Third degree equation

Question

While solving a problem on sequences and series, I got the following cubic equation $$ 8x^3−16x−85=0 $$ I cannot figure out how to solve it. I have tried to factor the L.H.S., but did not succeed. Please help. Are these type of equations solvable by factorization or is using a particular formula the only way?

Answer 1

Let $2x=t$.

Thus, $$8x^3-16x-85=t^3-8t-85=t^3-25t+17t-85=$$ $$(t-5)(t(t+5)+17)=(t-5)(t^2+5t+17)=(2x-5)(4x^2+10x+17).$$ After first step we can see that $5$ is a root of the polynomial $t^3-8t-85$,

which gives a factor $t-5$.

Answer 2

By the rational root theorem, the only possible rational roots are a divisor of $85$ divided by a divisor of $8$. The divisors of $85$ are $\pm 1, \pm 5, \pm 17, \pm 85$. The divisors of $8$ are $\pm 1$, $\pm 2$, $\pm 4$, and $\pm 8$. (Notice that $-p/-q = p/q$, so we really only need both signs for the numerators and can just use the positive denominators.)

Descartes' rule of signs tells us there is one positive real root. We can see that $x = 0$ is not a root. The rule of signs tells us there are two or zero negative real roots. Since there is definitely one positive real root, let's try the positive rational candidates first. We don't calculate precisely when it is clear that the result cannot be close to zero. \begin{align*} 1/1 &:& 8 \cdot 1^3 - 16 \cdot 1 - 85 &< 0 \\ 5/1 &:& 8 \cdot 5^3 - 16 \cdot 5 - 85 &\approx 1000 \\ 17/1 &:& 8 \cdot 17^3 - 16 \cdot 17 - 85 &> 1000 \\ 85/1 &:& 8 \cdot 85^3 - 16 \cdot 85 - 85 &> 1000 \\ 1/2 &:& 8 \cdot (1/2)^3 - 16 \cdot (1/2) - 85 &< - 85 \\ 5/2 &:& 8 \cdot (5/2)^3 - 16 \cdot (5/2) - 85 &= 0 \end{align*} ... and we have found the only positive real root. Factor it out (long division or synthetic division) and attack the quadratic. (Both of its roots are complex.)