2. Common Factor and Difference of Squares

Factoring means writing an expression as the product of its simplest factors.

Example 1: Factoring a number

14 = 7 × 2

[7 and 2 are the simplest factors of 14. We can't break it down any more than this.]

Example 2: Factoring an algebraic expression

3x + 15 = 3(x + 5)

This means that the factors of 3x + 15 are

3, and

(x + 5)

To be able to factor successfully, we need to recognise the formulas from Section 1. So it's a good idea to learn those formulas well!

Factoring Difference of Two Squares

To factor the difference of 2 squares, we just apply the formula given in Section 1 - Special Products in reverse. That is:

x2y2 = (x + y)(xy)

Example 3: Factoring difference of 2 squares

Factor 36s2 − 121t2

Answer

We first recognize it is a difference of 2 squares, then we use the formula given above.

36s2 − 121t2

= (6s)2 − (11t)2

= (6s − 11t)(6s + 11t)

Exercises

Factor the following:

(1) 18p3 − 3p2

Answer

We note that 3p2 divides evenly into the 2 terms in the question. So the factorization is given by:

18p3 − 3p2 = 3p2(6p − 1)

(2) 5a + 10ax − 5ay + 20az

Answer

5a + 10ax − 5ay + 20az = 5a(1 + 2xy + 4z)

Easy to understand math videos:
MathTutorDVD.com

(3) 36a2b 2 − 169c2

Answer

This is a difference of 2 squares.

36a2b2 − 169c2

= (6ab)2 − (13c)2

= (6ab + 13c)(6ab − 13c)

(4) (ab)2 − 1

Answer

Once again, we recognize this as a difference of 2 squares.

(ab)2 − 1 = (ab)2 − 12

Put X = a b and Y = 1

So (ab)2 − 1

= X2Y2

= (X + Y)(X Y)

= (a b + 1)(ab − 1)

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(5) y4 − 81

Answer

y4 − 81

= (y2)2 − (9)2

= (y2 + 9)(y2 − 9)

= (y2 + 9)(y + 3)(y − 3)

Easy to understand math videos:
MathTutorDVD.com

(6) r2 s2 + 2st t2

Answer

We recognize that this involves 2 differences of two squares. We group it as follows:

r2 s2 + 2st t2

= r2 − (s2 − 2st + t2)

We recognize that s2 − 2st + t2 is a square, and equals (st)2. So we can factor our expression as follows:

r2 s2 + 2st t2

= r2 − (s2 − 2st + t2)

= r2 − (st)2

[This is also a difference of 2 squares.]

= [r − (st)][r + (st)]

= (rs + t)(r + st)

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