Quick math factorise question

Yes, it's definetly solvable and has 3 real roots (2 positive, 1 negative).

Sounds like a homework question so I'll just suggest...

http://en.wikipedia.org/wiki/Cubic_equation
and Google for 'Cubic Roots' 'Cubic Factorize'.

Here's a hint:

10 = 4 * 2.5
15 = 6 * 2.5

Hopefully that gives you an idea.

is (2x-3)(5x^2-2) the kind of answer they are expecting?

No, the answer should be 3 binomials, (Ax + a)(Bx + b)(Cx + c)

To narrow down your possible factors, take into consideration that the factor will be in the form (x - a/b), so f(a/b) = 0, where 'a' is a factor of the constant (in your case, 6), and 'b' is a factor of one of the coefficients (10,-15,-4).

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Original post by Ra No, the answer should be 3 binomials, (Ax + a)(Bx + b)(Cx + c)

Not possible in general, unless you allow a, b, c to be complex. Factors correspond to roots of an equation, and the roots are either real, or conjugate pairs (hence the quadratic term in JohnnyBravo's answer - which is correct, BTW: easily verified by multiplying it out - which would decompose further to (sqrt(5)x + sqrt(2)i)(sqrt(5)x - sqrt(2)i)).

For sure there will be one real root so we first need to find that. Afterwards we end up with a 2nd degree polynomial which is easy to factorise.

One important thing to keep in mind is that, if you know the root is an integer, it t will always be a divisor of the term without x, in this case 6. So try the values -1,1,-2,2,-3,3,-6,6 in the polynomial to see for which of them you get 0. Then you can divide the polynomial by x-root and get the 2nd degree polynomial.

Unfortunately, this is not the case for this polynomial so you could try one of those methods from Wikipedia. Either that, or use a graphic calculator or graph software to find the roots.

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Original post by Zahlman Quote: Original post by Ra No, the answer should be 3 binomials, (Ax + a)(Bx + b)(Cx + c) Not possible in general, unless you allow a, b, c to be complex. Factors correspond to roots of an equation, and the roots are either real, or conjugate pairs (hence the quadratic term in JohnnyBravo's answer - which is correct, BTW: easily verified by multiplying it out - which would decompose further to (sqrt(5)x + sqrt(2)i)(sqrt(5)x - sqrt(2)i)).

The above multiplies out to 5x^2+2 (note the sign), but it's nearly correct.
Since the solution has already been given, I'll just repeat it for clarity...

(2x-3)(sqrt(5)x + sqrt(2))(sqrt(5)x - sqrt(2))

i.e 3 real roots of (3/2) and +/- (1/5)*(10^(1/2))

Could someone give me a quick runthrough of how to solve the OP's problem? Just as a reference :)

Tedious methods exist, if you like pain you may want to google for "Eisenstein's Irreducability criterion" or "Kronecker's method" (which involves Lagrange's interpolation formula). Note these methods are for rational solutions only, it is easy to construct a polynomial for which no rational roots (or complex real/imaginary rational) solutions exist (such as x^2 - 2 = 0).

The basic method for a cubic is to find a root, divide by the factor (which will be (x-r), where r is the root) and then solve the quadratic.

EDIT: Don't bother with Eisenstein's method, that just tells you if a polynomial can't be factorised with coefficients that are rational, and is a pain anyway. Gauss rated him highly though (as my book says, "Even Gauss could be wrong sometimes!").

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Original post by WillC Quote: Original post by Zahlman Quote: Original post by Ra No, the answer should be 3 binomials, (Ax + a)(Bx + b)(Cx + c) Not possible in general, unless you allow a, b, c to be complex. Factors correspond to roots of an equation, and the roots are either real, or conjugate pairs (hence the quadratic term in JohnnyBravo's answer - which is correct, BTW: easily verified by multiplying it out - which would decompose further to (sqrt(5)x + sqrt(2)i)(sqrt(5)x - sqrt(2)i)). The above multiplies out to 5x^2+2 (note the sign), but it's nearly correct. Since the solution has already been given, I'll just repeat it for clarity... (2x-3)(sqrt(5)x + sqrt(2))(sqrt(5)x - sqrt(2)) i.e 3 real roots of (3/2) and +/- (1/5)*(10^(1/2))

D'oh, lack of sleep. No i's, right. They don't give those to high school students anyway ;)