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I can't figure out how to multiply these polynominals $$(5x^2+3x^4-7x^3+5x+8)(2x^2-4x+9-6x^2+7x)$$

I tried multiplying like this $$(5x^2+3x^4-7x^3+5x+8)(2x^2-4x+9-6x^2+7x)$$ $$3x^4-7x^3+(5x^2)(2x^2)(-6x^2)+(5x)(7x)(-4x)+(8)(9)$$ $$3x^4-7x^3-60x^2-140x+72$$

It says the answer is $$-12x^6-19x^5-14x^4-68x^3+28x^2+69x+72$$ but how did they get it?

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First of all, collect like terms in the brackets on the right. After that you have to multiply every term in the brackets on the left by every term in the brackets on the right, and add it all up. –  Tom Oldfield Dec 8 '12 at 2:50
    
Is the problem written correctly? There is some grouping that can be done in the second set of parentheses otherwise. –  Amzoti Dec 8 '12 at 3:13
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A first step might be to arrange each of the polynomials in descending powers of $x$. That will help keep things more organized. –  André Nicolas Dec 8 '12 at 3:17
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2 Answers 2

up vote 7 down vote accepted

You have to multiply every term by every other term. A good way to make sure you don't miss any is to use a table.

First, combine like terms within each group of parentheses: $$(5x^2+3x^4-7x^3+5x+8)(2x^2-4x+9-6x^2+7x)\\ =(3x^4-7x^3+5x^2+5x+8)(-4x^2+3x+9)$$

Then form a table and multiply each term by multiplying the coefficients and adding the exponents: $$ \begin{array}{c|cc} \text{} & 3x^4 & -7x^3 & 5x^2 & 5x & 8 \\ \hline -4x^2 & -12x^6 & -28x^5 & -20x^4 & -20x^3 & -32x^2 \\ 3x & 9x^5 & -21x^4 & 15x^3 & 15x^2 & 24x \\ 9 & 27x^4 & -63x^3 & 45x^2 & 45x & 72 \\ \end{array} $$

Now take the new polynomial from the table and combine like terms: $$-12x^6-28x^5-20x^4-20x^3-32x^2+9x^5-21x^4+15x^3\\ +15x^2+24x+27x^4-63x^3+45x^2+45x+72\\$$$$ =-12x^6+(9-28)x^5+(27-21-20)x^4+(-63+15-20)x^3\\+(45+15-32)x^2+(45+24)x+72\\$$$$ =-12x^6-19x^5-14x^4-68x^3+28x^2+69x+72$$

This method will also work with negative and non-integer exponents, as it is not restricted to polynomials.

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+1 for the interesting use of a table. I've never done that before, but I can easily see where it would have saved a lot of headache... –  anorton Dec 8 '12 at 3:17
    
You could note that to write the final polynomial you start with the top left corner and add across the diagonals. –  000 Dec 8 '12 at 5:05
    
@Limitless That only works if the exponents line up nicely like this. Suppose you have something like $(3x^{-3}-x^{\pi/4})(\frac{\pi}{2}x^\sqrt{5}+x^{1/8}-7x^{-4})$. In that case, adding along the diagonals would not work. –  ctype.h Dec 8 '12 at 5:27
    
@ctype.h That's true! I didn't think about that. –  000 Dec 8 '12 at 5:29
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@AustinMohr While it is not a polynomial, I did mention that "this method will also work with negative and non-integer exponents," as the table method is not restricted to polynomials. Perhaps I should have stated that more clearly in the answer. –  ctype.h Dec 8 '12 at 5:42
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To multiply two polynomials, distribute each term in the left set of parentheses over the entire collection of terms in the right set of parentheses. Your example would start out like:

$$ \begin{align*} & (\color{red}{5x^2+3x^4-7x^3+5x+8})(2x^2-4x+9-6x^2+7x)\\ = & \color{red}{5x^2}(2x^2-4x+9-6x^2+7x)\\ & \color{red}{+ 3x^4}(2x^2-4x+9-6x^2+7x)\\ & \color{red}{- 7x^3}(2x^2-4x+9-6x^2+7x)\\ & \color{red}{+ 5x}(2x^2-4x+9-6x^2+7x)\\ & \color{red}{+ 8}(2x^2-4x+9-6x^2+7x). \end{align*} $$

Can you take it from here?

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I copied your expression exactly, but notice the righthand polynomial can be further simplified to $-4x^2 + 3x + 9$. –  Austin Mohr Dec 8 '12 at 2:54
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