How to factor 8x^2+x^3-8-x?

x³ + 8x² - x - 8
=> x² (x + 8) -1(x + 8)
=> (x + 8) (x² - 1)
=> (x + 8) (x + 1) (x - 1)

Got it!!!!!

8x^2+x^3-8-x

Write in standard form:

x^3+8x^2-x-8

Factor by grouping:

x^2(x + 8) - (x+8)


Factor a diff. of squares: (x^2 - 1) = (x+1)(x-1)

(x+8)(x^2 - 1)

(x+8)(x+1)(x-1) ---> answer

8x^2 + x^3 - 8 - x
= x^3 + 8x^2 - x - 8
= (x^3 + 8x^2) - (x + 8)
= x^2(x + 8) - 1(x + 8)
= (x + 8)(x^2 - 1)
= (x + 8)(x + 1)(x - 1)

f (x) = x ³ + 8 x ² - x - 8

f (1) = 1 + 8 - 1 - 8 = 0

Thus (x - 1) is a factor of f (x)

To find other factors use synthetic division:-

1 | 1____8____-1____-8
_ |_____ 1____ 9____ 8
_ | 1____9_____8____0

(x - 1)(x² + 9x + 8)

(x - 1)(x + 8)(x + 1)

(x+1)(x-1)(x+8)

the terms are written incorrect, lets arrange it first
x^3 - x + 8x^2 - 8
then bracket the similar terms
(x^3 - x) + (8x^2 - 8)
take out common factor of each term
x(x^2-1) + 8 (x^2-1)
then group them
we get
(x+8)(x^2-1)
further factorization
(x+8)(x+1)(x-1)

8x^2+x^3-8-x

rearrange 8x^2 + x^3 - 8 - x
x^3 + 8x^2 - x - 8
x(x^2 + 8x - 1) - 8