how would you find the quotient and remainder when 3x^4-x^2 is divided by x^3-x^2+2. I know we do long division but can someone help me go thought steps PLEASE!

Question

whpalmer4 · Answer

Have you tried the problem yet?

whpalmer4 · Answer

Look at the highest order term in both the dividend and the divisor.  They are respectively $3x^4$ and $x^3$.  What is the largest factor you can multiply the latter by without exceeding the former?

anonymous · Answer

well the quotient be 3x+3 and remainder be 2x^2-6x-6??

whpalmer4 · Answer

Well, we can check that: the quotient * the divisor + the remainder should give us the dividend.

$$(3x+3)(x^3-x^2+2) = 3x^4 -3x^3 + 6x + 3x^3 -3x^2+6 = 3x^4-3x^2+6x+6$$
$$3x^4-3x^2+6x+6 - (2x^2-6x-6) = 3x^4-3x^2+6x+6-2x^2+6x+6=$$
$$3x^4-5x^2+12x+12$$
Does that look like the right answer?

anonymous · Answer

no i guess it doesn't look right

whpalmer4 · Answer

Hmm, but it is right, where did I make a mistake?

whpalmer4 · Answer

$$(3x+3)(x^3-x^2+2) = 3x^4-3x^2+6x+6$$

Oh, I subtracted the remainder, instead of adding it.  Oops.

$$3x^4-3x^2+6x+6 + 2x^2-6x-6 = 3x^4 -3x^2 + 2x^2 + 6x - 6x + 6 - 6 =$$$$ 3x^4-x^2$$

anonymous · Answer

oh okay thank you super much!

whpalmer4 · Answer

Sorry about the wrong turn there!  I was still thinking of doing the division, I guess.

anonymous · Answer

Its okay! thank you for helping!!

whpalmer4 · Answer

Got another one?

anonymous · Answer

yes but should i make a new question or should i post here?

whpalmer4 · Answer

New question is always good, I think.