 The difference of two squares:  2nd Level

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The form (a + b)(ab)

Factoring by grouping

The sum and difference of odd powers

The difference of even powers

Example 2.  The form (a + b)(ab).

The following has the form (a + b)(ab):

(x + y + 8)(x + y − 8)

We may view  x + y  as the first term. 8 is the second.  Therefore it will produce the difference of two squares:

 (x + y + 8)(x + y − 8) = (x + y)2 − 64 = x2 + 2xy + y2 − 64

-- upon applying the rule for the square of a binomial.

Problem 7.    Each of these will produce the difference of two squares. Multiply.

 a)  (p + 3q + 2)(p + 3q − 2) = (p + 3q)2 − 4 = p2 + 6pq + 9q2 − 4
 b)  (x − y − 1)(x − y + 1) = (x − y)2 − 1 = x2 − 2xy + y2 − 1

Example 3.  Factoring by grouping.

x3 + 2x2 − 25 x − 50.

Let us factor that by grouping (Lesson 15) -- and then recognize the difference of two squares:

 x3 + 2x2 − 25x − 50 = x2(x + 2) − 25(x + 2) = (x2 − 25)(x + 2) = (x + 5)(x − 5)(x + 2)

Problem 8.   Factor by grouping.

 a)  x3 + 3x2 − 4x − 12 = x2(x + 3) − 4(x + 3) = (x2 − 4)(x + 3) = (x + 2)(x − 2)(x + 3)
 b)  x3 − 4x2 − 9x + 36 = x2(x − 4) − 9(x − 4) = (x2 − 9)(x − 4) = (x + 3)(x − 3)(x − 4)
 c)  2x3 − 3x2 − 50x + 75 = x2(2x − 3) − 25(2x − 3) = (x2 − 25)(2x − 3) = (x + 5)(x − 5)(2x − 3)
 d)  3x3 + x2 − 3x − 1 = x2(3x + 1) − (3x + 1) = (x2 − 1)(3x + 1) = (x + 1)(x − 1)(3x + 1)

The sum and difference of odd powers

The sum and difference of 5th powers can be factored as follows:

a5 + b5 = (a + b)(a4a3b + a2b2ab3 + b4)

a5 b5 = (ab)(a4 + a3b + a2b2 + ab3 + b4)

It is possible to verify that those are the factors by multiplying the right-hand sides.  If you do, then upon multiplying the two factors of a5 + b5, you will find ten terms:  five upon multiplication by a, and five upon multiplication by b.  Eight of them, however, must cancel in order to be left only with a5 + b5.

In fact, they cancel in pairs -- which is why the signs in the second factor must alternate.

For example, when a multiplies −a3b, the product is −a4b.  But when b multiplies a4, the product is +a4b, and they cancel.

Terms will thus cancel in pairs, leaving a5 + b5.

The important thing however is to study the form of the factors.
(a + b) is a factor of  a5 + b5; while (a b) is a factor of  a5 b5 .

In each second factor, the first term is a4.  (Multiplication by a then produces a5.)  The exponent of a then decreases as the exponent of b increases -- but the sum of the exponents in each term is 4.  (We say that the degree of each term is 4.)

In the second factor of a5b5, all the signs are +.  That insures the canceling.

For a proof of that factoring based on the Factor Theorem, see Topic 13 of Precalculus.

Problem 9.   Factor the following.

[Hint: 32 = 25.  Therefore, x5 + 32 has the form a5 + b5, with b = 2.,]

 a)  x5 + 32 = (x5 + 25) = (x + 2)(x4 − 2x3 + 4x2 − 8x + 16)
 b)  x5 − 32 = (x − 2)(x4 + 2x3 + 4x2 + 8x + 16)

xn − 1  can always be factored, because 1 = 1n, and all powers of 1 equal 1.

Problem 10.   Factor the following.

 a)  x5 − 1 = (x5 − 15) = (x − 1)(x4 + x3 + x2 + x + 1)
 b)  x5 + 1 = (x + 1)(x4 − x3 + x2 − x + 1)

Problem 11.  The sum and difference of two cubes.  Factor.

a)  a3 + b3 =  (a + b)(a2ab + b2)

b)  a3b3 =  (ab)(a2 + ab + b2)

In practice, it is mainly these that tend to come up.

Problem 12.   Factor.

 a)  x3 + 8 = x3 + 23 = (x + 2)(x2 − x· 2 + 22) = (x + 2)(x2 − 2x + 4)
 b)  x3 − 1 = x3 − 13 = (x − 1)(x2 + x· 1 + 12) = (x − 1)(x2 + x + 1)
 c)  x3 + 1 = (x + 1)(x2 − x + 1)

The difference of even powers

So much for the sum and difference of odd powers.  As for even powers, only their difference can be factored.  (If you doubt that, then try to factor a2 + b2 or a4 + b4.  Verify your attempt by multiplying out.)

If the exponent is even, then we can always recognize the difference of two squares:

a4b4 = (a2 + b2)(a2b2).

But also when n is even,  anbn  can be factored either with (ab) as a factor or (a + b).

a4b4 = (ab)(a3 + a2b + ab2 + b3)

a4b4 = (a + b)(a3a2b + ab2b3)

[If n is odd, then  anbn  can be factored only with the factor
(ab).]

When n is even, then for the following reason it is not possible to factor an + bn.

The factoring would begin:

(a + b)(an−1an−2b + . . .).

Upon continuing to alternate signs, the last term would be −bn−1. Therefore, upon multiplying that by +b, the product will be −bn, not +bn.

Problem 13.   Factor  x4 − 81  with (x + 3) as a factor.

 x4 − 81 = x4 − 34 = (x + 3)(x3 − x2· 3 + x· 32 − 33) = (x + 3)(x3 − 3x2 + 9x − 27).

Problem 14.   x4 − 1 = (x − 1)(x3 + x2 + x + 1)

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Next Lesson:  Exponents II

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