# How do you find the rest of the zeros given one of the zero #c=-1# and the function #f(x)=x^5+2x^4-12x^3-38x^2-37x-12#?

##### 1 Answer

The zeros of

#### Explanation:

#f(x) = x^5+2x^4-12x^3-38x^2-37x-12#

#f(-1) = -1+2+12-38+37-12 = 0#

So

#x^5+2x^4-12x^3-38x^2-37x-12=(x+1)(x^4+x^3-13x^2-25x-12)#

Substituting

#1-1-13+25-12 = 0#

So the quartic also has

#x^4+x^3-13x^2-25x-12 = (x+1)(x^3-13x-12)#

Substituting

#-1+13-12 = 0#

So the cubic also has

#x^3-13x-12 = (x+1)(x^2-x-12)#

To factor the remaining quadratic, note that

#x^2-x-12 = (x-4)(x+3)#

hence zeros

So the zeros of