# How do you divide #(5x^4+2x^3-9x+12)/(x^2-3x+4)#?

##### 2 Answers

The remainder is

#### Explanation:

We perform a long division

The remainder is

Pull out factors of

The answer is

#### Explanation:

Let's start by getting rid of our

#5x^2*(x^2-3x+4) = 5x^4 - 15x^3 + 20x^2#

So we can re-write the polynomial as:

#(5x^2(x^2-3x+4))/(x^2-3x+4)+(5x^4+2x^3-9x+12)/(x^2-3x+4) - (5x^4-15x^3+20x^2)/(x^2-3x+4)#

#= 5x^2 + (17x^3-20x^2-9x+12)/(x^2-3x+4)#

Now, let's divide the remaining part again. This time, to get rid of the

#(17x*(x^2-3x+4))/(x^2-3x+4)+ (17x^3-20x^2-9x+12)/(x^2-3x+4)-(17x^3-51x^2+68x)/(x^2-3x+4)#

#=17x + (31x^2-77x+12)/(x^2-3x+4)#

Finally, let's divide the remaining part one last time. This time, to get rid of the

#(31(x^2-3x+4))/(x^2-3x+4)+(31x^2-77x+12)/(x^2-3x+4) - (31x^2-93x+124)/(x^2-3x+4)#

#= 31 + (16x-112)/(x^2-3x+4)#

We can't divide any more since the numerator is smaller than the denominator. So, our final answer is:

#5x^2+17x+31+(16x-112)/(x^2-3x+4)#