Question

How do you find the real solutions of the polynomial `x^3-7x^2=7-x`?

Answer 1

`x=7`

Explanation:

`x^3-7x^2=7-x`

`=>x^3-7x^2+x-7=0`

let`" "f(x)=x^3-7x^2+x-7`

for cubics there is a least one real root.

finding this root by the factor theorem. The root(s) must be a factor of `7`.

`f(7)=7^3-7^2xx7+7-7`

`f(7)=cancel(7^3-7^3)+cancel(7-7)=0`

`:. x=7" is a root"`

to see if there are any other real root we factorise

`x^3-7x^2+x-7=(x-7)(x^2+bx+c)`

comparing constants

`=>c=1`

comparing coefficients of`" " x^2`

`LHS=-7`

`RHS=-7+b=>b=0`

giving

`x^3-7x^2+x-7=(x-7)(x^2+1)`

other roots `" "x^2+1=0`

no real roots.

`:.` the only real root is `x=7`