Question

The polynomial $x^3-3x^2-9x+c$ can be written in the form $(x-\alpha {)}^2(x-\beta )$, if value of $c$ is:

• A. -$5$
• B. $-7$
• C. $25$
• D. $27$

Answer 1

Answer: (A), (D)

$27$
-$5$

Given that, the polynomial $x^3-3x^2-9x+c$ can be written in the form of $(x-\alpha {)}^2(x-\beta )$.

This means, the equation $x^3-3x^2-9x+c$ has two equal roots.
Here, the roots are $\alpha ,\alpha ,\beta$
We know that, for a cubic polynomial of the form $a{x}^3+b{x}^2+cx+d$,

Sum of roots $=\frac{-\text{Coefficient of}{x}^2}{\text{Coefficient of}{x}^3}=\frac{-b}{a}$,

Product of roots taken two at a time $=\frac{\text{Coefficient of}x}{\text{Coefficient of}{x}^3}=\frac{c}{a}$

and product of roots $=\frac{-\text{Constant term}}{\text{Coefficient of}{x}^3}=\frac{-d}{a}$

$\therefore \alpha +\alpha +\beta =3$
or $2\alpha +\beta =3$ ........... (1)
Also, $\alpha \alpha +\alpha \beta +\alpha \beta =-9$
or $2\alpha \beta +{\alpha }^2=-9$ .............. (2)
Using (1) & (2), we get
$2\alpha (3-2\alpha )+{\alpha }^2=-9[\beta =3-2\alpha ]$
$\Rightarrow 6\alpha -3{\alpha }^2=-9$
$\Rightarrow {\alpha }^2-2\alpha -3=0$
$\Rightarrow (\alpha -3)(\alpha +1)=0$
$\Rightarrow \alpha =3,-1$

When $\alpha =-1,\beta =5[\beta =3-2(-1)=5]$
and when $\alpha =3,\beta =-3[\beta =3-2(3)=-3]$

Also, product of roots $={\alpha }^2\beta =-c$
When $\alpha =-1,\beta =5$:

$-c=(-1{)}^2(5)=5$

$\Rightarrow c=-5$

Also, when $\alpha =3,\beta =-3$:

$-c=\left(3{)}^2\right(-3)=-27$

$\Rightarrow c=27$

Hence, there are two possible values of $c:-5$ and $27$.

Therefore, both options A and D are correct.