How do you use the remainder theorem to evaluate $f(x)=x^5-47x^3-16x^2+8x+52$ at x=7?

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Answer 1 · 2017-01-16T14:35:20

$f(7)=10$

Remainder theorem states that if a polynomial $f(x)$ is divided by $(x-a)$ , the the remainder is $f(a)$ .

Hence to evaluate $f(x)=x^5-47x^3-16x^2+8x+52$ at $x=7$ , we need to divide $f(x)=x^5-47x^3-16x^2+8x+52$ by $(x-7)$ .

Now, $f(x)=x^5-47x^3-16x^2+8x+52$

$=x^4(x-7)+7x^3(x-7)+2x^2(x-7)-2x(x-7)-6(x-7)+10$

$=(x^4+7x^3+2x^2-2x-6)(x-7)+10$

As such the remainder on dividing $f(x)$ by $(x-7)$ is $10$

hence $f(7)=10$

Check $f(6)=7^5-47xx7^3-16xx7^2+8xx7+52$

$=16807-16121-784+56+52=10$

How do you use the remainder theorem to evaluate f(x)=x^5-47x^3-16x^2+8x+52f(x)=x^5-47x^3-16x^2+8x+52 at x=7?