Question

If one of the zeros of the cubic polynomial x^3+ax^2+bx+c is -1, then find the product of other two zeros.

Answer 1

Product of other two zeroes in the equation is –b.  

Solution:

To find the zeroes of the cubic equation $x^{3}+a x^{2}+b x+c$, let us say that the roots are $\alpha,\beta,\gamma$

The expressions of roots are a $x^{3}+b x^{2}+c x+d$

$\begin{array}{l}{\alpha+\beta+\gamma=-\frac{b}{a}} \\ \\{\alpha \beta \gamma=\frac{c}{a}} \\ \\{\alpha \beta+\beta \gamma+\gamma \alpha=-\frac{d}{a}}\end{array}$

Now one of the root is -1, putting α=-1 and a=1,b=a,c=b,d=c in the expression,

$\begin{array}{l}{-1+\beta+\gamma=-a} \\ \\{-1 \beta \gamma=b} \\ \\{-1 \beta+\beta \gamma-1 \gamma=-c}\end{array}$

To find the product of other zeroes we use $-1 \beta \gamma=b$

we get $\bold{\beta y=-b}$.

Answer 2

Answer:

Product of other two zeroes =c

Step-by-step explanation:

$Let \:p(x)=x^{3}+ax^{2}+bx+c , \: \\\alpha, \beta\: and \: \gamma \: are\:\\ three\: zeroes \:of\: p(x)$

$\alpha = -1\: (given)$

Compare p(x) with Ax³+Bx²+Cx+D , we get

A=1, B=a , C = b , D = c

$We\: know \: that\\ \alpha\beta\gamma= \frac{-D}{A}$

$\implies (-1)\beta\gamma = \frac{-c}{1}$

$\implies \beta\gamma = \frac{-c}{-1}$

$\implies \beta\gamma = c$

Therefore,

Product of other two zeroes =c

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