MHB Evaluating $f(15)$ without a Calculator

[anemone]

Let $f(x)= x^{11}-16x^{10}+16x^9-16x^8+16x^7-16x^6+16x^5-16x^4+16x^3-16x^2+16x-1$.

Evaluate $f(15)$ without using the calculator.

 

[kaliprasad]

Let $f(x)= x^{11}-16x^{10}+16x^9-16x^8+16x^7-16x^6+16x^5-16x^4+16x^3-16x^2+16x-1$.

Evaluate $f(15)$ without using the calculator.

Spoiler

we have

$f(x) = x^{10}(x-15) - x^9(x-15) + x^8(x-15) - x^7(x-15) + x^6(x-15) - x^5(x-15) + x^4(x-15) - x^3(x-15) + x^2(x-15) - x(x-15) + x - 1$

so f(15) = 14

 

[anemone]

Spoiler

we have

$f(x) = x^{10}(x-15) - x^9(x-15) + x^8(x-15) - x^7(x-15) + x^6(x-15) - x^5(x-15) + x^4(x-15) - x^3(x-15) + x^2(x-15) - x(x-15) + x - 1$

so f(15) = 14

Very well done, kaliprasad, and thanks for participating!:)

 

[Greg]

Spoiler

By inspection, $$x-1$$ is a factor of $$f(x)$$. After polynomial long division, we have

$$f(x)=(x-1)(x^{10}-15x^9+x^8-15x^7+x^6-15x^5+x^4-15x^3+x^2-15x+1)$$

so

$$f(15)=14\cdot1=14$$

 

[anemone]

Spoiler

By inspection, $$x-1$$ is a factor of $$f(x)$$. After polynomial long division, we have

$$f(x)=(x-1)(x^{10}-15x^9+x^8-15x^7+x^6-15x^5+x^4-15x^3+x^2-15x+1)$$

so

$$f(15)=14\cdot1=14$$

Very good job, greg1313! And thanks for participating! :)