How to solve this?
If (ax^3+bx^2-3x+2b)/(x^2-x-2)ax3+bx2−3x+2bx2−x−2 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=-3/5−35 ,(3(2x+1))/53(2x+1)5 , but i dont know how to solve.)
If
(tha answer provided is a=
1 Answer
Explanation:
Given:
(ax^3+bx^2-3x+2b)/(x^2-x-2)ax3+bx2−3x+2bx2−x−2
Note that:
x^2-x-2 = (x-2)(x+1)x2−x−2=(x−2)(x+1)
So for the quotient to be a polynomial, the numerator must be divisible by both
Hence it must be zero when
So we find:
0 = a(color(blue)(2))^3+b(color(blue)(2))^2-3(color(blue)(2))+2b0=a(2)3+b(2)2−3(2)+2b
color(white)(0) = 8a+4b-6+2b0=8a+4b−6+2b
color(white)(0) = 8a+6b-60=8a+6b−6
color(white)(0) = 2(4a+3b-3)0=2(4a+3b−3)
and:
0 = a(color(blue)(-1))^3+b(color(blue)(-1))^2-3(color(blue)(-1))+2b0=a(−1)3+b(−1)2−3(−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
5a-6=0
and hence:
a = 6/5
Putting this value for
0 = -6/5+3b+3 = 3b+9/5 = 3(b+3/5)
and hence:
b = -3/5
Substituting these values for
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)