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I can't figure out how to simplify this polynominal $$5x^2+3x^4-7x^3+5x+8+2x^2-4x+9-6x^2+7x$$

I tried combining like terms $$5x^2+3x^4-7x^3+5x+8+2x^2-4x+9-6x^2+7x$$ $$(5x^2+5x)+3x^4-(7x^3+7x)+2x^2-4x-6x^2+(8+9)$$ $$5x^3+3x^4-7x^4+2x^2-4x-6x^2+17$$

It says the answer is $$3x^4-7x^3+x^2+8x+17$$ but how did they get it?

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Did the color help?! :-D –  000 Dec 5 '12 at 2:08
    
Yes it did, thanks. –  Cynea Osodgt Dec 5 '12 at 2:18
    
Mission success! :-) Please click the checkmark to whatever answer you think is best; it will improve your accepted answer percentage. –  000 Dec 5 '12 at 2:34
    
I voted for the two that helped –  Cynea Osodgt Dec 5 '12 at 2:40
    
Can't i choose more then 1 answer??? I want to choose both anwers that helped me –  Cynea Osodgt Dec 5 '12 at 3:06
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3 Answers

up vote 3 down vote accepted

You cannot combine terms like that, you have to split your terms by powers of $x$.

So for example $$5x^2+5x+2x^2 = (5+2)x^2+5x = 7x^2+5x$$ and not $5x^3+2x^2$. Using this, you should end up with your answer.

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Group the terms with the same power of $x$ together.

$5x^2+3x^4−7x^3+5x+8+2x^2−4x+9−6x^2+7x$

$=3x^4−7x^3+5x^2+2x^2−6x^2+5x−4x+7x+8+9$

$=3x^4−7x^3+x^2+8x+17$

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Observe the magical power of color:

$$\color{blue}{5}x^\color{blue}{2}+3x^4-7x^3+\color{green}{5}x+\color{orange}{8}+\color{blue}{2}x^\color{blue}{2}+(\color{green}{-4})x+\color{orange}{9}+(\color{blue}{-6})x^\color{blue}{2}+\color{green}{7}x.$$

Instead of Color-Me-Elmo, we have Color-Me-Like-Terms-And-Combine (not as catchy, I know): $$3x^4-7x^3+(\color{blue}{5}+\color{blue}{2}+(\color{blue}{-6}))x^\color{blue}{2}+(\color{green}{5}+(\color{green}{-4})+\color{green}{7})x+(\color{orange}{8}+\color{orange}{9}).$$

Presto-simplification-o!


Combining Like Terms

In a polynomial $p(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$ and $q(x)=b_nx^n+b_{n-1}x^{n-1}+\dots+b_1x+b_0$, they are added thusly: $$ \begin{align} p(x)+q(x)&=a_nx^n+b_nx^n+a_{n-1}x^{n-1}+b_{n-1}x^{n-1}+\cdots+a_1x+b_1x+a_0+b_0\\ &=(a_n+b_n)x^n+(a_{n-1}+b_{n-1})x^{n-1}+\cdots+(a_1+b_1)x+(a_0+b_0). \end{align} $$

In other words, add the coefficients of terms with the same power.

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Did you mean $q(x)=b_nx^n+\color{red}{b}_{n-1}x^{n-1}+\dots+\color{red}{b}_1x+{\color{red}{b}‌​}_{0}$? –  ctype.h Dec 5 '12 at 2:20
    
Yes, actually . . . @ctype.h. I get a little too excited sometimes. –  000 Dec 5 '12 at 2:23
    
@JasperLoy I am getting you a Color-Me-Elmo for Christmas. –  000 Dec 5 '12 at 2:58
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