How do you find all the zeros of f(x)= 6x^3 +25x^2 +3x -4?

Aug 5, 2016

$f \left(x\right)$ has zeros: $\frac{1}{3}$, $- \frac{1}{2}$ and $- 4$

Explanation:

$f \left(x\right) = 6 {x}^{3} + 25 {x}^{2} + 3 x - 4$

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ 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.

So the only possible rational zeros of $f \left(x\right)$ are:

$\pm \frac{1}{6} , \pm \frac{1}{3} , \pm \frac{1}{2} , \pm \frac{2}{3} , \pm 1 , \pm \frac{4}{3} , \pm 2 , \pm 4$

Since the coefficients of $f \left(x\right)$ have one change of sign, we can use Descartes' rule of signs to deduce that it has exactly one positive Real zero.

Note that:

$f \left(0\right) = - 4$ and $f \left(1\right) = 6 + 25 + 3 - 4 = 30$

So there is a Real root somewhat closer to $0$ than to $1$.

$f \left(\frac{1}{6}\right) = 6 \left(\frac{1}{216}\right) + 25 \left(\frac{1}{36}\right) + 3 \left(\frac{1}{6}\right) - 4$

$= \frac{1 + 25 + 18 - 144}{36} = - \frac{100}{36} = - \frac{25}{9}$

$f \left(\frac{1}{3}\right) = 6 \left(\frac{1}{27}\right) + 25 \left(\frac{1}{9}\right) + 3 \left(\frac{1}{3}\right) - 4$

$= \frac{2 + 25 + 9 - 36}{9} = 0$

So $x = \frac{1}{3}$ is a zero and $\left(3 x - 1\right)$ a factor:

$6 {x}^{3} + 25 {x}^{2} + 3 x - 4 = \left(3 x - 1\right) \left(2 {x}^{2} + 9 x + 4\right)$

We can factor the remaining quadratic using an AC method:

FInd a pair of factors of $A C = 2 \cdot 4 = 8$ with sum $B = 9$. The pair $8 , 1$ works. Use this pair to split the middle term and factor by grouping:

$2 {x}^{2} + 9 x + 4$

$= 2 {x}^{2} + 8 x + x + 4$

$= 2 x \left(x + 4\right) + 1 \left(x + 4\right)$

$= \left(2 x + 1\right) \left(x + 4\right)$

Hence the other two zeros are: $x = - \frac{1}{2}$ and $x = - 4$