Question

If $1$ is zero of $3x^3-x^2-3x+1$, then the other two zeros are

• A. $1$ and $\frac{-2}{3}$
• B. $1$ and $-1$
• C. $-1$ and $\frac{-1}{3}$
• D. $-1$ and $\frac{1}{3}$

Answer 1

The correct option is D $-1$ and $\frac{1}{3}$

Let $\alpha ,\beta$ be the zeros of the cubic polynomial

$3x^3-x^2-3x+1$.
$1+\alpha +\beta =-\frac{\text{coefficient of}{x}^2}{\text{coefficient of}{x}^3}=\frac{-(-1)}{3}=\frac{1}{3}$

$\Rightarrow \alpha +\beta =\frac{1}{3}-1=\frac{(-2)}{3}$ (i)

Also,
$1\times \alpha \times \beta =-\frac{d}{a}=-\frac{\text{constant term}}{\text{coefficient of}{x}^3}$

$\alpha \beta =-\frac{1}{3}$
$\Rightarrow \alpha =-\frac{1}{3\beta }$

Putting the value $\alpha =-\frac{1}{3\beta }$ in equation (i)
$\Rightarrow \frac{-1}{3\beta }+\beta =-\frac{2}{3}$

$\Rightarrow \frac{-1+3{\beta }^2}{3\beta }=-\frac{2}{3}$

$\Rightarrow -1+3{\beta }^2=-2\beta$

$\Rightarrow 3{\beta }^2+2\beta -1=0$

$\Rightarrow 3{\beta }^2+3\beta -\beta -1=0$

$\Rightarrow 3\beta (\beta +1)-1(\beta +1)=0$

$\Rightarrow (\beta +1)(3\beta -1)=0$

$\beta =-1,\beta =\frac{1}{3}$

Now,
$\Rightarrow \alpha =-\frac{1}{3\beta }$

Putting the value of $\beta$
$\Rightarrow \alpha =-\frac{1}{3(-1)}=\frac{1}{3}$ or,
$\alpha =-\frac{1}{3\left(\frac{1}{3}\right)}=-1$

Therefore, $\alpha ,\beta =(\frac{1}{3},-1)or(-1,\frac{1}{3})$

Hence, option $D$ is correct.