How to solve this?
If #(ax^3+bx^2-3x+2b)/(x^2-x-2)# can be simplified to a polynomial in #x# , find the values of #a# and #b# . Then find the polynomial.
(tha answer provided is a=#6/5# , b=#-3/5# ,#(3(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)#
Note that:
#x^2-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))+2b#
#color(white)(0) = 8a+4b-6+2b#
#color(white)(0) = 8a+6b-6#
#color(white)(0) = 2(4a+3b-3)#
and:
#0 = a(color(blue)(-1))^3+b(color(blue)(-1))^2-3(color(blue)(-1))+2b#
#color(white)(0) = -a+b+3+2b#
#color(white)(0) = -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)#