Section 13.5 special factoing techniques

Slide - 1Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
2
Factoring and
Applications
13

1. Factor the difference of squares.
2. Factor a perfect square trinomial.
3. Factor a difference of cubes.
4. Factor a sum of cubes.
Objectives
13.5 Special Factoring Techniques

Factor a Difference of Squares
Factoring a Difference of Squares
x2 – y2 = (x + y)(x – y)
The following conditions must be true for a binomial to be a
difference of squares.
1. Both terms of the binomial must be squares, such as
x2, 9y2, 25, 1, m4
2. The terms of the binomial must have different signs (one
positive and one negative).

y yy
(a) x2 – 64
Example
Factor each binomial, if possible.
x2 y2
= (x + 8)(x – 8)= (x)2 – (8)2
x x x
(b) y2 + 36 Since y2 + 36 is a sum of squares, it cannot be
factored. It is a prime polynomial.
CAUTION
After any common factor is removed, a sum of squares
cannot be factored.

Example
Factor each binomial, if possible.
(a) 9y2 – 121 = (3y + 11)(3y – 11)= (3y)2 – (11)2
(b) 25z2 – 12 Because 12 is not the square of an integer,
this binomial is not a difference of squares.
It is a prime polynomial.

Example
Factor completely.
(a) y4 – 81
= (y2 + 9)(y2 – 9)
= (y2)2 – (9)2
= (y2 + 9)(y + 3)(y – 3)

(b) 48m4 – 243n4
Example (cont)
Factor completely.
= 3(16m4 – 81n4) Factor out the GCF of 3.
= 3[(4m2)2 – (9n2)2]
= 3(4m2 + 9n2)(4m2 – 9n2) Difference
of squares
Sum of squares
cannot be factored.
= 3(4m2 + 9n2)(2m + 3n)(2m – 3n)
CAUTION
Factor again when any of the factors is a difference of
squares. Check by multiplying.

Factor a Perfect Square Trinomial
Factoring Perfect Square Trinomials
x2 + 2xy + y2 = (x + y)2
x2 – 2xy + y2 = (x – y)2
A perfect square trinomial is a trinomial that is the
square of a binomial.

Example
(a) Factor x2 – 20x + 100.
This looks like the second form of a
perfect square trinomial. So, we
will guess that this factors as
(x – 10)2
To check, take twice the product of the two terms in the squared
binomial.
2 · x · 10 = 20x
Since 20x is the middle term of the trinomial, the trinomial is a
perfect square.
x2 –20x + 100 = (x – 10)2
x y

Example
(a) Factor 9z2 + 16z + 64.
This looks like the first form of a
perfect square trinomial. So, we will
guess that this factors as
(3z + 8)2
To check, 2 · 3z · 8 = 48z should be the middle term of the
trinomial. That is not the case. This trinomial is not a perfect
square, and cannot be factored. It is a prime polynomial.
x y

Example (cont)
(b) Factor 16y3 + 56y2 + 49y.
First, factor out the GCF of y.
16y3 + 56y2 + 49y = y(16y2 + 56y + 49) It looks like we have a
perfect square trinomial.
Guess the factors.= y(4y + 7)2
To check, 2 · 4y · 7 = 56y should be the middle term of the
trinomial. Since that is the case, our guess was right, and the
original polynomial factors.
16y3 + 56y2 + 49y = y(4y + 7)2

Note
1. The sign of the second term in the squared binomial
is always the same as the sign of the middle term in
the trinomial.
2. The first and last terms of a perfect square trinomial
must be positive, because they are squares. For
example, the polynomial x2 – 2x – 1 cannot be a
perfect square because the last term is negative.
3. Perfect square trinomials can also be factored using
grouping or FOIL, although using the method of this
section is often easier.

Difference of Cubes
Difference of Cubes
x3 – y3 = (x – y)(x2 + xy + y2)

–5a
a3
–125
Factoring Difference of Cubes
Factor each polynomial. Recall, x3 – y3 = (x – y)(x2 + xy + y2).
(a) a3 – 125
Check: = (a – 5)(a2 + 5a + 25)
Opposite of the product of the cube
roots gives the middle term.
Example
 3 3
5a
     2 2
5 5 5a a a
     2
5 5 25a a a

(b) 8g3 – h3
Continued.
  
3 3
2g h
          
 
2 2
2 2 2g h g g h h
     2 2
2 4 2g h g gh h

(c) 64m3 – 27n3
Continued.
    
3 3
4 3m n
            
 
2 2
4 3 4 4 3 3m n m m n n
     2 2
4 3 16 12 9m n m mn n

Sum of Cubes
Sum of Cubes
x3 + y3 = (x + y)(x2 – xy + y2)

Note on Signs
NOTE
Difference of Cubes x3 – y3 = (x – y)(x2 + xy + y2)
Sum of Cubes x3 + y3 = (x + y)(x2 – xy + y2)
The sign of the second term in the binomial factor of a sum or difference
of cubes is always the same as the sign in the original polynomial.
In the trinomial factor, the first and last terms are always positive;
the sign of the middle term is the opposite of the sign of the second term
in the binomial factor.

Factoring Sums of Cubes
Factor each polynomial. Recall, x3 + y3 = (x + y)(x2 – xy + y2).
(a) n3 + 8
(b) 64v3 + 27g3
Example
 3 3
2n
     2 2
2 2 2n n n
     2
2 2 4n n n
    
3 3
4 3v g
            
 
2 2
4 3 4 4 3 3v g v v g g
     2 2
4 3 16 12 9v g v gv g

Factoring Sums of Cubes
Factor each polynomial. Recall, x3 + y3 = (x + y)(x2 – xy + y2).
(c) 2k3 + 250
Continued.
  3
2 125k
  3 3
2 5k
     2
2 5 5 25k k k

Section 13.5 special factoing techniques

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