How do you divide #(3x^2+7x-20) / (x+4)#?

1 Answer
Jul 30, 2015

Answer:

# 3x-5 #

Explanation:

Hint: If you see the question carefully, you might notice that the question was intended so that you may have to factor the numerator in such a way that # x + 4 # is a factor, and gets cancelled with the denominator. Let's try doing that.

Let # N = 3 x^2 + 7 x - 20 #.
Note that # 3x(x+4) = 3x^2 + 12x #

Thus, we write:
# N = 3 x^2 + 12 x - 5x - 20 = 3x (x+4) -5 (x+4) = (3x-5) (x+4) #

Now we come to the question itself
# ( 3 x^2 + 7 x - 20 ) / ( x + 4 ) = ( (3x-5) cancel((x+4)) ) / cancel((x+4)) = 3x-5#