How do you factor #x^3-2x^2+4x-8#?

2 Answers
Aviv S.
Mar 12, 2018

The factored form is #(x^2+4)(x-2)#.

Explanation:

You have to factor by grouping:

# color(white)=x^3-2x^2+4x-8 #

# =color(blue)(x^2)*x-color(blue)(x^2)*2+4x-8 #

# =color(blue)(x^2)(x-2)+4x-8 #

# =color(blue)(x^2)(x-2)+color(red)4*x+color(red)4*-2 #

# =color(blue)(x^2)(x-2)+color(red)4(x-2) #

# =(color(blue)(x^2)+color(red)4)(x-2) #

Ben
Mar 12, 2018

This factors to #(x^2+4)(x-2)#

Explanation:

Starting with the left-hand side, we see #x^3#; lets factor that first.

Based on the middle terms, I would believe that the factors are uneven (NOT #x^(3/2)#) This is because that if it WAS even, there wouldn't be two terms in the middle of the equation. I begin to write the factors as such:

#(x^2+a)(x+b)#

Now, expand the above and back-substitute the values of a & b from the starting equation:

#(x^2+a)(x+b)=x^3+bx^2+ax+ab#

Comparing the middle coefficients, it would seem that b=-2 and a=4. This is confirmed with the final term #ab=4*(-2)=-8#

Therefore, the factored solution is:
#(x^2+4)(x-2)#

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