Chapter Five 5.3

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§ 5.3
Greatest Common Factors and Factoring by
Grouping
Factoring
Factoring a polynomial means finding an equivalent
expression that is a product. For example, when we take the
polynomial
3a 2 x  6a 2 y  2bx  4by
And write it as

 x  2 y  3a 2  2b

we say that we have factored the polynomial. In factoring, we write a
sum as a product.
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 5.3
Factoring
Factoring a Monomial from a Polynomial
1) Determine the greatest common factor of all terms in the polynomial.
2) Express each term as the product of the GCF and its other factor.
3) Use the distributive property to factor out the GCF.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 5.3
Factoring
EXAMPLE
Factor: 49x 4 y3 z 5  70x3 y 5 z 4  35x 4 y 4 z 3 .
SOLUTION
First, we determine the greatest common factor of the three
terms.
49x 4 y3 z 5  70x3 y5 z 4  35x4 y 4 z 3
Notice that the greatest integer that divides into 49, 70 and 35
(the coefficients of the terms) is 7. The variables raised to the
smallest exponents are x3 , y3 and z 3 .
The GCF is 7 x3 y 3 z 3 .
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 5.3
P 332
Factoring
CONTINUED
49x 4 y3 z 5  70x3 y5 z 4  35x 4 y 4 z 3
 7 x3 y3 z 3  7 xz2  7 x3 y3 z 3 10y 2 z  7 x3 y3 z 3  5xy Express each term
as the product of
the GCF and its
other factor

 7 x3 y 3 z 3 7 xz2 10y 2 z  5xy

Factor out the
GCF
P 332
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 5.3
A Strategy for Factoring Polynomials,
page 363
1.
If there is a common factor, factor out the GCF or factor out a common factor with
a negative coefficient.
2.
Determine the number of terms in the polynomial and try factoring as follows:
(a) If there are two terms, can the binomial be factored by using one of the
following special forms.
Difference of two squares:
A2  B 2   A  B A  B
///////////////////////////////////////////
Sum and Difference of two cubes:
(b) If there are three terms,
2
2
2


A

2
AB

B

A

B
If is the trinomial a perfect square trinomial
2
A2  2 AB  B2   A  B
use one of the adjacent forms:
If the trinomial is not a perfect square trinomial,
If a is equal to 1, use the trial and error
If a is > than 1, use the grouping method
(c) If there are four or more terms, try factoring by grouping.
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 5.6
Factoring
EXAMPLE
P 333
Factor:  5 x 3  50x 2  10x .
SOLUTION
The GCF is 5x. Because the leading coefficient, -5, is negative,
we factor out a common factor with a negative coefficient. We
will factor out the negative of the GCF, or -5x.
 5x3  50x 2  10x
 
 5x x2  5x10x  5x2

 5x x 2 10x  2

Important to
factor the
negative sign
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 5.3
Express each term as
the product of the
GCF and its other
factor
Factor out the GCF
Factoring
Check Point 1
Factor
20x  30x
2
10x
10x2x  3
Check Point 2a
Factor
9 x  21x
4
3x 2

3x2 3x2  7
2

P 332-333
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 5.2
Factoring
Check Point 2b
15x3 y 2  25x4 y3
Factor
5x3 y 2
5x y 3 - 5xy
3
2
Check Point 2c
Factor
16x y  8x y  4 x y
4
5
3
4
2
3
2 3

4x y

4 x 2 y 3 4 x 2 y 2  2 xy  1
P 332-333
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 5.2
Factoring
Check Point 3
Factor
 2 x3  10x 2  6 x

- 2x

- 2x x - 5x  3
2
Want a to be
positive
P 334
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 5.2
Factoring by Grouping
EXAMPLE
Factor: 7 xx  y   x  y  .
SOLUTION
Let’s identify the common binomial factor in each part of the
problem.
7 x x  y    x  y 
The GCF, a
binomial, is x + y.
The GCF, a
binomial, is x + y.
P 334
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 5.3
Factoring by Grouping
CONTINUED
We factor out the common binomial factor as follows.
7 x x  y    x  y 
 7 xx  y  1 x  y 
 x  y 7 x  1
This step, usually omitted, shows
each term as the product of the
GCF and its other factor, in that
order.
Factor out the GCF
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 5.3
Factoring
Check Point 4a
Factor
3x  4  7ax  4
x - 4
x  43  7a
Check Point 4b
Factor
7 xa  b  a  b
a  b
a  b7x -1
P 334
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 5.2
Factoring by Grouping
Factoring by Grouping
1) Group terms that have a common monomial factor. There
will usually be two groups. Sometimes the terms must be
rearranged.
2) Factor out the common monomial factor from each group.
3) Factor out the remaining common binomial factor (if one
exists).
P 334
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 5.3
Factoring by Grouping
EXAMPLE
P 334
Factor: 3a 2 x  6a 2 y  2bx  4by .
SOLUTION
There is no factor other than 1 common to all terms. However,
we can group terms that have a common factor:
3a 2 x  6a 2 y
Common factor is
: 3a 2
3a 2 x  6a 2 y  3a 2 x  2 y 
+
 2bx  4by
Use -2b, rather than 2b, as the common
factor: -2bx – 4by = -2b(x + 2y). In this
way, the common binomial factor, x + 2y,
appears.
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 5.3
Factoring by Grouping
CONTINUED
P 334
The voice balloons illustrate that it is sometimes necessary to
use a factor with a negative coefficient to obtain a common
binomial factor for the two groupings. We now factor the
given polynomial as follows:
3a 2 x  6a 2 y  2bx  4by


 3a 2 x  6a 2 y   2bx  4by
 3a 2 x  2 y   2bx  2 y 

 x  2 y  3a 2  2b

Group terms with common
factors
Factor out the common
factors from the grouped
terms
Factor out the GCF
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 5.3
Factoring
Check Point 5
Factor
x 3  4 x 2  5 x  20
x  4x
2

5
Check Point 6
Factor
4x2  20x  3xy 15y
x  54x - 3y
P 335-6
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 5.2
DONE
Factoring by Grouping
EXAMPLE
Your local electronics store is having an end-of-the-year sale.
The price on a large-screen television had been reduced by
30%. Now the sale price is reduced by another 30%. If x is
the television’s original price, the sale price can be represented
by
(x – 0.3x) – 0.3(x – 0.3x)
(a) Factor out (x – 0.3x) from each term. Then simplify the
resulting expression.
(b) Use the simplified expression from part (a) to answer these
questions. With a 30% reduction followed by a 30% reduction,
is the television selling at 40% of its original price? If not, at
what percentage of the original price is it selling?
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 5.3
Factoring by Grouping
CONTINUED
SOLUTION
(a) (x – 0.3x) – 0.3(x – 0.3x)
= 1(x – 0.3x) – 0.3(x – 0.3x)
= (x – 0.3x)(1 – 0.3)
= (x – 0.3x)(0.7)
This step, usually omitted,
shows each term as the
product of the GCF and its
other factor, in that order.
Factor out the GCF
Subtract
= 0.7x – 0.21x
= 0.49x
Distribute
Subtract
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 5.3
Factoring by Grouping
CONTINUED
(b) With a 30% reduction, followed by another 30% reduction,
the expression that represents the reduced price of the television
simplifies to 0.49x. Therefore, this series of price reductions
effectively gives a new price for the television at 49% its
original price, not 40%.
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 5.3
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