X3 - 5x2 - 13x - 7?

Answer 1

This does readily factor!!!

So by trial and error.

Discount '0' because the you are left with '-7'

Try '-1

(-1)^(3) - 5(-1)^(2) - 13(-1) - 7 =>

-1 - 5 + 13 - 7 adding = '0'

Hence x = -1

Therefore ( x + 1) is a factor.

(x+1)

( x^(3) - 5(x^(2) - 13x - 7 = (x^(2)

-x^(3) -x^(2) =

-6x^(2) - 13x = ( x^(2) -6x

6x^(2) + 6x =

-7(x + 1) ( x^(2) - 6x + 7)

(x + 1)(x^(2) - 6x + 7) is fully factored.

If you factor the Quadratic, you will move into the region of IMGINARY (i) numbers.

Answer 2

So the roots are -1 and 7

(x-7)(x+i)(x-i)