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

1 Answer
Jun 16, 2017

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