# If $4x^{10}-x^9-3x^8+5x^7+kx^6+2x^5-x^3+kx^2+5x-5$ when divided by $(x+1)$ gives a remainder of -14, then the value of k equals?

If $4x^{10}-x^9-3x^8+5x^7+kx^6+2x^5-x^3+kx^2+5x-5$ when divided by $(x+1)$ gives a remainder of -14, then the value of k equals?

I got this and similar type of question in a book and I don't really know how to exactly solve it. Any help will be appreciated.

• @egreg why did you censor your comment ? was it wrong ? Commented Apr 29, 2017 at 17:21
• @Evariste what do you mean? Commented Apr 29, 2017 at 17:22

Let $f(x)=4x^{10}-x^9-3x^8+5x^7+kx^6+2x^5-x^3+kx^2+5x-5$. Then $f(-1)=-14$ since the remainder when we divide by $x+1$ is $-14$.
So, $4+1-3-5+k-2+1+k-5-5=-14$ Thus, $2k=4$and $k=2$
Let $$f(x) = 4x^{10}-x^9-3x^8+5x^7+kx^6+2x^5-x^3+kx^2+5x-5$$. Now, since it is given that $$(x+1)$$ gives a remainder $$-14$$ when $$f(x)$$ is divided by it, then $$f(-1) = -14$$. So, placing $$-1$$ in place of $$x$$ we get $$4+1-3-5+k-2+1+k-5-5= -14 \implies 2k-14 = -14 \implies 2k=0.$$
So $$k=0$$.