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    Let’s use this as an example:
    f(x)=2x^3 -5x^2 -23x -10.

    So we want the last 2 numbers in our brackets to multiply to -10. We can use (plus or minus) 10,1,5,2. What I’m wondering is if there is any strategical methods/ways of spotting which value could be substituted into f(x) to get 0 or is it just luck of the draw? Thanks
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    (Original post by Y11_Maths)
    Let’s use this as an example:
    f(x)=2x^3 -5x^2 -23x -10.

    So we want the last 2 numbers in our brackets to multiply to -10. We can use (plus or minus) 10,1,5,2. What I’m wondering is if there is any strategical methods/ways of spotting which value could be substituted into f(x) to get 0 or is it just luck of the draw? Thanks
    With this particular example, you can rule out + or - 10 immediately.

    You can also rule out + or - 1 as 2-5-23-10 is not zero and -2-5+23-10 is also not zero.

    Another option, if you're feeling lazy, is to use the table mode on your calculator.
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    (Original post by Y11_Maths)
    Let’s use this as an example:
    f(x)=2x^3 -5x^2 -23x -10.

    So we want the last 2 numbers in our brackets to multiply to -10. We can use (plus or minus) 10,1,5,2. What I’m wondering is if there is any strategical methods/ways of spotting which value could be substituted into f(x) to get 0 or is it just luck of the draw? Thanks
    Cubics are not as straight-forward to factorise like quadratics are. At the end of the day, you can always use the general formula but no sane person would use the cubic one.

    Also, it's a cubic. It will have at most 3 real roots whose product must be equal to NOT -10, but in fact -\dfrac{-10}{2}=5 (*). So try the +ve and -ve factors of 5.

    Once you have at least one real root, say it's x=\alpha, then you can just divide f by (x-\alpha) and obtain a quadratic. This should be much simpler to factorise into (x-\beta)(x-\gamma) where \beta,\gamma are its roots.
    Hence you can say that f(x)=(x-\alpha)(x-\beta)(x-\gamma)

    (*) This is quite simply because if we choose to denote the roots of f as \alpha,\beta,\gamma then we know that

    f(x)=2(x-\alpha)(x-\beta)(x-\gamma)=2x^3-2(\alpha+\beta+\gamma)x^2+2( \alpha \beta + \beta \gamma + \gamma \alpha)x-2\alpha\beta\gamma

    Since the constant term we got is -10, we get that -2\alpha \beta \gamma = -10, hence the product is 5.
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    (Original post by BuryMathsTutor)
    With this particular example, you can rule out + or - 10 immediately.

    You can also rule out + or - 1 as 2-5-23-10 is not zero and -2-5+23-10 is also not zero.

    Another option, if you're feeling lazy, is to use the table mode on your calculator.
    I didn’t think of that! Can I use a calculator in both papers at as?
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    (Original post by Y11_Maths)
    Let’s use this as an example:
    f(x)=2x^3 -5x^2 -23x -10.

    So we want the last 2 numbers in our brackets to multiply to -10. We can use (plus or minus) 10,1,5,2. What I’m wondering is if there is any strategical methods/ways of spotting which value could be substituted into f(x) to get 0 or is it just luck of the draw? Thanks
    You want the last 3 numbers need to multiply to give -(-10/2) = 5.
    So I basically notice that since there's odd and even coefficients, it may be worth trying an even number that is a factor of 10.
    Also, the 3 numbers also need to add up to -(-5/2) = 5/2, so there must be a fractional root, so try x = ±1/2
    That should start you off with a few ideas.
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    hence the product is 5.
    Thank you for explaining this so clearly!
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    (Original post by 3pointonefour)
    You want the last 3 numbers need to multiply to give -(-10/2) = 5.
    So I basically notice that since there's odd and even coefficients, it may be worth trying an even number that is a factor of 10.
    Also, the 3 numbers also need to add up to -(-5/2) = 5/2, so there must be a fractional root, so try x = ±1/2
    That should start you off with a few ideas.
    Ok thank you
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    (Original post by Y11_Maths)
    I didn’t think of that! Can I use a calculator in both papers at as?
    You can.
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    (Original post by BuryMathsTutor)
    You can.
    Are you required to have one or can the papers be done in the absence of a calculator?
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    (Original post by 3pointonefour)
    You want the last 3 numbers need to multiply to give -(-10/2) = 5.
    So I basically notice that since there's odd and even coefficients, it may be worth trying an even number that is a factor of 10.
    Also, the 3 numbers also need to add up to -(-5/2) = 5/2, so there must be a fractional root, so try x = ±1/2
    That should start you off with a few ideas.
    How would they want me to set out my answer?
    1) f(x) = (x-5)(2x+1)(x+2)
    2) f(x) = 2(x-5)(x+1/2)(x+2)
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    (Original post by Y11_Maths)
    How would they want me to set out my answer?
    1) f(x) = (x-5)(2x+1)(x+2)
    2) f(x) = 2(x-5)(x+1/2)(x+2)
    The first way is how you'd usually set it out since it's much tidier and standard notation to leave linear factors like that
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    (Original post by 3pointonefour)
    The first way is how you'd usually set it out since it's much tidier and standard notation to leave linear factors like that
    Ok thanks
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    (Original post by Y11_Maths)
    Are you required to have one or can the papers be done in the absence of a calculator?
    You will need one.
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    Are you going to rename yourself Y12_Maths??
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    (Original post by Y11_Maths)
    Are you required to have one or can the papers be done in the absence of a calculator?
    At A Level you’ll often find that one factor is easy to find e.g. (x-1) and the others can be found by division.
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    (Original post by BuryMathsTutor)
    You will need one.
    Ok cheers
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    (Original post by Your Local Cat)
    Are you going to rename yourself Y12_Maths??
    I can? How do I change my name?
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    (Original post by Your Local Cat)
    Are you going to rename yourself Y12_Maths??
    Can you even change your username?
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    (Original post by Notnek)
    At A Level you’ll often find that one factor is easy to find e.g. (x-1) and the others can be found by division.
    Yeah, that takes longer imo so I prefer to use (x-1)(ax^2 +bx + c) and fill in my a and c values to which I can easily find out b by comparing to my original f(x)
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    (Original post by 3pointonefour)
    Can you even change your username?
    I hope you can 😂
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