F(x)=x3+5x2+11x+15

Question

Find the zeros

Answer 1

f(x) = x³+5x²+11x+15

 

Since all of the coefficients are positive, there are no positive roots.  

 

So, using that fact together with the Rational Zero Theorem, the only possible rational zeros are -1, -3, -5, and -15.

 

Checking, we find that -3 is a zero.  So, x - (-3) is a factor of f(x).

 

Divide f(x) synthetically by x-(-3) = x + 3.

 

-3⌋   1      5      11      15

    ______-3___-6____-15

       1      2       5       0

 

So, f(x) = (x+5)(x²+2x+5)

 

Setting x²+2x+5 = 0, we get x = [-2 ± √-16]/2 = [-2 ± 4i]/2

 

                                                                   = -1 ± 2i

 

The roots of f(x) are -3, -1+2i, and -1-2i.