How do you factor #x^3+7x^2+14x+8#?

2 Answers
Mar 5, 2018

Answer:

#color(magenta)(=(x+1)(x+4)(x+2)#

Explanation:

#x^3+7x^2+14x+8#

#=x^3-x+7x^2+15x+8# [Adding and subtracting #x#]

#=x(x^2-1)+7x^2+15x+8#

Identity:
#color(red)(a^2-b^2=(a+b)(a-b) # &
#color(red)((x+a)(x+b)=x^2+(a+b)x+ab#

#= x(x+1)(x-1)+[7x(x+1)+8(x+1)#

#= x(x+1)(x-1)+(x+1)(7x+8)#

#= (x+1)[x(x-1)(7x+8)]#

#= (x+1)[x^2-x+7x+8]#

#= (x+1)[x^2+6x+8]#

Identity:
#color(red)((x+a)(x+b)=x^2+(a+b)x+ab#

#color(magenta)(=(x+1)(x+4)(x+2)#

~Hope this helps! :)

Mar 5, 2018

Answer:

(x + 1)(x + 2)(x + 4)

Explanation:

#f(x) = x^3 + 7x^2 + 14x + 8#
First, we note that f(-1) = - 1 + 7 - 14 + 8 = 0.
So, one factor is (x - 1).
After division, or guest -->
#f(x) = (x + 1)(x^2 + 6x + 8)#
Find 2 numbers knowing the sum (6) and the product (8). They are 2 and 4. Finally,
f(x) = (x + 1)(x + 2)(x + 4)