1. ## Polynomial Division

Hi,

How do I go about the following polynomial division? Particularly with the fact that the coefficients of the numerator are smaller than the denominator.

h(x) = 6x^4 + 3x^3 + 6x^2 + 3
g(x) = 12x^3 - 30x^2 + 18x - 12

What is h(x) / g(x)?

I know there is a remainder.

Thanks

2. ## Re: Polynomial Division

I don't know how to show long division in latex.. annoying.

$6x^4 + 3x^3 + 6x^2 + 3 = \dfrac 1 2 x (12x^3-30x^2+18x-12) + (12x^3-3x^2+6x+3)$

$(12x^3-3x^2+6x+3) = 1 (12x^3-30x^2+18x-12) + (27x^2-12x+15)$

so

$\dfrac{6x^4 + 3x^3 + 6x^2 + 3}{12x^3-30x^2+18x-12} = \left(\dfrac 1 2 x + 1\right) + \dfrac{27x^2-12x+15}{12x^3-30x^2+18x-12}$

3. ## Re: Polynomial Division

Thank you.

But your first substitution equates to

6x^4 - 3x^3 + 6x^2 +3.

Which is not the original expression.

4. ## Re: Polynomial Division

I get x/2 + 3/2 with remainder 14x^2 - 7x + 7
Too lazy to check if correct!!

5. ## Re: Polynomial Division

Thanks, that's pretty much inline with what I managed.

6. ## Re: Polynomial Division

Originally Posted by Stormin63
h(x) = 6x^4 + 3x^3 + 6x^2 + 3
g(x) = 12x^3 - 30x^2 + 18x - 12
Dividing each by 3 makes process a little easier:
(long division arrangement):
4x^3 - 10x^2 + 6x - 4 / 2x^4 + x^3 + 2x^2 + 0x + 1

7. ## Re: Polynomial Division

h(x) = 6x^4 + 3x^3 + 6x^2 + 3
g(x) = 12x^3 - 30x^2 + 18x - 12

What is h(x) / g(x)?

The way I was taught to do this:
12x^3 divides into 6x^4, (1/2) x times.

Multiplying all of 12x^3- 30x^2+ 18x- 12 gives 6x^4- 15x^3+ 9x^2- 6x. Subtract that from 6x^4- 3x^3+ 6x^2+ 0x+ 3 (don't forget that "0x"!) leaves 12x^3- 3x^2+ 6x+ 3. Now, of course, 12x^3 divides into 12x^3 exactly once. Subtracting 1(12x^3- 30x^2+ 18x- 12) from 12x^3- 3x^2+ 6x+ 3 leaves 27x^2- 12x+ 15.

h(x)/g(x) is (1/2)x+ 1 with remainder 27x^2- 12x+ 15.

8. ## Re: Polynomial Division

I did the equivalent of that you did, Hall;
how cum we don't get same results?