If [tex]x^{3}-6x^{2} +ax+b[/tex] is exactly divisible by [tex]x^{2} -3x+2[/tex], then 12a+22b=?

2 thoughts on “If [tex]x^{3}-6x^{2} +ax+b[/tex] is exactly divisible by [tex]x^{2} -3x+2[/tex], then 12a+22b=?”

1. (2 – b)b = 48

   2b -b^2= 48

   now we have a quadratic equation :-

   b^2 – 2b – 48=0

   also, an multiplication the product will be -48

   this will be in the

   b^2 -8b +6b -48 =0

   now by factorizing we have :-

   b(b – 8) + 6(b – 8) = 0

   (b+6)(b-8) = 0

   b+6 = 0

   b= -6

   b – 8= 0

   b=8

   so the value of b can be 8 or -6

   substituing the value of be in equation (i) we have :-

   a + b = 2

   a + 8 = 2

   a = 2–8

   a = -6

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2. Answer:

   The answer for this sum is a=-6

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