Choose the correct answer from the given four options in the foll…

MCQ

Choose the correct answer from the given four options in the following questions:
If one of the zeroes of the cubic polynomial $x^3+a x^2+b x+c$ is $-1$ , then the product of the other two zeroes is:
.

✓
$b-a+1$.
B
$b-a-1$.
C
$a-b+1$.
D
$a-b-1$

✓

Answer

Correct option: A.

$b-a+1$.

Let $p ( x )= x ^3+ a x ^2+ bx + c$
Let $a, p$ and $y$ be the zeroes of the given cubic polynomial $p(x)$.
$\therefore \alpha=-1 \text { [given] }$
$\text { and } p(-1)=0$
$\Rightarrow(-1)^3+a(-1)^2+b(-1)+c=0$
$\Rightarrow-1+a-b+c=0$
$\Rightarrow c=1-a+b \ldots . .(i)$
We know that,
$\text { Product of all zeroes }=(-1)^3 \frac{\text { Constant term }}{\text { Coefficient of } x^3}=-\frac{c}{1}$
$\alpha \beta \gamma=-c$
$\Rightarrow(-1) \beta \gamma=-c[\therefore \alpha=-1]$
$\Rightarrow \beta \gamma=c$
$\Rightarrow \beta \gamma=1 a+b[\text { From Eq. (i) }]$
Hecne, product of the other two roots is $1-a+b$.
Alternate Answer
Since, -1 is one of the zeroes of the cubic polynomial $f(x)=x^2+a x^2+b x+c$ i.e., $(x+1)$ is a factor of $f(x)$,
Now, using division algorithm,
$x + 1 = { x ^ { 2 } + ( a - 1 ) x + ( b - a + 1 ) }$
$\frac{x^3+x^2}{(a-1) x^2+b x}$
$\frac{(a-1) x^2+(a-1) x}{(b-a+1) x+c}$
$\frac{(b-a+1) x(b-a+1)}{(c-b+a-1)}$
$\Rightarrow x^2+a x^2+b x+c=(x+1) x\left\{x^2+(a-1) x+(a-a+1)>+(c-b+a-1)\right.$
$\Rightarrow x^2+a x^2+b x+(b-a+1)=(x+1)\left\{x^2+(a-1) x+(b-a+1)\right.$
Let a and p be the other two zeroes of the given polynomial, then
$\text{Product of zeroes}=(-1)\alpha.\beta=\frac{\text{-Constant term}}{\text{Coefficient of x}^3}$
$\Rightarrow\ -\alpha.\beta=\frac{-(\text{b}-\text{a}+1)}{1}$
$\Rightarrow\ \alpha\beta=-\text{a}+\text{b}+1$
Hence, the required product of other two roots is $(-a + b + 1).$