How do you factor #x^3 + 10x^2 + 33x + 34#?

1 Answer
Jul 8, 2016

Answer:

#x^3+10x^2+33x+34=^=(x^2+8x+17)(x+2)#

Explanation:

As putting #x=-2#, we get #x^3+10x^2+33x+34=(-2)^3+10xx2^2+33xx(-2)+34=-8+40-66+34=0#. Hence, #-2# is a zero of the function and #(x+2)# is a factor of #x^3+10x^2+33x+34#.

Now dividing the function by #(x+2)#, as #x^3+10x^2+33x+34=^=x^2(x+2)+8x(x+2)+17(x+2)=(x^2+8x+17)(x+2)#, we get #x^2+8x+17#.

Now discriminant of #x^2+8x+17# is #8^2-4*1*17=64-68=-4# is negative, as such it cannot be factorized in real factors.

And hence factors of the function are

#x^3+10x^2+33x+34=^=(x^2+8x+17)(x+2)#