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I had this equation as a bonus question on a quiz today. I got to a certain extend and was completely unsure how to continue.

  1. $$(10x^4-18x^3-94x^2-8x+2) \div (10x+2)$$

I figured the easiest first step would be to divide. (Actually, I assumed it was factorable, but I have no idea how to go about that)

$$\frac{(10x^4-18x^3-94x^2-8x+2)} {(10x+2)}$$ $$\frac{10x^4}{10x+2} + \frac{-18x^3}{10x+2} + \frac{-94x^2}{10x+2} + \frac{-8x}{10x+2} + \frac{2}{10x+2}$$

Simplifying some more (multiplying by $10x+2$)

$$10x^4-18x^3-94x^2-8x-2 = 0$$

But what do I do from here? Do I factor? Do do something else?


If I do factor How do I go about that? If not, what do I do instead?

  • 2
    Why did you remove the $+2$ from all the denominators? – 2017-01-19
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    And what is your actual purpose? Do you want to solve that $=0$? – 2017-01-19
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    @RSerrao, I "multiplied" by 2, because the right hand side is 0 – 2017-01-19
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    [Polynomial long division](https://en.wikipedia.org/wiki/Polynomial_long_division) – 2017-01-19
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    @Travis Please note that you cannot remove the $+2$ from the denominator like you did. You can multiply both sides directly by $10x + 2$, removing all the denominators at once, but not removing only the $+2$. Multiplying by 2 won't make the $+2$ disappear. It will turn the denominator into $5x + 1$ or it will double the numerator – 2017-01-19
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    @RSerrao I figured I made a mistake somewhere, I'll fix that – 2017-01-19
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    Have you come across the process called [polynomial division](https://en.wikipedia.org/wiki/Polynomial_long_division)? – 2017-01-19
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    @RobArthan I just read about it, as Zoran recommended, but I'm still not 100% sure how to apply it to this equation – 2017-01-19
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    It's not an operation on equations. It's an operation on polynomials. You have an answer now, so I leave you to look at that. – 2017-01-19

3 Answers 3

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Polynomial long division:

\begin{array}{c|cc} & &\color{red}{x^3} & \color{green}{-2x^2}& \color{blue}{-9x}& \color{orange}{+1}& \\ \hline 10x+2 & 10x^4 & -18x^3 & -94x^2& -8x& +2 \\ & \color{red}{10x^4} &\color{red}{+2x^3}\\ & & -20x^3 & -94x^2& \vdots & \vdots\\ & & \color{green}{-20x^3} & \color{green}{-4x^2}\\ & & & -90x^2& -8x & \vdots\\ & & & \color{blue}{-90x^2}& \color{blue}{-18x} \\ & & & & 10x & +2\\\ & & & & \color{orange}{10x} & \color{orange}{+2}\ \end{array}

$\therefore (10x^4-18x^3-94x^2-8x+2) =(10x+2)(x^3-2x^2-9x+1)$

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multiples of $10 x + 2$

$$ \begin{array}{rrrrrr} x^3 \mapsto & 10 x^4 & + 2 x^3 &&& \\ -2x^2 \mapsto & & -20 x^3 & -4 x^2 && \\ -9x \mapsto & & & -90 x^2 & -18 x & \\ +1 \mapsto & & & & 10 x & + 2 & \\ &&&&& \\ & 10x^4 & - 18 x^3 & -94 x^2 & -8 x & + 2 & \\ \end{array} $$

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One idea would be to guess that $10x+2$ is a factor.

We suppose

$$10x^4 - 18x^3 -94x^2 -8x +2 = (10x+2)(ax^3+bx^2+cx+d).$$

What does this tell us? We immediately know by considering the $x^4$ term, that $a=1$. By considering the $x^3$ term we have $-18 = 10b +2a = 10b+2$. This tells us that $b=-2$. The $x^2$ term tells us that $-94=2b+10c=10c-4$, so $c=-9$. The $x$ term tells us that $-8=2c+10d=-18+10d$ so $d=1$.

What have we shown? We have shown that if $10x+2$ divides your polynomial, then

$$10x^4-18x^3-94x^2-8x+2 = (10x+2)(x^3-2x^2-9x+1).$$

To show that this holds we need to factorise it out and check! I'll leave that to you.

Hope this helps.