How to solve this?

If (ax^3+bx^2-3x+2b)/(x^2-x-2)ax3+bx23x+2bx2x2 can be simplified to a polynomial in xx, find the values of aa and bb. Then find the polynomial.

(tha answer provided is a=6/565, b=-3/535,(3(2x+1))/53(2x+1)5, but i dont know how to solve.)

1 Answer
Jul 28, 2018

a=6/5a=65, b=-3/5b=35 and (6/5x^3-3/5x^2-3x+6/5)/(x^2-x-2) = 3/5(2x+1)65x335x23x+65x2x2=35(2x+1)

Explanation:

Given:

(ax^3+bx^2-3x+2b)/(x^2-x-2)ax3+bx23x+2bx2x2

Note that:

x^2-x-2 = (x-2)(x+1)x2x2=(x2)(x+1)

So for the quotient to be a polynomial, the numerator must be divisible by both (x-2)(x2) and (x+1)(x+1).

Hence it must be zero when x=2x=2 and when x=-1x=1

So we find:

0 = a(color(blue)(2))^3+b(color(blue)(2))^2-3(color(blue)(2))+2b0=a(2)3+b(2)23(2)+2b

color(white)(0) = 8a+4b-6+2b0=8a+4b6+2b

color(white)(0) = 8a+6b-60=8a+6b6

color(white)(0) = 2(4a+3b-3)0=2(4a+3b3)

and:

0 = a(color(blue)(-1))^3+b(color(blue)(-1))^2-3(color(blue)(-1))+2b0=a(1)3+b(1)23(1)+2b

color(white)(0) = -a+b+3+2b0=a+b+3+2b

color(white)(0) = -a+3b+30=a+3b+3

So:

{ (4a+3b-3=0), (-a+3b+3=0) :}

Subtracting the second equation from the first (to eliminate the term in 3b), we find:

5a-6=0

and hence:

a = 6/5

Putting this value for a into the second equation, we get:

0 = -6/5+3b+3 = 3b+9/5 = 3(b+3/5)

and hence:

b = -3/5

Substituting these values for a and b in the numerator, we have:

6/5x^3-3/5x^2-3x+6/5 = 1/5(6x^3-3x^2-15x-6)

color(white)(6/5x^3-3/5x^2-3x+6/5) = 3/5(2x^3-x^2-5x-2)

color(white)(6/5x^3-3/5x^2-3x+6/5) = 3/5(x-2)(2x^2+3x+1)

color(white)(6/5x^3-3/5x^2-3x+6/5) = 3/5(x-2)(x+1)(2x+1)

So:

(6/5x^3-3/5x^2-3x+6/5)/(x^2-x-2) = (6/5x^3-3/5x^2-3x+6/5)/((x-2)(x+1)) = 3/5(2x+1)