X 5 32

w factor (x

(aõ bà µ) = (a b) (aà + a³b + a²b² + ab³ + bô)

(xà µ 32) = (x 2) (xô + x³ (2) + x² (2)  + x (2) ³ + (2) Â)

(xà  32) = (x 2) (xô + 2x³ + 4x² + 8x + 16)

X 5 32

X 5 32

This is a very good question. Here are some helpful things you should know:

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

This sounds scary, I know, but it helps when you look at some examples. Consider cases n = 2, 3, 4 and 5:

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, then:

x 5 32 = (x2) (x 4 + 2x 3 + 4x 2 + 8x + 16)

Quartic factor is irreparable. It can be shown that a major factor from x n to a n, where a is a number, cannot be reduced if and only if n is a prime number (e.g. n = 5). ۔ Tasting is trivial, but it's true, so I won't spend much time considering it in any other way.

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Another unconventional way of looking at things is the geometric series. Catch:

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

Or maybe imagine a geometric series with a long equation and a general ratio of the initial term x 4 and y / x. The first term is x 4, the second term is:

x 4 * y / x = x 3y

The third term is:

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

Etc. We know the formula of the geometric series combination 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) / (xy)

= (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)

Chewing gum.

X5 32

x 5 32 = x 5 2 5, so we need to add (x2) (x 4 + ... + 2 4). If you multiply what we get.

x 5 2x 4 + 16x 32 y ... We need to add the term in ... to cancel 2x 4 + 16x.

To cancel 2x 4, we need to add 2x 3 ... so if (x 2) we get (x 4 + 2x 3 + ... + 16)

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 to ..., so if (x2) (x 4 + 2x 3 + 4x 2 + ... + 16)

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

x 5 + 16x 8x 2 32

To cancel 8x 2, we have to add 8x to ... if (x 2) we get (x 4 + 2x 3 + 4x 2 + 8x + ... + 16)

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 more terms ...

x 5 32 = (x2) (x 4 + 2x 3 + 4x 2 + 8x + 16)

X 5 32

X 5 32

(x 5 32) = (x2) (x 4 + 2x³ + 4x² + 8x + 16)

X 5 32