one root of $f(x)=x^{3}-9x^{2}+26x - 24$ is $x = 2$. what are all the roots of the function? use the…

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.\n$x = 2,x = 3$, or $x = 4$\n$x=-2,x=-3$, or $x=-4$\n$x = 1,x = 2,x = 3$, or $x = 13$\n$x=-1,x=-2,x=-3$, or $x=-13$

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.\n$x = 2,x = 3$, or $x = 4$\n$x=-2,x=-3$, or $x=-4$\n$x = 1,x = 2,x = 3$, or $x = 13$\n$x=-1,x=-2,x=-3$, or $x=-13$

Answer

Explanation:

Step1: Use polynomial long - division

Since (x = 2) is a root of (f(x)=x^{3}-9x^{2}+26x - 24), divide (x^{3}-9x^{2}+26x - 24) by (x - 2). [ \begin{align*} \frac{x^{3}-9x^{2}+26x - 24}{x - 2}&=x^{2}-7x + 12 \end{align*} ]

Step2: Factor the quotient

Factor the quadratic (x^{2}-7x + 12). We need two numbers that multiply to (12) and add up to (7). The numbers are (3) and (4). So (x^{2}-7x + 12=(x - 3)(x - 4)).

Step3: Find all roots

The roots of (f(x)) are found by setting (f(x)=(x - 2)(x - 3)(x - 4)=0). Then (x=2), (x = 3), or (x=4).

Answer:

A. (x = 2,x = 3), or (x = 4)