Question

If –4 is one of the zeroes of the cubic polynomial $x^3-21x-20,$ then the other two zeroes are

• A. 2 and 5
• B. -2 and 5
• C. 5 and -1
• D. -5 and 1

Answer 1

The correct option is C 5 and -1

Given that $\alpha =-4$ is one of the zeroes of $x^3-21x-20.$

Let the other two zeroes be $\beta \mathrm{and}\gamma$.

For cubic polynomial ${\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+d,$ sum of zeroes is given by $\frac{-b}{a}$

Now, $\alpha +\beta +\gamma =\frac{-b}{a}$

$\therefore -4+\beta +\gamma =0$

$\Rightarrow \beta +\gamma =4$ ... (i)

Also, $\alpha \times \beta \times \gamma =\frac{-d}{a}$

$\Rightarrow (-4)\times \beta \times \gamma =20$

$\Rightarrow \mathrm{\beta \gamma }=-5$

$\Rightarrow \gamma =\frac{-5}{\beta }$ ...(ii)

Put $\gamma =\frac{-5}{\beta }\mathrm{in}\left(i\right)$

$\Rightarrow \beta =-\frac{5}{\beta }=4$

$\Rightarrow {\beta }^2-5=4\beta$

$\Rightarrow {\beta }^2-4\beta -5=0$

$\Rightarrow {\beta }^2-5\beta +\beta -5=0$

$\Rightarrow \beta (\beta -5)+(\beta -5)=0$

$\Rightarrow (\beta -5)(\beta +1)=0$

$\Rightarrow \beta =5\mathrm{or}-1$

$\therefore \mathrm{If}\beta =5,\gamma =-1$

$\mathrm{Or},\mathrm{If}\beta =-1,\gamma =5$

Thus, the other two zeroes are 5 and –1.

Hence, the correct answer is option (c).