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

1 Answer
Jun 16, 2017

Answer:

#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#