Question

If one of the zeros of the polynomial f(x)=x^2+px+12 is 4,find the value of constant p

Answer 1

`p=-7`

Explanation:

Given
`f(x)=x^2+px+12`, when
`f(x)=0`, the right-hand side will break down to
`(x+a)(x+b)` where
`a+b=p` and
`ab=12`.

Because we know one of the zeroes is
`x=4`, then either
`a=-4` or
`b=-4`. If we decide that
`a=-4`, for example, and we expand
`(x-4)(x+b)`, we get
`x^2+(b-4)x-4b`, which means
`b=-3` and
`p=-7`.

We know this because
`-4(-3)=12` and
`-4+(-3)=-7`.