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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.)

1 Answer
George C.
Jul 28, 2018

#a=6/5#, #b=-3/5# and #(6/5x^3-3/5x^2-3x+6/5)/(x^2-x-2) = 3/5(2x+1)#

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 #(x-2)# and #(x+1)#.

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

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 #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)#

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