+0

Factorise.

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361
3

Factorise.

iv) x^4 - ( x- z)^4

i neeed helppppp!

Jan 18, 2015

#3
+5

x^4 - ( x- z)^4 =

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

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

[2x - z ] [z ] [2x^2 - 2xz + z^2 ]   Jan 19, 2015

#1
+5

$$\ x^4 - ( x-z)^4 \; \hspace{3ex}\leftarrow \text {Factor} \\ \; \\ \ = \; (x^2)^2 - ((x-z)^2)^2 \\ \ = \; (x^2 - (x-z)^2)(x^2 + (x-z)^2)\\ \ = \; (x - (x-z))(x + (x-z)) (x^2 + x^2)(x^2 + x^2 + z^2 - 2xz)\\ \ = \; (x - x + z)(2x-z)(x^2 + x^2 +z^2 - 2xz)\\ \ = \; z(2x-z)(2x^2 + z^2 - 2xz)$$

_

It is outside my normal nature to post something without a snarky comment.

Rosala, you better check it carefully because I may have a made an intentional mistake. I am a troll, you know!

It is easy enough for any of the top answerers to note my “error” but it doesn’t seem  they are around at the moment.

Jan 18, 2015
#2
+5

Thanks Nauseated, your answer looks good to me I thought your snarky comment about being a troll was quite funny :)

Rosala, only one small suggestion, when you do questions like this it makes it easier if you use square brackets as well as round brackets.  It makes it easier to seperate everything properly visually and in your head.

Jan 18, 2015
#3
+5

x^4 - ( x- z)^4 =

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

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

[2x - z ] [z ] [2x^2 - 2xz + z^2 ]   CPhill Jan 19, 2015