what must be added to x3-3×2-12x+19 so that the result is exactly divisible by x2+x-6?

Question

what must be added to x3-3×2-12x+19 so that the result is exactly divisible by x2+x-6?

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Thu Cúc 8 months 2021-07-18T05:34:53+00:00 1 Answers 8 views 0

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    2021-07-18T05:36:00+00:00

    Answer:

    x^3-3x^2-12x+19 is divisible by x^2+x-6  when 2x+5 is added to it.

    Step-by-step explanation:

    Let p(x)=x^3-3x^2-12+19  &  q(x)=x^2+x-6

    By division algorithm, when p(x) is divided by q(x), the remainder is a linear expression in x

    So, let r(x)=ax+b is added to p(x) so that p(x)+r(x) is divisible by q(x)

    Let, f(x)=p(x)+r(x)

    f(x)=x^3-3x^2-12x+19+ax+b

    f(x)=x^3-3x^2+x(a-12)+b+19

    We have,

    q(x)=x^2+x-6=(x+3)(x-3)

    Clearly, q(x) is divisible by (x-2) and (x+3)

    { (x-2) and (x+3) are factors of q(x) }

    We have,

    f(x) is divisible by q(x)

    (x-2) and (x+3) are factors of f(x)

    From factors theorem,

    If (x-2) and (x+3) are factors of f(x)

    then f(2)=0 and f(-3)=0 respectively,

    f(2)=02^3-3(2)^2+2(a-12)+b+19=0

    8-12+2a-24+b+19=0

    2a+b-9=0

    Similarly,

    f(-3)=0(-3)^3-3(-3)^2+(-3)(a-12)+b+19=0

    -27-27-3a+36+b+19=0

    b-3a+1=0

    Subtract (1) from (2)

    b-3a+1-(2a+b-9)=0-0

    b-3a+1-2a-6+9=0

    -5a+10=05a=10a=2

    Put a=2 in equation 2

    b-3 × 2+1=0b-6+1=0b-5=0b=5

    r(x)=ax+br(x)=2x+5

    Hence, x^3-3x^2-12x+19 is divisible by x^2+x-6 when 2x+5 is added to it.

    Hope this helps…

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