Question

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

Answer 1

Let $p\left(x\right)=x^3+3x^2-x-3$

$p\left(1\right)=(1{)}^3+3(1{)}^2-1-3=0$

$p(-1)=(-1{)}^3+3(-1{)}^2+1-3=0$

$p(-3)=(-3{)}^3+3(-3{)}^2+3-3=0$

Hence, $1,-1$ and $-3$ are the zeroes of the given polynomial.

If $\alpha ,\beta ,\gamma ,$ are roots of a cubic equation $a{x}^3+b{x}^2+cx+d=0$, then

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

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

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

$\Rightarrow -3=-\frac{b}{a}-3=-\frac{3}{1}=-3\Rightarrow -1=-\frac{1}{1}-1=-1\Rightarrow 3=-\frac{(-3)}{1}3=3$

Hence the relationship between zeroes and coefficients is also satisfied.