In Section 5.2, you saw that (a ! b)(a " b) # a2 " b2. You can factor a
difference of squares as a2 " b2 # (a ! b)(a " b).
You also saw that (a ! b)2 # a2 ! 2ab ! b2 and
(a " b)2 # a2 " 2ab ! b2. You can factor a perfect square
trinomial as a2 ! 2ab ! b2 # (a ! b)2 or a2 " 2ab ! b2 # (a " b)2.
Example 1 Difference of Squares
Factor.
a) x2 " 100
b) 98a2 " 450b2
Solution
a)
a2 " b2 # (a ! b)(a " b)
" 100 # (x)2 " 102
# (x ! 10)(x " 10)
Use the pattern for a difference of squares.
x2
b) 98a2 " 450b2 # 2(49a2 " 225b2)
# 2[(7a)2 " (15b)2]
# 2(7a ! 15b)(7a " 15b)
Remove the greatest common factor.
Factor the difference of squares.
Example 2 Perfect Square Trinomials
Verify that each trinomial is a perfect square. Then, factor.
a) x2 ! 6x ! 9
b) x2 " 12x ! 36
Solution
a) Since x2 # (x)2 and 9 # 32, the first and last terms are perfect squares.
Since 6x # 2(x)(3), the middle term is twice the product of the
square roots of the first and last terms.
Therefore, x2 ! 6x ! 9 is a perfect square trinomial.
a2 ! 2ab ! b2 # (a ! b)2
Use the appropriate perfect
2
2
2
x ! 6x ! 9 # (x) ! 2(x)(3) ! 3
square trinomial pattern.
# (x ! 3)2
b) Since x2 # (x)2 and 36 # 62, the first and last terms are
perfect squares.
Twice the product of these square roots is 2(x)(6) # 12x.
Therefore, x2 " 12x ! 36 is a perfect square trinomial.
Use the appropriate perfect
a2 " 2ab ! b2 # (a " b)2
2
2
2
x " 12x ! 36 # (x) " 2(x)(6) ! 6
square trinomial pattern.
# (x " 6)2
The middle term of the
trinomial is —12x, so a
difference has been
squared.
5.6 Factor a Perfect Square Trinomial and a Difference of Squares • MHR 251
Example 3 More Complex Perfect Square Trinomials
Verify that each trinomial is a perfect square. Then, factor.
a) 4x2 ! 28x ! 49
b) 25k2 " 60km ! 36m2
Solution
a) Since 4x2 # (2x)2 and 49 # 72, the first and last terms are
perfect squares.
Since 28x # 2(2x)(7), the middle term is twice the product
of the square roots of the first and last terms.
Therefore, 4x2 ! 28x ! 49 is a perfect square trinomial.
4x2 ! 28x ! 49 # (2x)2 ! 2(2x)(7) ! 72
# (2x ! 7)2
The middle term of the
trinomial is —60km, so
a difference has been
squared.
b) Since 25k2 # (5k)2 and 36m2 # (6m)2, the first and last terms
are perfect squares.
Twice the product of these square roots is 2(5k)(6m) # 60km.
Therefore, 25k2 " 60km ! 36m2 is a perfect square trinomial.
25k2 " 60km ! 36m2 # (5k)2 " 2(5k)(6m) ! (6m)2
# (5k " 6m)2
Example 4 Area of a Region
a) Find an algebraic expression
for the area of the shaded region.
b) Write the area expression in
factored form.
Solution
a) The area of the shaded region is the
difference in the areas of the two squares.
Area # (3x ! 8)2 " (x " 2)2
b) Method 1: Expand, Then Factor
#
#
#
#
#
#
#
#
#
252 MHR • Chapter 5
(3x ! 8)2 " (x " 2)2
9x2 ! 48x ! 64 " (x2 " 4x ! 4)
9x2 ! 48x ! 64 " x2 ! 4x " 4
8x2 ! 52x ! 60
4(2x2 ! 13x ! 15)
4(2x2 ! 10x ! 3x ! 15)
4[(2x2 ! 10x) ! (3x ! 15)]
4[2x(x ! 5) ! 3(x ! 5)]
4[(x ! 5)(2x ! 3)]
4(x ! 5)(2x ! 3)
3x + 8
x— 2
Method 2: Factor as a Difference of Squares
This is a difference of squares, a2 ! b2, with a " (3x # 8)
and b " (x ! 2).
(3x # 8)2 ! (x ! 2)2
" [(3x # 8) # (x ! 2)][(3x # 8) ! (x ! 2)]
" (3x # 8 # x ! 2)(3x # 8 ! x # 2)
" (4x # 6)(2x # 10)
" 2(2x # 3)[2(x # 5)]
" 4(2x # 3)(x # 5)
Key Concepts
!
Always look for a common factor first when factoring a trinomial.
!
You can factor a difference of squares as a2 ! b2 " (a # b)(a ! b).
!
You can factor a perfect square trinomial as
a2 # 2ab # b2 " (a # b)2 or a2 ! 2ab # b2 " (a ! b)2.
Communicate Your Understanding
C1
Use words and diagrams to explain why x2 # 9 cannot be factored
over the integers.
C2
When her classmate showed Barbara the first step in Example 3b),
25k2 ! 60km # 36m2 " (5k)2 ! 2(5k)(6m) # (6m)2, Barbara asked,
“Where did the 2 come from?” Answer Barbara’s question.
Practise
For help with questions 1 and 2, see Example 1.
1. Factor.
a)
x2
For help with question 3, see Example 2.
3. Verify that each trinomial is a perfect
! 16
b)
y2
square. Then, factor.
! 100
c) 9k2 ! 36
d) 4a2 ! 121
a) x2 # 12x # 36
b) k2 # 18k # 81
e) 36w2 ! 49
f) 144p2 ! 1
c) y2 ! 6y # 9
d) m2 ! 14m # 49
g) 16n2 ! 25
h) 100g2 ! 81
e) x2 # 20x # 100
f) 64 ! 16r # r 2
For help with question 4, see Example 3.
2. Factor.
a) m2 ! 49n2
c) 100 !
9c2
b) h2 ! 25d2
d)
169a2
!
49b2
4. Verify that each trinomial is a perfect
square. Then, factor.
e) 25x2 ! 36y2
f) 16c2 ! 9d2
a) 4c2 # 12c # 9
b) 16k2 ! 8k # 1
g) 162 ! 8s2
h) 75h2 ! 27g2
c) 25x2 # 70x # 49
d) 9y2 ! 30y # 25
e) 100c2 ! 180c # 81
f) 25 # 80y # 64y2
5.6 Factor a Perfect Square Trinomial and a Difference of Squares • MHR 253
Connect and Apply
9. Determine two values of k so that
5. Each of the following is not factorable
over the integers. Why not?
a) 9x2 ! 16y
c) 10w2 ! 70wz " 49z2
d) 25n2 " 36m2
c) 49c2 ! k
10. Factor, if possible.
6. Factor fully, if possible.
a) 4x2 " 28xy " 49y2
b) 9k2 ! 24km " 16m2
c) 25p2 " 60pq " 144q2
d) 9y2 ! 7x2
e)
a) m2 ! kn2
b) kx2 ! 9
b) 36a2 " 107a " 81
2a2
each trinomial can be factored as a
difference of squares.
a) 9a2b2 ! 24abcd " 16c2d2
b) 225 ! (x " 5)2
c) (3c " 2)2 ! (3c ! 2)2
d) 4x2 " 26x " 9
11. The area of an unknown shape is
! 28ab "
98b2
f) 196n2 ! 144m2
g) 25x2 " 70xy " 14y2
h) 100f2 ! 120fg " 36g2
i) 400p3 ! 900pq2
For help with question 7, see Example 4.
7. a) Find an algebraic expression for
the area of the shaded region.
b) Write the area expression in
factored form.
2x + 5
x— 3
represented by 9x2 " 30x " 25. If x must
be an integer, what shape(s) could this
figure be?
12. A box is in the shape of a rectangular
prism. Its volume is given as x3 ! 2x2 " x.
a) Determine algebraic expressions
for the dimensions.
b) Describe the faces of the box.
13. Chapter Problem In
Section 5.3, question
12, you found an
algebraic expression
for the total of the
top surface areas of
2x + 5
the three prisms
used to make the pedestal.
x
x
x
2x + 5
a) Write algebraic expressions for the
8. Determine all values of b so that
each trinomial is a perfect square.
a)
y2
b)
4x2
c)
9n2
d)
w2
e)
81m2
f)
16x2
" by " 121
49p2
" 10w " b
exposed surface areas when x represents
5 cm.
14. The radius of a circle has been decreased
! 90m " b
! 88xy "
b) Factor each expression from part a).
c) Compare the expressions for the
" bx " 25
" bnp "
exposed top surface areas of the middle
and bottom layers of the pedestal.
b2y2
by a certain amount. Its area is now given
as #r 2 ! 14#r " 49#, where r was the
original radius, in centimetres.
a) What was the decrease in radius?
b) What was the decrease in area?
254 MHR • Chapter 5