.
25a3b = 5 • 5 • a3 • b
40a2 = 2 • 2 •2 • 5 • a2
GCF = 5 •a2 = 5a2
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Slide 13- 6
Factoring a Monomial GCF Out of a Polynomial
To factor a monomial GCF out of a given polynomial:
1. Find the GCF of the terms that make up the polynomial.
2. Rewrite the given polynomial as a product of the GCF and parentheses
that contain the result of dividing the given polynomial by the GCF.
Given polynomial = GCF
Example: Factor:
 Given polynomial 


GCF


9x yz  15x y 18x y
2
3
4
2
9 x yz, 15x y, and 18x y .
Solution
2
3
4 2
Find the GCF of
Because the first term in the polynomial is negative, we will factor out the
negative of the GCF to avoid a negative first term inside the parentheses.
We will factor out 3x2 y.
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Slide 13- 7
continued
2.
Write the given polynomial as the product of the GCF and the parentheses
containing the quotient of the given polynomial and the GCF.
2
3
4 2


9
x
yz

15
x
y

18
x
y 
2
2
3
4 2
9x yz  15x y 18x y  3x y 

2
3x y


2
3
4 2


9
x
yz
15
x
y
18
x
y 
2
 3x y 



2
2
2

3
x
y

3
x
y

3
x
y


 3 x 2 y  3 z  5 x  6 x 2 y 
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Slide 13- 8
Example
Factor.
a b  5  8 b  5
Solution: Notice that this expression is a sum of
two products, a and (b + 5), and 8 and (b + 5).
Further, note that (b + 5) is the GCF of the two
products.
 a  b  5  8  b  5 
a b  5  8 b  5   b  5 

b5


 a  b  5 8  b  5 
  b  5 


b5 
 b5
 b  5 a  8
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Slide 13- 9
Factoring by grouping.
To factor a four-term polynomial by grouping:
1. Factor out any monomial GCF (other than 1) that is common to all four
terms.
2. Group together pairs of terms and factor the GCF out of each pair.
3. If there is a common binomial factor, then factor it out.
4. If there is no common binomial factor, then interchange the middle two
terms and repeat the process. If there is still no common binomial factor,
then the polynomial cannot be factored by grouping.
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Slide 13- 10
Example
Factor. 8 p3  24 p2  3 pq  9q
Solution: First we look for a monomial GCF (other
than 1). This polynomial does not have one.
Because the polynomial has four terms, we now
try to factor by grouping.
8 p3  24 p2  3 pq  9q
  8 p 3  24 p 2    3 pq  9q 
 8 p2  p  3  3q  p  3
  p  3  8 p 2  3q 
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Slide 13- 11
Factoring Trinomials of the
Form x2 + bx + c
To factor a trinomial of the form x2 + bx + c :
1. Find two numbers with a product equal to c
and a sum equal to b.
2. The factored trinomial will have the form:
(x + first number) (x + second number).
Note: The signs in the binomial factors
can be minus signs, depending on the signs of
b and c.
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Example
Factor. x2 – 6x + 8
Solution: We must find a pair of numbers whose
product is 8 and whose sum is –6. If two
numbers have a positive product and negative
sum, they must both be negative. Following is a
table listing the products and sums:
Product
Sum
(–1)(–8) = 8
–1 + (–8) = –9
(–2)(–4) = 8
–2 + (–4) = –6
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This is the
correct
combination.
Slide 13- 13
continued
Answer
x2 – 6x + 8 = (x – 2)(x – 4)
Check We can check by multiplying the binomial
factors to see if their product is the
original polynomial.
(x – 2)(x – 4) = x2 – 4x – 2x + 8
= x2 – 6x + 8
Multiply the factors using
FOIL.
The product is the original
polynomial.
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Slide 13- 14
Example
Factor. a2 – ab – 20b2
Solution: We must find a pair of terms whose
product is 20b2 and whose sum is –1b. These
terms would have to be –5b and 4b.
Answer a2 – ab – 20b2 = (a – 5b)(a + 4b)
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Slide 13- 15
Factor out a monomial GCF, then factor the trinomial of the form
x2 + bx + c.
Example: Factor. 4xy3 + 12xy2 – 72xy
Solution
First, we look for a monomial GCF (other than 1). Notice that the GCF of the terms
is 4xy.
Factoring out the monomial, we have
4xy3 + 12xy2 – 72xy = 4xy(y2 + 3y – 18)
Now try to factor the trinomial to two binomials. We must find a pair of numbers whose
product is
–18 and whose sum is 3.
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Slide 13- 16
continued
Product
Sum
(–1)(18) = –18
–1 + 18 = 17
(–2)(9) = – 18
–2 + 9 = 7
(–3)(6) = – 18
–3 + 6 = 3
This is the correct
combination.
Answer
4xy3 + 12xy2 – 72xy = 4xy(y – 3)(y + 6)
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Slide 13- 17
Factoring Trinomials of the Form
ax2 + bx + c, where a  1
Factoring by Trial and Error
To factor a trinomial of the form ax2 + bx + c, where a 1, by trial
and error:
1. Look for a monomial GCF in all the terms. If there is one, factor it
out.
2. Write a pair of first terms whose product is ax2.
ax2





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3. Write a pair of last terms whose product is c.
c





4. Verify that the sum of the inner and outer products is bx (the middle term
of the trinomial).





Inner
+ Outer
bx
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Slide 13- 19
If the sum of the inner and outer products is not bx, then try the
following:
a. Exchange the first terms of the binomials from step 3, then repeat step
4.
b. Exchange the last terms of the binomials from step 3, then repeat step
4.
c. Repeat steps 2 – 4 with a different combination
of first and last terms.
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Slide 13- 20
Example
Factor. 6 x2  13x  5
Solution
The first terms must multiply to equal 6x2.
These could be x and 6x, or 2x and 3x.
6 x2  13x  5  
+



The last terms must multiply to equal –5.
Because –5 is negative, the last terms in the
binomials must have different signs. This factor
pair must be 1 and 5.
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Slide 13- 21
continued
Now we multiply binomials with various
combinations of these first and last terms until we find
a combination whose inner and outer products
combine to equal 13x.
 x  56x 1  6x22  x  30x  5  6x22  29x  5
 x 16x  5  6x  5x  6x  5  6x  x  5
3x 1 2x  5  6x2 15x  2x  5  6x2 13x  5 Incorrect
combinations.
2
2
3x  5 2x 1  6x  3x 10x  5  6x  7x  5
 2x 13x  5  6x2 10x  3x  5  6x2  7x  5
2
2
2
x

5
3
x

1

6
x

2
x

15
x

5

6
x
13x  5 Correct



combination.
Answer 6x2 13x  5   2x  53x 1
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Slide 13- 22
Example
21x3  60 x2  9 x
Factor.
Solution First, we factor out the monomial GCF, 3x.
21x3  60 x2  9 x  3 x  7 x 2  20 x  3
Now we factor the trinomial within the parentheses.
The first terms must multiply to equal 7x2.
These could be x and 7x.
3 x  7 x 2  20 x  3   3x 


+

The last terms must multiply to equal 3.
Because 3 is a prime number, its factors are 1
and 3.
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Slide 13- 23
continued
Now we multiply binomials with various
combinations of these first and last terms until we find
a combination whose inner and outer products
combine to equal –20x.
3x  x  1 7 x  3  3x  7 x 2  3x  7 x  3  3x  7 x 2  4 x  3
3x  x  1 7 x  3  3x  7 x 2  3x  7 x  3  3 x  7 x 2  4 x  3
3x  x  3 7 x  1  3x  7 x 2  x  21x  3  3x  7 x 2  20 x  3
3x  x  3 7 x  1  3x  7 x 2  x  21x  3  3x  7 x 2  20 x  3 
Correct combination.
Answer 21x3  60x2  9x  3x  x  3 7x 1
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Slide 13- 24
Factoring ax2 + bx + c, where a 1, by Grouping

2
To factor a trinomial of the form ax + bx + c, where a 1, by grouping:
1.
2.
3.
4.
Look for a monomial GCF in all the terms. If there is one, factor it out.
Multiply a and c.
Find two factors of this product whose sum is b.
Write a four-term polynomial in which bx is written as the sum of two like
terms whose coefficients are the two factors you found in step 3.
5. Factor by grouping.
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Slide 13- 25
Example
Factor. 2 x  15x  7
Solution
Notice that for this trinomial, a = 2, b = –15, and
c = 7. We begin my multiplying a and c: (2)(7) = 14.
2
Now we find two factors of 14 whose sum is –15.
Notice that these two factors must both be negative.
Factors of ac
Sum of Factors of ac
(–2)(–7) = 14
–2 + (–7) = –9
(–1)(–14) = 14
–1 + (– 14) = –15
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Correct
Slide 13- 26
continued
2 x2 –15x  7  2 x 2 –x – 14x  7
 x  2x 1  7  2x 1
  2x 1   x  7 
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Slide 13- 27
Factoring Special Products
1.
2.
3.
4.
Factor perfect square trinomials.
Factor a difference of squares.
Factor a difference of cubes.
Factor a sum of cubes.
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Factoring Perfect Square Trinomials
a2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (a – b)2
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Slide 13- 29
Example
Factor. 9a2 + 6a + 1
Solution
This trinomial is a perfect square because the first
and the last terms are perfect squares and twice the
product of their roots is the middle term.
9a2 + 6a + 1
The square root of 9a2 is 3a.
The square root of 1 is 1.
Twice the product of 3a and 1 is (2)(3a)(1) = 6a,
which is the middle term.
Answer 9a2 + 6a + 1 = (3a + 1)2
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Slide 13- 30
Example
Factor. 16x2 – 56x + 49
Solution
This trinomial is a perfect square.
16x2 – 56x + 49
The square root of 16x2 is 4x.
The square root of 49 is 7.
Twice the product of 4x and 7 is (2)(4x)(7) = 56x,
which is the middle term.
Answer 16x2 – 56x + 49 = (4x – 7)2
Use a2 – 2ab + b2 = (a – b)2,
where a = 4x and b = 7.
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Slide 13- 31
Factoring a Difference of Squares
a2 – b2 = (a + b)(a – b)
Warning: A sum of squares a2 + b2 is
prime and cannot be factored.
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Slide 13- 32
Example
Factor. 9x2 – 16y2
Solution
This binomial is a difference of squares because
9x2 – 16y2 = (3x)2 – (4y)2 .
To factor it, we use the rule a2 – b2 = (a + b)(a – b).
a2 – b2 = (a + b)(a – b)
9x2 – 16y2 = (3x)2 – (4y)2 = (3x + 4y)(3x – 4y)
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Slide 13- 33
Example
Factor. n4 – 625
Solution
This binomial is a difference of squares, where
a = n2 and b = 25.
n4 – 625 = (n2 + 25)(n2 – 25)
Use a2 – b2 = (a + b)(a – b).
= (n2 + 25)(n + 5)(n – 5)
Factor n2 – 25, using
a2 – b2 = (a + b)(a – b)
with a = n and b = 5.
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Slide 13- 34
Factoring a Difference of Cubes
a3 – b3 = (a – b)(a2 + ab + b2)
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Slide 13- 35
Example
Factor. 216x3 – 64
Solution
This binomial is a difference of cubes.
a3 – b3 = (a – b) (a2 + a b + b2)
216x3 – 64 = (6x)3 – (4)3 = (6x – 4)((6x)2 + (6x)(4) + (4)2)
= (6x – 4)(36x2 + 24x + 16)
Note: The trinomial may seem like a perfect
square. However, to be a perfect square, the
middle term should be 2ab. In this trinomial,
we only have ab, so it cannot be factored.
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Slide 13- 36
Factoring a Sum of Cubes
a3 + b3 = (a + b)(a2 – ab + b2)
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Slide 13- 37
Example
Factor. 6x +162xy3
Solution
The terms in this binomial have a monomial GCF, 6x.
6x +162xy3 = 6x(1 + 27y3)
= 6x(1 + 3y)((1)2 – (1)(3y) + (3y)2)
= 6x(1 + 3y)(1 – 3y + 9y2)
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Slide 13- 38
Strategies for Factoring
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Factoring a Polynomial
To factor a polynomial, first factor out any monomial GCF, then
consider the number of terms in the polynomial. If the polynomial
has:
I. Four terms, then try to factor by grouping
II. Three terms, then determine if the trinomial is a perfect square or
not.
A. If the trinomial is a perfect square, then
consider its form.
1. If in the form a2 + 2ab + b2, then the factored
form is (a + b)2.
2. If in the form a2  2ab + b2, then the
factored form is (a  b)2.
B. If the trinomial is not a perfect square, then consider its form.
1. If in the form x2 + bx + c, then find two factors of c whose
sum is b, and write the factored form as
(x + first number)(x + second number).
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Slide 13- 40
Factoring a Polynomial continued
2. If in the form ax2 + bx + c, where a  1, then use trial and
error. Or, find two factors of ac whose sum is b; write these
factors as coefficients of two like terms that, when combined,
equal bx; and then factor by grouping.
III. Two terms, then determine if the binomial is a difference of
squares, sum of cubes, or difference of cubes.
A. If given a binomial that is a difference of squares, a2 – b2,
then the factors are conjugates and the factored form is
(a + b)(a – b). Note that a sum of squares cannot be factored.
B. If given a binomial that is a sum of cubes, a3 + b3, then the
factored form is (a + b)(a2 – ab + b2).
C. If given a binomial that is a difference of cubes, a3 – b3, then
the factored form is (a – b)(a2 + ab + b2).
Note: Always look to see if any of the factors can be factored.
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Slide 13- 41
Example
Factor. 12x2 – 8x – 15
Solution
There is no GCF.
Not a perfect square, since the first and last terms are not perfect squares.
Use trial and error or grouping.
(x – 3)(12x + 5) = 12x2 + 5x – 36x – 15 No
(6x – 3)(2x + 3) = 12x2 + 18x – 6x – 9 No
(6x + 5)(2x – 3) = 12x2 – 18x + 10x – 15
= 12x2 – 8x – 15 Correct
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Slide 13- 42
Factor. 5x3 – 10x2 – 120x
Example
Solution
5x(x2 – 2x – 24)
Factored out the monomial GCF, 5x.
Look for two numbers whose product is –24 and whose sum is
2.
Product
Sum
(1)(24) = 24
4(6) = 24
1 + 24 = 23
4 + (6) = 2
Correct combination.
5x3 – 10x2 – 120x =5x(x + 4)(x – 6)
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Slide 13- 43
Example
Factor. 8a4 – 72n2
Solution
8a4 – 72n2 = 8(a4 – 9n2)
Factor out the monomial GCF, 8.
a4 – 9n2 is a difference of squares
= 8(a2 – 3n)(a2 + 3n)
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Slide 13- 44
Example
Factor. 12y5 + 84y3
Solution
12y3(y2 + 7)
Factor out the monomial GCF, 12y3.
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Slide 13- 45
Example
Factor. 150x3y – 120x2y2 + 24xy3
Solution
6xy(25x2 – 20xy + 4y2) Factor out the monomial GCF, 6xy.
6xy(5x – 2y)2
Factor the perfect square trinomial.
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Slide 13- 46
Example
Factor. x5 – 2x3 – 27x2 + 54
Solution
No common monomial, factor by grouping.
(x5 – 2x3)(– 27x2 + 54)
x3(x2 – 2)–27(x2 – 2)
(x2 – 2)(x3 – 27)
Difference of cubes
(x2 – 2)(x – 3)(x2 + 3x + 9)
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Slide 13- 47
Solving Quadratic Equations by Factoring
.
Zero-Factor Theorem
If a and b are real numbers and ab = 0, then a = 0 or b = 0.
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Solve. (x + 4)(x + 5) = 0
Example
Solution
According to the zero-factor theorem, one of the two factors, or both factors, must
equal 0.
x + 4 = 0 or
x + 5 = 0 Solve each equation.
x = 4
x = 5
Check
For x = 4:
For x = 5:
(x + 4)(x + 5) = 0
(x + 4)(x + 5) = 0
(4 + 4)(4 + 1) = 0
(5 + 4)(5 + 5) = 0
0(3) = 0
(1)(0) = 0
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Solving Equations with Two or More Factors Equal to 0
To solve an equation in which two or more factors are equal to 0, use the
zero-factor theorem:
1. Set each factor equal to zero.
2. Solve each of those equations.
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Slide 13- 50
Example
Solve.
a. y(5y + 2) = 0
Solution
a. y(5y + 2) = 0
y = 0 or 5y + 2 = 0
5y = 2
This equation
is already
solved.
b. x(x + 2)(5x – 4) = 0
2
y
5
b.
x( x  2)(5x  4)  0
x  0 x  2  0 5x  4  0
x  2
5x  4
4
x
5
To check, we verify that the solutions satisfy the
original equations.
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Slide 13- 51
Solve quadratic equations by factoring.
Quadratic equation in one variable: An equation that can be
written in the form ax2 + bx + c = 0, where a, b, and c are all
real numbers and a  0.
Solving Quadratic Equations Using Factoring
To solve a quadratic equation:
1. Write the equation in standard form
(ax2 + bx + c = 0).
2. Write the variable expression in factored form.
3. Use the zero-factor theorem to solve.
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Slide 13- 52
Example
Solve. 2x2 – 5x – 3 = 0
Solution
The equation is in standard form, so we can factor.
2x2 – 5x – 3 = 0
(2x + 1)(x – 3) = 0 Use the zero-factor theorem to solve.
2x + 1 = 0 or x – 3 = 0
2 x  1
1
x
2
x3
To check, we verify that the
solutions satisfy the original
equations.
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Slide 13- 53
Example
Solve. 6y2 + 11y = 10 + 4y
Solution
Write the equation in standard form.
6y2 + 11y = 10 + 4y
6y2 + 7y = 10
Subtract 4y from both sides.
6y2 + 7y – 10 = 0
Subtract 10 from both sides.
(6y – 5)(y + 2) = 0
Factor.
6y – 5 = 0 or y + 2 = 0 Use the zero-factor theorem.
6y  5
5
y
6
y  2
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Slide 13- 54