# Can you use the remainder theorem when f(x) is divided by x-(1/3)?

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### Question Description

Use the remainder theorem to find the remainder when f(x) is divided by x-(1/3). Then use the factor theorem to determine whether x-(1/3) is a factor of f(x).

f(x)= 3x^4-x^3+15x-5

Then is x-(1/3) a factor of f(x)?

Really appreciate your help!! Thank you!

shyamnair3
School: University of Virginia

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The remainder theorem states that the remainder of the division of a polynomial $f(x)$ by a linear polynomial $x-a$ is equal to $f(a) .$

$f(x)= 3x^4-x^3+15x-5$

To find the remainder when f(x) is divided by x-(1/3) using remainder theorem,

substitute x = 1/3 in f(x).

$\\ f\bigg(\frac{1}{3}\bigg)= 3\bigg(\frac{1}{3}\bigg)^4-\bigg(\frac{1}{3}\bigg)^3+15\times\bigg(\frac{1}{3}\bigg)-5\\ \\ f\bigg(\frac{1}{3}\bigg)=\frac{1}{27}-\frac{1}{27}+5-5\\ \\ f\bigg(\frac{1}{3}\bigg)=0$

So the remainder is 0.

The factor theorem states that a polynomial $f(x)$ has a factor $(x - k)$ if and only if $f(k)=0$.

We showed that $\dpi{120} f\bigg(\frac{1}{3}\bigg)=0$.

Hence, by factor theorem $\dpi{120} x-\frac{1}{3}$ is a factor of f(x).

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Anonymous
Excellent job

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