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Here I am subtracting two cubic functions: $ax^3 +bx^2+cx+d -2(a(x+1)^3 +b(x+1)^2+c(x+1)+d) = -x^3-3x^2-3x-1$

Then I equate co-efficients to get:

$-1 = -a$

$-6a-b = -3$

$-6a-4b-c = -3$

$-6a-2b-2c-d = -1 $

Does this make sense so far?

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In the last equation you need $-2a$ –  Maesumi Jan 26 '13 at 19:16
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2 Answers 2

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Can you now solve for $a, b , c, d$? Your work this far looks good, by equating the coefficients of each side of the expanded equation, knowing the equality holds.

Just double check the coefficient for $a$ in the your last equation. (The coefficient of $a$ in your last equation should be $-2$, and not $-6$.)

But you might want to show your expansion of the subtracted polynomial, and your simplification, in your final solution (if handing this in) so you can justify your conclusions.

Do you know how to solve a system of four equations in four unknowns?

You can simply use substitution, $a = 1$ substituted in the second equation to solve for $b$...etc. But since you have "linear algebra" as a tag, you may be able to use techniques (Gaussian elimination, or solve the augmented coefficient matrix) to solve for $a, b, c, d$.

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Two polynomials $$f(x)=a_0+a_1x+\cdots+a_nx^n$$ and $$g(x)=b_0+b_1x+\cdots+b_nx^n$$ (having the same degree $n$, here) are equal if and only if $a_i=b_i$, for all $i=1,\ldots,n$.

So, you have to expand both to compare the coefficients.

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I did bexpand both –  fosho Jan 26 '13 at 19:04
    
So, if you didn't miss anything (even a sign) the values should be right. –  Sigur Jan 26 '13 at 19:06
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