Question

Find the product of the zeroes of a polynomial p(x)=x2+2x+8​

Answer 1

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Correct Equation :

${x}^{2} - 2x - 8$

GiVen:

Quadratic polynomial $x^2-2x-8$

To Find:

The zeros of the quadratic polynomial relationship between zeroes and coefficients ?

SolutiOn:

First we solve the quadratic polynomial to get the roots of the polynomial.

Applying Middle term split,

$x^2-2x-8=0$

$x^2-4x+2x-8=0$

$x(x-4)+2(x-4)=0$

$(x-4)(x+2)=0$

$(x-4)=0,(x+2)=0$

$x=4,x=-2</p><p>$

So, The roots of the quadratic polynomial are $\alpha=4$,$\beta=-2$

The zeros of the polynomial are

$\alpha+\beta=4-2=2$

$\alpha \beta=4(-2)=-8$

The zeros of the quadratic polynomial relationship between zeroes and coefficients is

Let a is the coefficient of x², b is the coefficient of x and c is the constant

i.e. Substituting, a=1,b=-2 and c=-8

Sum of zeros is

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

$\alpha+\beta=-\frac{-2}{1}$

$\alpha+\beta=2$

It is verified.

Product of zeros is

$\alpha\beta=\frac{c}{a}$

$\alpha\beta=\frac{-8}{1}$

$\alpha\beta=-8$

It is verified.

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