Question

Find the sum of zeroes of the polynomial x2 - 2x - 8
(a) 2 (b) -2 (c) 4 (d) -4

Answer 1

Answer:

Step-by-step explanation:

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

Applying Middle term split,

x^2-2x-8=0x2−2x−8=0

x^2-4x+2x-8=0x2−4x+2x−8=0

x(x-4)+2(x-4)=0x(x−4)+2(x−4)=0

(x-4)(x+2)=0(x−4)(x+2)=0

(x-4)=0,(x+2)=0(x−4)=0,(x+2)=0

x=4,x=-2x=4,x=−2

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

The zeros of the polynomial are

$$\begin{lgathered}\alpha+\beta=4-2=2\\\alpha \beta=4(-2)=-8\end{lgathered}$$

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.

And your answer is -2