Question

if p is the root of x^2+2x+6=0 then find the value of(p+2)(p+3)(p+4)(p+5)​

Answer 1

Answer:

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P(x)=x2+px-6 =0

Putting the value of x as -2

P(-2)=(-2)2+(-2)p-6 =0

P(-2)=4-2p-6=0

P(-2)= -2-2p=0

P(-2)= -2p=2

P(-2)= p=-2/2

p=-1

We got the value of p

Now putting the value of p in the equation to find the zeroes of it.

4x2-2(-1)x-6=0

4x2+2x-6=0

4x2+6x-4x-6=0

2x(2x+3)-2(2x+3)=0

(2x+3) , (2x-2)

(x=-3/2) ,(x=2/2 =1) ARE THE TWO zeroes

We have given that the second equation has the same factors as the first one therefore we will substitute the value of x as 1 and p as -1 .

G(x) =x2-px+q=0

G(1)=(1)2-(-1)(1)+q=0

G(1)=1+1+q=0

=2+q=0

q=-2

Therefore , p=-1 & q=-2

Answer 2

please see the attachment