Question

If $x^3+3x^2-9x+c$ is in the form $(x-\alpha {)}^2(x-\beta )$, then $c$ is equal to

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

Answer 1

Answer: (B), (C)

The correct options are
B $-27$
C $5$

$f\left(x\right)=x^3+3x^2-9x+c$ is of the form.

${(x-\alpha )}^2(x-\beta )$ , showing that $\alpha$ is a double root so that ${f\left(x\right)}^{{}^{\prime }}=0$ has also one root $\alpha$,

i.e.,

${f\left(x\right)}^{{}^{\prime }}=3x^2+6x-9=0$ has one root $\alpha$,

Hence, $x^2+2x-3=0$ or $(x+3)(x-1)=0$ has the root $\alpha$.

Which can be either $-3$ or $1$.

if $\alpha$ $=1$,

then $f\left(x\right)=0$

$⟹1+3-9+c=0⟹c-5=0⟹c=5$

If $\alpha =3$ then,

$f\left(x\right)=0⟹-27+27+27+c=0⟹c=-27$