Question

find the value of p and q so that 1, -2 are the zeros of the polynomial x^3+10x^2+px+q

Answer 1

Answer:

Step-by-step explanation:

p(x)=x^3+10x^2+px+q.

x=1 and -2(1 and-2 are zeroes of p(x)).

p(1)=0 and p(-2)=0.

So,p(1)=1^3+10×1^2+p×1+q.

=1+10+p+q.

=11+p+q. -eq(1).

Now,p(-2)=(-2)^3+10×(-2)^2+p×(-2)+q.

=(-8)+40-2p+q.

=32-2p+q. -eq(2).

We know that p(1)=p(-2).

So,11+p+q=32-2p+q.

2p+p=32-11+q-q.

3p=21.

p=21/3

p=7.

Put the value of p in eq(1),

11+p+q=0 (p(1)=0)

11+7+q=0

18+q=0

q=-18.

Hence,p=7 and q=-18.