Grade 8 Mathematics Quarter 1

1. FACTORS OF
POLYNOMIALS
Factoring the Difference
of Two Squares

2. Jessebel G. Bautista
Antonio J. Villegas Voc’l High School
My
Profile

3. OBJECTIVES
• Identify the perfect squares
• Factor the difference of two squares
• Factor with accuracy

4. Factoring a polynomial involves writing
it as a product of two or more
polynomials. It reverses the process of
polynomial multiplication.
Every polynomial that is a difference of squares can
be factored by applying the following formula:
a2 – b2 = (a + b) (a – b)

5. Every polynomial that is a
difference of squares can be
factored by applying the
following formula:
a2 – b2 = (a + b) (a – b)
Note that a and b in the pattern
can be any algebraic expression.
For example, for a = x , and b=2
we get the following:
x2 – 22 = (x – 2) (x + 2)
The polynomial x2 – 4 is now expressed in factored form, (x+2)(x−2).
We can expand the right-hand side of this equation to justify the
factorization:
(x + 2) (x – 2) = x(x – 2) + 2(x – 2)
= x2 – 2x + 2x – 4
= x2 – 4
Now that we understand the pattern, let's use it to factor a few more
polynomials.

6. Both x2 and 16 are perfect square since x2 = (x)2 and 16 = (4)2. In other
words.
x2 – 16 = (x)2 – (4)2
EXAMPLE 1: Factoring x2 - 16
Since the two squares are being subtracted, we can see that this
polynomial represents a difference of squares. We can use
the difference of squares pattern to factor this expression:
a2 – b2 = (a + b) (a – b)
In our case, a = x and b = 4. Therefore, our polynomial factors as
follows:
(x)2 – (4)2 = (x – 4) (x + 4)
We can check our work by ensuring the product of these two factors is
x2 – 16 .

7. Both 9x2 and 100 are perfect square since 9x2 = (3x)2 and 100 = (10)2.
In other words.
9x2 – 100 = (3x)2 – (10)2
EXAMPLE 2: Factoring 9x2 - 100
Since the two squares are being subtracted, we can see that this
polynomial represents a difference of squares. We can use
the difference of squares pattern to factor this expression:
a2 – b2 = (a + b) (a – b)
In our case, a = 3x and b = 10. Therefore, our polynomial factors as
follows:
(3x)2 – (10)2 = (3x – 10) (3x + 10)
We can check our work by ensuring the product of these two factors is
9x2 – 100 .

8. Both 36m2 and 49n4 are perfect square since 36m2 = (6m)2 and
49n4 = (7n2)2. In other words.
36m2 – 49n4 = (6m)2 – (7n2)2
EXAMPLE 3: Factoring 36m2 – 49n4
Since the two squares are being subtracted, we can see that this
polynomial represents a difference of squares. We can use
the difference of squares pattern to factor this expression:
a2 – b2 = (a + b) (a – b)
In our case, a = 6m and b = 7n2. Therefore, our polynomial factors as
follows:
(6m)2 – (7n2)2 = (6m – 7n2) (6m + 7n2)
We can check our work by ensuring the product of these two factors is
36m2 – 49n4 .

9. 1. 4x2 – 25
2. 9y2 – 100
3. 49m2 – 121n2
4. 144 – 16n2
5. m2n2 – 196
TO DO …
Factor the following:
6. 169a2 – b2c2
7. 9x2 – 81
8. 289m2 – 256n2
9. 16b2 – 9c2
10. 121c4 – 169d2

10. REFERENCES:
https://www.khanacademy.org/math/algebra/x2f8bb11
595b61c86:quadratics-multiplying-
factoring/x2f8bb11595b61c86:factor-difference-
squares/a/factoring-quadratics-difference-of-squares
Grade 8 Math Time
k-to-12-grade-8-math-learner-module