Question

`p(x)=3x^2-x^2+2x-5`, what is `p(x)` when `x=-2` and `x=3`?

Answer 1

`p(-2)=-37" and " p(3)=73`

Explanation:

To evaluate p( -2 ) and p( 3 ) substitute x = - 2 and x = 3
into p( x )

`rArrp(color(red)(-2))=3(color(red)(-2))^3-(color(red)(-2))^2+2(color(red)(-2))-5`

`color(white)(rArrp(-2))=(3xx-8)-(+4)+(2xx-2)-5`

`color(white)(rArrp(-2))=-24-4-4-5`

`color(white)(rArrp(-2))=-37`

`rArrp(color(magenta)(3))=3(color(magenta)(3))^3-(color(magenta)(3))^2+2(color(magenta)(3))-5`

`color(white)(rArrp(3))=(3xx27)-9+6-5`

`color(white)(rArrp(3))=81-9+6-5`

`color(white)(rArrp(3))=73`

Answer 2

`p(-2)=-37 and p(3)=73`

Explanation:

This is an function, which means that all values of `x` would satisfy the equation.

When `x=-2`,

`p(-2)=3(-2)^3-(-2)^2+2(-2)-5`
`color(white)(xxx//.)=-24-4-4-5`
`color(white)(xxx//.)=-37`

When `x=3`,

`p(3)=3(3)^3-(3)^2+2(3)-5`
`color(white)(x//.)=81-9+6-5`
`color(white)(x//.)=73`