Question

Verify that $1,-1,$ and $-3$ are the zeroes of the cubic polynomial, $x^3+x^2-x-3$ and verify the relationship between zeroes and the coefficients.

Answer 1

Given polynomial is $x^3+3x^2-x-3$

Let zeros be $\alpha =1$, $\beta =-1$, $\gamma =-3$

a) ${\alpha }^3+3{\alpha }^2-\alpha -3$

$1+3-1-3=0$

$⟹\alpha$ is a zero

b) ${\beta }^3+3{\beta }^2-\beta -3$

$-1+3+1-3=0$

$⟹\beta$ is a zero

c) ${\gamma }^3+3{\gamma }^2-\gamma -3$

$-27+27+3-3-=0$

$⟹\gamma$ is a zero

We know that

$1)\alpha +\beta +\gamma =\frac{-b}{a}$

LHS$=1-1-3=-3$

RHS$=\frac{-3}{1}$

$=-3$

$\therefore$ LHS=RHS

$2)\alpha \beta +\beta \alpha +\alpha \gamma =\frac{c}{a}$

LHS$=-1+3-3$

$=-1$

RHS=$-1$

$\therefore$ LHS=RHS

$3)\alpha \beta \gamma =\frac{-d}{a}$

LHS$=1(-1)(-3)$$=3$

RHS$=-\left(\frac{-3}{1}\right)$$=3$

$\therefore$ LHS=RHS