# How do you find all the zeros of 3x^4+7x^3+x^2-7x-4?

Jun 29, 2016

$- 1 , - 1 , 1 \mathmr{and} - \frac{4}{3}$

#### Explanation:

Let f(x) denote the given biquadratic.

$f \left(\pm 1\right) = 0$. So,$\pm 1$ are two of the roots.

$f ' = 12 {x}^{3} + 21 {x}^{2} + 2 x - 7 \mathmr{and} f \left(- 1\right) = 0$

Therefore, $- 1$ is a double root.

Sum of the roots

= -(coefficient of ${x}^{3}$)/(coefficient of ${x}^{4}$)

= #-7/3.

So, the 4th root is $- \frac{7}{3} - \left(- 1 - 1 + 1\right) = - \frac{4}{3}$

Thus, the zeros are $- 1 , - 1. 1 \mathmr{and} - \frac{4}{3}$.

Jun 29, 2016

Zeros: $\textcolor{b l u e}{- 1}$ (with multiplicity $2$), $\textcolor{b l u e}{1}$ and $\textcolor{b l u e}{- \frac{4}{3}}$.

#### Explanation:

$f \left(x\right) = 3 {x}^{4} + 7 {x}^{3} + {x}^{2} - 7 x - 4$

First note that the sum of the coefficients is $0$.

That is:

$3 + 7 + 1 - 7 - 4 = 0$

So $f \left(1\right) = 0$, $\textcolor{b l u e}{x = 1}$ is a zero and $\left(x - 1\right)$ a factor:

$3 {x}^{4} + 7 {x}^{3} + {x}^{2} - 7 x - 4 = \left(x - 1\right) \left(3 {x}^{3} + 10 {x}^{2} + 11 x + 4\right)$

Looking at the remaining cubic factor, note that if you reverse the signs of the coefficients of the terms of odd degree, then the sum is $0$.

That is:

$- 3 + 10 - 11 + 4 = 0$

Hence $\textcolor{b l u e}{x = - 1}$ is a zero and $\left(x + 1\right)$ a factor:

$3 {x}^{3} + 10 {x}^{2} + 11 x + 4 = \left(x + 1\right) \left(3 {x}^{2} + 7 x + 4\right)$

Looking at the remaining quadratic, reversing the sign of the coefficient of the term of odd degree, the sum of the coefficients is $0$.

That is:

$3 - 7 + 4 = 0$

Hence $\textcolor{b l u e}{x = - 1}$ is a zero again and $\left(x + 1\right)$ a factor:

$3 {x}^{2} + 7 x + 4 = \left(x + 1\right) \left(3 x + 4\right)$

Finally, from the remaining linear factor, we find the last zero $\textcolor{b l u e}{x = - \frac{4}{3}}$.

So the zeros of our quartic are:

$\textcolor{b l u e}{- 1}$ with multiplicity $2$, $\textcolor{b l u e}{1}$ and $\textcolor{b l u e}{- \frac{4}{3}}$.