Question

Use the Remainder Theorem to completely factor p(x) = x 3 - 6x 2 + 11x - 6. can someone help me please!!!!!!

Answer 1

so by using remainder remainder theorm ,
we want the remainder zero here to completely factor p(x)

lets check

for p(1) here

\(\large \begin{align} \color{black}{
p(x) = x^3 - 6x^2 + 11x - 6\\~\\

p(1) = (1)^3 - 6(1)^2 + 11(1) - 6\\~\\
p(1) = 1 - 6 + 11- 6\\~\\
p(1) =0
}\end{align}\)

as p(1) =0 therefore \(\large \begin{align} \color{black}{ (x-1)}\end{align}\)

is a factor of p(x)

so divide \(\large \begin{align} \color{black}{p(x) = x^3 - 6x^2 + 11x - 6}\end{align}\)

by \(\large \begin{align} \color{black}{(x-1)}\end{align}\) here using Polynomiial

long division

U will get quotient \(\large \begin{align} \color{black}{(x^2-5x+6)}\end{align}\)

which further factors in \(\large \begin{align} \color{black}{(x-2)(x-3)}\end{align}\)


so \(\large \begin{align} \color{black}{p(x)=(x-1)(x-2)(x-3)}\end{align}\)