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    I'm having some trouble with this question to do with factorising a cubic equation... I have no idea what to do with the given information.

    Name:  functions.png
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    Thanks in advance
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    The factor theorem is; if f(a)=0, then (x-a) is a factor of f(x), this would mean (x-a)(some quadratic factor) = 0. You can use algebraic long division to solve for the quadratic factor.
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    (Original post by Gabzinc)
    I'm having some trouble with this question to do with factorising a cubic equation... I have no idea what to do with the given information.

    Name:  functions.png
Views: 19
Size:  7.2 KB

    Thanks in advance
    If f(-4) =0 you know that x+4 is a factor of it. Start by factorising that out

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    (Original post by NotNotBatman)
    The factor theorem is; if f(a)=0, then (x-a) is a factor of f(x), this would mean (x-a)(some quadratic factor) = 0. You can use algebraic long division to solve for the quadratic factor.
    (Original post by BobBobson)
    If f(-4) =0 you know that x+4 is a factor of it. Start by factorising that out
    I've never studied factor theorem or factorising cubics before. I'll give it some practice tonight.

    Thanks guys.
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    (Original post by Gabzinc)
    I'm having some trouble with this question to do with factorising a cubic equation... I have no idea what to do with the given information.

    Name:  functions.png
Views: 19
Size:  7.2 KB

    Thanks in advance
    Since you've been given that x=-4 is a root that must mean you can write the polynomial in the form:

    (x + 4) times a quadratic. Simple log division will give you the quadratic:

    f(x) = (x+4)(2x^2 -x -3)

    Now you just have to factor the quadratic. Try a few things. You should quickly discover that the full factorization is:

    f(x) = (x+4)(x+1)(2x-3)

    So the roots of f(x) = 0 are -4, -1, 3

    Hopefully this helps.
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    (Original post by sameehaiqbal)
    Hopefully this helps.
    Unfortunately providing a full solution rarely helps and that is why it is frowned upon here. Please read the Forum Rules pinned to the top of the Maths Forum.
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    (Original post by Mr M)
    Unfortunately providing a full solution rarely helps and that is why it is frowned upon here. Please read the Forum Rules pinned to the top of the Maths Forum.
    Lol, sorry. I wasn't aware. I was just trying to help.
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    (Original post by sameehaiqbal)
    Lol, sorry. I wasn't aware. I was just trying to help.
    I know you meant well.
 
 
 
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