<s>[INST] How do I continue solving this polynomial division? 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.


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


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?
 [/INST] 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.
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