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

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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