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very complicated maths

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Hey guys, need some help with a question ive been stuck on for a looooooong time!

The question is:
Given that (x+4) is a factor of the expression 3x^3+x^2+px+24, find the constant value of p. Hence factorise and solve the equation 3x^3+x^2+px+24=0

Any ideas??????

  • very complicated maths -

    If x+4 is a factor of a polynomial
    q(x), then you have:

    q(x) = (x+4)r(x)

    for some polynomial r(x). If you put
    x = -4, you get:

    q(-4) = 0


    Putting x = -4 in

    3x^3+x^2+px+24

    and equating to zero gives:

    -3 2^6 + 2^4 - 4 p + 3*8 = 0 ----->

    p = -3 2^4 + 4 + 3*2 = -38

  • very complicated maths -

    im really lost and really want to understand this question. Is there any other way of explaining it. Thank you for your time

  • very complicated maths -

    you could do a synthetic division. The values along the bottom row are

    3 -11 p+44 -4p-152

    that means that

    3x^3+x^2+px+24 = (x+4)(3x^2-11x+p+44) + (-4p-152)

    the remainder is -4p-152, which we want to be zero if (x+4) is a factor.

    so, p = -38, and our cubic is

    3x^3 + x^2 - 38x + 24 = (x+4)(3x^2-11x+6)

  • very complicated maths -

    If (x+4) is a factor then if you use -4 for x the original polynomial is zero.

    Just plug in -4 for x like they did.

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