What are all the zeroes of g(x) = 3x^3 + 2x^2 - 7x - 6?

Jul 1, 2016

The reqd. zeroes are $- 1 , \frac{1 + \sqrt{73}}{6} , \frac{1 - \sqrt{73}}{6.}$

Taking $\sqrt{73} \cong 8.544$, the zeroes are $- 1 , 1.591 , - 1.257 .$

Explanation:

Observe that the sum of the co-effs. of odd-powered terms of $g \left(x\right) = 3 - 7 = - 4 ,$ & that of even-powered $= 2 - 6 = - 4.$ Hence, $\left(x + 1\right)$ is a factor of $g \left(x\right) .$

Now we arrange the terms of $g \left(x\right)$ in such a way that $\left(x + 1\right)$ can be taken out as a common factor from $g \left(x\right)$ as shown below :-

$g \left(x\right) = 3 {x}^{3} + 2 {x}^{2} - 7 x - 6 ,$
$= 3 {x}^{3} + 3 {x}^{2} - {x}^{2} - x - 6 x - 6 ,$
$= 3 {x}^{2} \left(x + 1\right) - x \left(x + 1\right) - 6 \left(x + 1\right) ,$
$= \left(x + 1\right) \left(3 {x}^{2} - x - 6\right) .$

To work out the zeroes of the quadr. poly. $3 {x}^{2} - x - 6 = f \left(x\right) , s a y ,$ we compare it with the std. qudr. poly. $a {x}^{2} + b x + c ,$ to give us,

$a = 3 , b = - 1 , c = - 6.$

We will use the formula to find the zeroes of $f \left(x\right)$.

If $\alpha , \beta$ are the zeroes of the std. qudr. poly., by the formula,

$\alpha , \beta = \frac{- b \pm \sqrt{\Delta}}{2 a} ,$ where, $\Delta = {b}^{2} - 4 a c = 1 - 4 \cdot 3 \cdot \left(- 6\right) = 1 + 72 = 73.$

Hence, $\alpha , \beta = \frac{1 \pm \sqrt{73}}{6.}$

Altogether, the zeroes are $- 1 , \frac{1 + \sqrt{73}}{6} , \frac{1 - \sqrt{73}}{6.}$

Taking $\sqrt{73} \cong 8.544$, the zeroes are $- 1 , 1.591 , - 1.257 .$