Question

7. If one of the zeroes of the cubic polynomial x3 + ax² + bx + c is -1, then the product of The other two zeroes are
(a) b – a + 1
(b) b – a – 1
(c) a – b + 1
(d) a – b – 1

Answer 1

option d

Step-by-step explanation:

You can just put the value of first zero and find the answer

Answer 2

$\text { Let } p(x)=x^{3}+a x^{2}+b x+c$

$\text { Let a, } p \text { and } y \text { be the zeroes of the given cubic polynomial } p(x) \text {. }$

$\therefore \alpha=-1$   (given)

$\text { and } p(-1)=0$

$\begin{array}{l}\Rightarrow(-1)^{3}+a(-1)^{2}+b(-1)+c=0 \\\Rightarrow-1+a-b+c=0 \\\Rightarrow c=1-a+b\end{array}$

$\text { Product of all zeroes }=(-1)^{3} \cdot \frac{\text { Constant term }}{\text { Coefficient of } x^{3}}=-\frac{c}{1}$

$\begin{array}{l}a \beta Y=-c \\\Rightarrow(-1) \beta Y=-C \\\Rightarrow \beta \mathrm{Y}=c \\\Rightarrow \beta \mathrm{y}=1-\mathrm{a}+\mathrm{b}\end{array}$

$\text { Hence, product of the other two roots is } 1-\mathrm{a}+\mathrm{b} \text {. }$