How do you divide (2x^4+12x^3-5x^2+9x+3)/(x^2-4) 2x4+12x35x2+9x+3x24?

1 Answer
May 11, 2018

2x^2+12x+32x2+12x+3

Explanation:

Using Division Euclid's with said to us is always possible to divide a number for other same him having rest, then:
b = aq + rb=aq+r , com r < a.r<a.

b = 2x^4+12x^3−5x^2+9x+3b=2x4+12x35x2+9x+3
a = x^2−4 a=x24

using qq for to know how to go in b, we'll have q = x^2+12x+3q=x2+12x+3

and r = 57x+15r=57x+15

Like this, 2x^4+12x^3−5x^2+9x+3 = (x^2−4)(x^2+12x+3)+ (57x+15)2x4+12x35x2+9x+3=(x24)(x2+12x+3)+(57x+15)

Simplifying, for to find q you'll have to test the possibilities for each term of b, for example:
(x^2-4) * 2x^2 = 2x^4-8x^2(x24)2x2=2x48x2 , this result is the first term is b and more something each we'll subtraction in b, and this process will to repeat until us can't do it more.