How do you factor given that #f(-3)=0# and #f(x)=x^3+x^2+2x+24#?

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Answer 1 · 2017-05-24T14:39:46

# f(x)=(x+3)(x^2-2x+8).#

We have, for, #f(x)=x^3+x^2+2x+24,#

#f(-3)=-27+9-6+24=0.#

From the Factor Theorem, hence, we conclude that,

#x-(-3)=x+3# is a factor of #f(x).#

One way, then, is to divide #f(x)# by #(x+3)# by Long Division.

Instead, have a look at the following :

#f(x)=x^3+x^2+2x+24,#

#=ul(x^3+3x^2)-ul(2x^2-6x)+ul(8x+24),#

#=x^2(x+3)-2x(x+3)+8(x+3),#

# rArr f(x)=(x+3)(x^2-2x+8).#

For the Quadr. Poly. #x^2-2x+8,# we have,

#Delta=b^2-4ac=4-4(1)(8)=-28 <0.#

Hence, it can not be factored further in #RR.#

Enjoy Maths.!