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

1 Answer
Oct 31, 2015

#frac{2x^4+4x^3-5x^2+3x-2}{x^2+2x-3}=2x^2+1+frac{x+1}{x^2+2x-3}#

Explanation:

Note that if

#frac{a(x)}{b(x)}=Q(x)+frac{R(x)}{b(x)}#,

then

#a(x)=b(x)Q(x)+R(x)#.

In this case,

#a(x)=2x^4+4x^3-5x^2+3x-2# and #b(x)=x^2+2x-3#.

We are interested in finding #Q(x)# and #R(x)#.

First, observe that the degree of the polynomial in the numerator is greater than the degree of the polynomial in the denominator.

Divide the leading term of the numerator by the leading term of the denominator to get the first term of #Q(x)#.

#frac{2x^4}{x^2}=2x^2#

Multiply the result back with #b(x)#.

#(x^2+2x-3)(2x^2)=2x^4+4x^3-6x^2#

Now, subtract the result from the numerator.

#(2x^4+4x^3-5x^2+3x-2)-(2x^4+4x^3-6x^2)=x^2+3x-2#

So now we can say that

#frac{2x^4+4x^3-5x^2+3x-2}{x^2+2x-3}=2x^2+frac{x^2+3x-2}{x^2+2x-3}#.

Now observe that the degree of the numerator is still not less than that of the denominator.

As done previously, divide the leading term of the numerator by the leading term of the denominator to get the second term of #Q(x)#.

#frac{x^2}{x^2}=1#

Multiply the result back with #b(x)#.

#(x^2+2x-3)(1)=x^2+2x-3#

Now, subtract the result from the numerator.

#(x^2+3x-2)-(x^2+2x-3)=x+1#

So now we can say that

#frac{2x^4+4x^3-5x^2+3x-2}{x^2+2x-3}=2x^2+1+frac{x+1}{x^2+2x-3}#.

Observe that the degree of the numerator is less than that of the denominator. We are done.