The difference of two squares: 2nd Level
The sum or difference of any powers
Example 2. The form (a + b)(a − b).
The following has the form (a + b)(a − b):
(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 this 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 or difference of any powers
The sum and difference of 5th powers can be factored as follows:
a5 + b5 = (a + b)(a4 − a3b + a²b² − ab3 + b4)
a5 − b5 = (a − b)(a4 + a3b + a²b² + ab3 + b4)
It is possible to verify that those are the factors by multiplying the right-hand sides. The important thing however is to study the form of the factors. A factor of a5 + b5 is (a + b), while a factor of a5 − b5 is
(a − b).
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 a5 + b5, the signs alternate. (If they did not, then on multiplying, nothing would cancel to produce only those two terms.)
In the second factor of a5 − b5, all the signs are + . (That insures the canceling.)
For a proof of that factoring based on the Factor Theorem, see Topic 13 of Precalculus.
In particular, xn − 1 can always be factored for any positive integer n,
because 1 = 1n, and all powers of 1 are 1.
Problem 9. Factor the following.
| a) x5 − 1 | = | (x − 1)(x4 + x3 + x² + x + 1) |
| b) x5 + 1 | = | (x + 1)(x4 − x3 + x² − x + 1) |
Problem 10. The sum and difference of two cubes. Factor.
a) a3 + b3 = (a + b)(a² − ab + b²)
b) a3 − b3 = (a − b)(a² + ab + b²)
In practice, it is mainly these that tend to come up.
Problem 11. 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 the sum and difference of 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 n is even, then we can always recognize the difference of two squares:
a4 − b4 = (a2 + b2)(a2 − b2).
But also when n is even, an − bn can be factored either with (a − b) as a factor or (a + b).
a4 − b4 = (a − b)(a3 + a2b + ab2 + b3)
a4 − b4 = (a + b)(a3 − a2b + ab2 − b3)
[If n is odd, then an − bn can be factored only with the factor
(a − b).]
For, with n even and the factor (a + b), the right-hand factor will have an even number of terms. And since those terms alternate in sign, the final term will be −bn−1. Therefore, −bn will be correctly produced upon multiplication with +b. But when n is odd, the right-hand factor will have an odd number of terms. Therefore the final term would be +bn−1. Hence it will be impossible to produce −bn upon multiplication with +b.]
Problem 12. 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). | |
Next Lesson: Algebraic fractions
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