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-60x^5+77x^4+6119x^3-4044x^2-99952x-24640

how to form into factored equation

We want to factor the quintic polynomial p(x) = - 60x^5 + 77x^4 + 6119x^3 - 4044x^2 - 99952x - 24640.

We will make use of the following fact.  If p(x) is a polynomial, and a is a real number such that p(a) = 0, then (x-a) is a factor of p(x).

With this in mind, we set a to be a few small integers, and compute p(a), i.e. p(0), p(±1), p(±2), p(±3), p(±4), ... , p(±10).

We find that p(5) = p(-4) = 0.  So (x-5) and (x-(-4)) = (x+4) must be factors of p(x), i.e. p(x) = (x-5)(x+4) q(x), where q(x) is some polynomial.

Compute q(x) = p(x) / [(x-5)(x+4)] = - 60 x^3 + 17 x^2 + 4936 x + 1232.

Now set a to be a few simple rational numbers, and compute q(a), i.e. q(±1/2), q(±1/3), q(±2/3), q(±3/2), q(±1/4).

We find that q(-1/4) = 0.  So (4x + 1) is a factor of q(x), i.e. q(x) = (4x + 1)r(x), where r(x) is some polynomial.

Compute r(x) = q(x) / (4x + 1) = - 15x^2 + 8 x + 1232, which factors to - (5x + 44)(3x - 28).

So p(x) = (x-5)(x+4) q(x) = (x-5)(x+4)(4x + 1)r(x) = - (x-5)(x+4)(4x + 1)(5x + 44)(3x - 28).

Note: The reason this approach worked is that of the 5 possible complex roots of the quintic p(x), at least 3 were rational.

Simple. Use the rational-root test. It says that for polynomials with integer coefficients, if the leading coefficient is L and the constant coefficient is C, and Q is a rational root (if one exists) then Q=K/N where K is a factor of C and N is a factor of L. In the case of this quintic polynomial, we have L=-60 and C=-24640. Thus the candidates are "+/- (factor of -24640) / (factor of -60)". Unfortunately, there are a lot of factors for -24640, and -60 so there are lot's of candidates.