Question

factor completely s3-2s2-4s+8

Answer 1

\[s^3-2s^2-4s+8\]

Answer 2

how can i factor this completly or is it prime ?

Answer 3

i just say the form \[SOMTHING1^2 \times SOMMTHING2\]of factor is

Answer 4

Factor by grouping honey,
I'll show ya

s^3-2s^2 -4s+8
GCF is s^2 GCF is 4
So, So,
s^2(s-2s) 4(s-2)

Answer 5

Look wrong :P because...it's 2s..anyway..@amistre64 will help.

Answer 6

@amistre64

Answer 7

lol...

Anyway let us use factor by grouping as @uri suggested. Firstly factor out the \(\bf s^2\):\[\bf \implies s^3-2s^2-4s+8=s^2(s-2)-4s+8\]Now let's factor out \(\bf -4 \) from the rest of the unfactored expression:\[\bf s^2(s-2)-4s+8=s^2(s-2)-4(s-2)\]Now we can factor out the \(\bf s-2\) since we have two terms that share it:\[\bf =(s^2-4)(s-2)\]We can now factor \(\bf s^2-4\) using difference of squares which says that \(\bf (a+b)(a-b)=a^2-b^2\):\[\bf \implies (s^2-4)(s-2)=(s-2)(s+2)(s-2)\]Now we have two \(\bf s-2\) so we can re-write our final factored expression as:\[\bf =(s-2)^2(s+2)\]

Answer 8

@claudiaa794

Answer 9

you are the best thankyou c: