Find the sum of zeroes of the polynomial x2 - 2x - 8 1 (a) 2 (b) -2 (c) 4 (d) -4
Answers
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.
The answer is -2