Question

pleae factorise step by step

Answer 1

\[x ^{4}-(x-z)^{4}\]

Answer 2

\[[x^2-(x-z)^2][x^2+(x-z)^2]\]

Answer 3

ya

Answer 4

k

Answer 5

can somone help me?!!!!!

Answer 6

\[[x-(x-z)][x+[x-z)][x^2+(x-z)^2]\]

Answer 7

the answer is \[z(2x-z)(2x ^{2}-2xz+z ^{2})\]

Answer 8

ima get confused if i type, so I'll just write it out on paper.. hold on

Answer 9

plzz do it step by step

Answer 10

So we're here right?
(x^2-(x-z)^2)(x^2+(x-z)^2)

Answer 11

that's by using a^2-b^2 = (a-b)(a+b)

Answer 12

k

Answer 13

where a = x^4, b = (x-z)^4

Answer 14

Right, so now expand the insides.

Answer 15

\[(x^2-(x-z)^2)(x^2+(x-z)^2)\]

Answer 16

(x^2-(x-z)^2)(x^2+(x-z)^2) = (x^2 - [x^2-2xz+z^2])(x^2 + [x^2-2xz+z^2])

Answer 17

\[(x^2-(x-z)^2)(x^2+(x-z)^2) = (x^2 - [x^2-2xz+z^2])(x^2 + [x^2-2xz+z^2])\]

Answer 18

(x^2 - [x^2-2xz+z^2])(x^2 + [x^2-2xz+z^2]) =
(x^2-x^2+2xz-z^2)(x^2+x^2-2xz+z^2)

Answer 19

= (2xz-z^2)(2x^2-2xz+z^2)

Answer 20

now pull out a z from the first parenthesis:
z(2x-z)(2x^2-2xz+z^2)

Answer 21

k

Answer 22

\[ x ^{4}-(x-z)^{4} = (x^2 +(x-z)^2) \times (x^2 - (x-z)^2)\]

using, \( a^2-b^2= (a+b)(a-b) \)

Applying the same thing on \( (x^2 - (x-z)^2) \) we will get \( (x - (x-z))\times (x + (x-z)) = z(2x-z) \)

now, \(x^2 +(x-z)^2 = 2 x^2-2 x z+z^2 \)

Now we have to put together the pieces we get \( x ^{4}-(x-z)^{4} = (2 x-z) z \left(2 x^2-2 x z+z^2\right) \)

Answer 23

Mertsj, I am not sure why I am wrong, but apparently wolfram is making a same mistake ( http://www.wolframalpha.com/input/?i=Factor%5Bx%5E4+-+%28x+-+z%29%5E4%5D )

Or am I missing something real silly? :)