How do you factor −16x3+3x2+5x−4?

1 Answer
George C.
Feb 20, 2018

−16x3+3x2+5x−4=−12x−1⋅13x2+3x+4

Explanation:

I wonder whether the question meant to factor:

−16x3+3x2+5x−4

since the question as given has three irrational real zeros only expressible in terms of trigonometric expressions or cube roots of complex numbers.

So let's factor 6x3+3x2+5x−4

By the rational roots theorem, the only possible rational zeros of 6x3+3x2+5x−4 are expressible in the form pq for integers p,q with p a divisor of the constant term −4 and q a divisor of the coefficient 6 of the leading term.

That means that the only possible rational zeros are:

±16,±13,±12,±23,±1,±43,±2,±4

In addition note that the pattern of the signs of the coefficients is +++−. With one change of signs, Descartes' Rule of Signs tells us that this cubic has exactly one positive real zero.

So, trying the positive possibilities first we find:

6(12)3+3(12)2+5(12)−4=34+34+52−4=0

So x=12 is a zero and (2x−1) a factor:

6x3+3x2+5x−4=(2x−1)(3x2+3x+4)

Note that the discriminant of the remaining quadratic factor is negative:

(3)2−4(3)(4)=9−48=−39

So this quadratic only has factors with non-real complex coefficients. Since this question is asked under Algebra, I will stop at the real factorisation to find:

−16x3+3x2+5x−4=−12x−1⋅13x2+3x+4

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