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

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(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

Answer:The answer for this sum is a=-6