How do you factor (x^5 - 32)

How do you factor binomials when ones raised to the fifth??
Please show

(a⁵ - b⁵) = (a - b) (a⁴ + a³b + a²b² + ab³ + b⁴)


(x⁵ - 32) = (x - 2) (x⁴ + x³(2) + x²(2)² + x(2)³ + (2)⁴)

(x⁵ - 32) = (x - 2) (x⁴ + 2x³ + 4x² + 8x + 16)

That's a very good question. Here's a helpful factorisation you should know:

x^n - y^n = (x - y)(x^(n - 1) + x^(n - 2) y + x^(n - 3) y^2 + ... + x^2 y^(n - 3) + xy^(n - 2) + y^(n - 1))

That looks truly frightening, I know, but it helps if you see some examples. Look at the n = 2, 3, 4, and 5 cases:

x^2 - y^2 = (x - y)(x + y)
x^3 - y^3 = (x - y)(x^2 + xy + y^2)
x^4 - y^4 = (x - y)(x^3 + x^2 y + xy^2 + y^3)
x^5 - y^5 = (x - y)(x^4 + x^3 y + x^2 y^2 + x y^3 + y^4)
etc.

So, in this case, we have y = 2, so:

x^5 - 32 = (x - 2)(x^4 + 2x^3 + 4x^2 + 8x + 16)

The quartic factor is actually irreducible. It can be proven that the large factor of x^n - a^n, where a is an integer, is irreducible exactly when n is prime (e.g. n = 5). Proving that is tricky, but it is true, so I wouldn't spend too much time trying to factorise it further.

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Another slightly off-the-wall way to look at this is as a geometric series. Take:

x^4 + x^3 y + x^2 y^2 + x y^3 + y^4

or perhaps a longer equivalent, and consider it as a geometric series with initial term x^4, and common ratio of y/x. The first term is x^4, the second term is:

x^4 * y/x = x^3 y

The third term is:

x^3 y * y/x = x^2 y^2

etc. We know a formula for the sum of a geometric series, which gives us:

x^4 + x^3 y + x^2 y^2 + x y^3 + y^4 = x^4 * (1 - (y/x)^5) / (1 - y/x)
= x^5 * (1 - (y/x)^5) / (x - y)
= (x^5 - y^5) / (x - y)
x^5 - y^5 = (x - y)(x^4 + x^3 y + x^2 y^2 + x y^3 + y^4)

Chew it over.

x^5 - 32 = x^5 - 2^5 so we need something like (x - 2)(x^4 + ... + 2^4). When you multiply this out we get

x^5 - 2x^4 + 16x - 32 and the ... We need to add terms in the ... to cancel the -2x^4 + 16x.

To cancel -2x^4 we need to add 2x^3 in the ... so when (x - 2)(x^4 + 2x^3 + ... + 16) we get

x^5 + 2x^4 + 16x - 2x^4 - 4x^3 - 32
x^5 + 16x - 4x^3 - 32

To cancel -4x^3 we need to add 4x^2 in the ... so when (x - 2)(x^4 + 2x^3 + 4x^2 + ... + 16) we get

x^5 + 2x^4 + 4x^3 + 16x - 2x^4 - 4x^3 - 8x^2 - 32
x^5 + 16x - 8x^2 - 32

To cancel the -8x^2 we need to add 8x in the ... so when (x - 2)(x^4 + 2x^3 + 4x^2 + 8x + ... + 16) we get

x^5 + 2x^4 + 4x^3 + 8x^2 + 16x - 2x^4 - 4x^3 - 8x^2 - 16x - 32
x^5 - 32

So we don't need any more terms in the ...

Therefore, x^5 - 32 = (x - 2)(x^4 + 2x^3 + 4x^2 + 8x + 16)

(x^5 - 32)=(x-2)(x^4+2x³+4x²+8x+16)