Question

One root of f(x)=x^3-9x^2+26x-24 is x = 2. What are all the roots of the function? Use the Remainder Theorem.

Answer 1

Answer:

$x=3$ or $x=4$

Step-by-step explanation:

the given function is;

$f(x)=x^3-9x^2+26x-24$

According to the rational roots theorem, the possible rational roots are;

$\pm1,\pm2\pm3,\pm4,\pm6,\pm8,\pm12,\24$.

According to the Remainder Theorem, if $f(a)=0$, then $x=a$ is a zero of the polynomial.

$f(3)=3^3-9(3)^2+26(3)-24$

$f(3)=27-81+78-24$

$f(3)=24-24=0$

Also,

$f(4)=4^3-9(4)^2+26(4)-24$

$f(4)=64-144+104-24$

$f(4)=64-64=0$

Therefore the other roots are;

$x=3,x=4$