How do you factor #x^5+x^3+8x^2+8#?

1 Answer
Nov 16, 2015

Answer:

Factor by grouping and the sum of cubes identity to find:

#x^5+x^3+8x^2+8 = (x^2+1)(x+2)(x^2-2x+4)#

Explanation:

The sum of cubes identity may be written:

#a^3+b^3=(a+b)(a^2-ab+b^2)#

We use this with #a=x# and #b=2# to find:

#x^5+x^3+8x^2+8#

#=(x^5+x^3)+(8x^2+8)#

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

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

#=(x^2+1)(x^3+2^3)#

#=(x^2+1)(x+2)(x^2-x(2)+2^2)#

#=(x^2+1)(x+2)(x^2-2x+4)#