Question

What is the sum of the roots of the polynomial shown below f(x)=x^3-8x^2-23x+30

Answer 1

F(1)=0 so x=1 is a zero => x-1 is a factor 1 | 1 -8 -23 30 | 1 -7 -30 --------------------- 1 -7 -30 | 0 x^2-7x-30 is another factor x^2-7x-30=(x-12)(x+5) x=12, x=-5 our roots too s0 1+12+(-5)=13-5=8

Answer 2

Answer: The sum of roots of the polynomial is 8.

Explanation:

The given polynomial is,

$f(x)=x^3-8x^2-23x+30$

Use hit and trial method to find the one root. Since $\pm1$ is the possible rational root for each polynomial.

$f(1)=(1)^3-8(1)^2-23(1)+30=0$

Sicen at x=1 the value of f(x) is 0 therefore 1 is a root of the polynomial and (x-1) is a factor of the polynomial.

Use synthetic division or long division method to find the other factor of the polynomial.

$f(x)=x^3-8x^2-23x+30=(x-1)(x^2-7x-30)$

$f(x)=(x-1)(x^2-10x+3x-30)$

$f(x)=(x-1)(x(x-10)+3(x-10))$

$f(x)=(x-1)(x-10)(x+3)$

Equation each factor of f(x) equal to 0.

$x=-3,1,10$

So, the roots are -3, 1, 10. Sum of roots is,

$-3+1+10=8$

Therefore, the sum of roots of the polynomial is 8.