Question

The sum of all the zeroes of polynomial p(x) = $5x^5-20x^4+5x^3+50x^2-20x-40$ is ___.

Answer 1

$p\left(x\right)=5x^5-20x^4+5x^3+50x^2-20x-40$ $=5(x+1{)}^2(x-2{)}^3$

In this case we do have a product of 3 terms, however, the first is a constant and does not make the polynomial zero. So from the final two terms, the polynomial is zero for x = –1 and x = 2. Therefore, the zeroes of the polynomial are

x = - 1,-1, 2, 2, 2

Sum of the zeros = -1-1 + 2 + 2 + 2 = 4

Product of zeros = (-1)$\times$(-1)$\times$2$\times$2$\times$2 = 8

Easier method:
For any polynomial, sum of zeroes $=\frac{-b}{a}=\frac{20}{5}=4$
Product of zeroes $=\frac{-z}{a}$ [This is the formula for odd degree polynomials. For even degree polynomials, product of zeroes $=\frac{z}{a}$, where z is the constant term]
$=\frac{40}{5}=8$