Question

Obtain all zeroes of p(x)=x3+13x2+32x+20 if one of its zero is -2

Answer 1

Here we are given the polynomial:
$p(x) = x^3+13x^2+32x+20$ 

We are given that -2 is a zero. 

That is, if we take p(x)=0, x = -2 is a solution. 

x = -2 is a solution. 
So, (x+2)=0 is a solution. This means that (x+2) is a factor of p(x). 

We can now factorize p(x) as below:

$x^3+13x^2+32x+20 \\ \\ =x^3+2x^2+11x^2+22x+10x+20 \\ \\ = x^2(x+2) + 11x(x+2) + 10(x+2) \\ \\ = (x+2)(x^2+11x+10) \\ \\ = (x+2)(x^2 + x + 10x + 10) \\ \\ = (x+2)(x(x+1)+10(x+1)) \\ \\ = (x+2)(x+1)(x+10)$


Now, we need zeros. 

$p(x)=0 \\ \\ x^3+13x^2+32x+20 = 0 \\ \\ (x+2)(x+1)(x+10)=0 \\ \\ x+2=0 \, \, \, OR x+1=0 \, \, \, OR \, \, \, x+10=0 \\ \\ \\ \implies \boxed{x=-2} \, \, \, OR \, \, \, \boxed{x=-1} \, \, \, OR \boxed{x=-10}$


Thus, -2, -1 and -10 are the zeros of the polynomial. 


Hope it helps
Purva
Brainly Community