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    I decided to work through some questions in my revision guide and I could figure this question out :
    Each of these expressions has a factor (x+/- p) find the value of p and hence factorise the expression completely :
    1) x^3-10x^2+19x+30
    2)x^3+x^2-4x-4
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    (Original post by Mf1999)
    I decided to work through some questions in my revision guide and I could figure this question out :
    Each of these expressions has a factor (x+/- p) find the value of p and hence factorise the expression completely :
    1) x^3-10x^2+19x+30
    2)x^3+x^2-4x-4
    Try to substitute values of x = \pm p into the expressions. You can assume that p is an integer (can be positive or negative).

    If the resulting expression is equal to zero, one of your factors will be (x \mp p) (note the sign reversal).

    You can then perform algebraic long division to solve for the other factors.
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    The factor theorem tells us that if (x-p) is a factor of f(x), then f(p) will equal zero. To use that for the given question, you should try putting x equal to a few obvious values, e.g. 1, -2, 2, -2, 3, -3 to see which of them makes f(x) zero. If you find, say, that f(1) = 0, then you can also say that (x-1) is a factor of f(x).

    To get the remaining factors, you must divide f(x) by (x-1) then factorise the resulting quadratic using any of the usual methods.
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    (Original post by old_engineer)
    The factor theorem tells us that if (x-p) is a factor of f(x), then f(p) will equal zero. To use that for the given question, you should try putting x equal to a few obvious values, e.g. 1, -2, 2, -2, 3, -3 to see which of them makes f(x) zero. If you find, say, that f(1) = 0, then you can also say that (x-1) is a factor of f(x).

    To get the remaining factors, you must divide f(x) by (x-1) then factorise the resulting quadratic using any of the usual methods.
    This, as well as you can cut down on how many values you check by using the helpful fact that any integer root of a polynomial (with integer coefficients, which is true in this case) will be a factor of the constant term. So you know exactly which values to try. For example, if the constant term is 5, the only integer roots possible are 1,-1,5 and -5. There will be no other integer roots. This fact is easy to prove if you're so inclined.

    You could take this one step further and use what is known as the Rational Root Theorem, which again, is easy to prove: if a polynomial has integer coefficients, then \frac{m}{n} (with m,n coprime integers) can only be a root if n divides the leading coefficient of the polynomial and m divides the constant term.

    Not sure why schools don't mention this; it's a very easy theorem and allows students to avoid checking integers/rational number randomly ad infinitum.
 
 
 
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