Question

If `P(x)=x^3+13x^2-8x+20` what is the value of `P(x0` when `x=1`?

Answer 1

`P(-1)=40`

Explanation:

If `P(color(red)x)=color(red)x^3+13color(red)x^2-8color(red)x+20`
then
`color(white)("XX")P(color(red)(-1))=(color(red)(-1))^3+13 * (color(red)(-1))^2 - 8 * (color(red)(-1))+20`

`color(white)("XXXXXX")=-1 +13+8 +20`

`color(white)("XXXXXX")=40`

Answer 2

`40`

Explanation:

To evaluate P(-1), substitute x = - 1 in P(x)

`rArrP(color(red)(-1))=(color(red)(-1))^3+13(color(red)(-1))^2-8(color(red)(-1))+20`

`=-1+13+8+20`

`=40`