Question

Find the zeroes of the polynomial f(x) = 4x² + 8x and verify the relationship between the zeroes and its coefficients. ​

Answer 1

F(x)= 4x²+8x

let f(x) be f(0)

f(0)= 4×0²+8×0= 0

please mark it as brainlist answer

Answer 2

Step-by-step explanation:

Given:-

• A quadratic polynomial f(x) = 4x² + 8x

To Find:-

• The zeroes of the polynomial

• And verify the relationship between the zeroes and coefficients.

Concept used:-

For a quadratic polynomial f(x) = ax² + bx + c

Sum of zeroes = $\rm \dfrac{-b}{a}$

Product of zeroes = $\rm \dfrac{c}{a}$

In the given quadratic equation 4x² + 8x

a = 4 , b = 8 and c = 0

Solution:-

f(x) = 4x² + 8x

$\sf \implies4 {x}^{2} + 8x = 0$

By factorization:-

$\sf \implies4x(x + 2) = 0$

$\sf \implies4x = 0 \: \: \: (or) \: \: \: x + 2= 0$

$\sf \implies x = \dfrac{0}{4} \: \: \: (or) \: \: \: x = 0 - 2$

$\sf \implies x = 0 \: \: \: (or) \: \: \: x = - 2$

Therefore, the zeroes are 0 and -2

Now, let us verify the relationship between the zeroes and coefficients.

$\sf Sum \: of \: zeroes = 0 + (-2) = -2$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{-b}{a} = \dfrac{ - 8}{4} = -2$

$\sf \therefore Sum \: of \: zeroes = \dfrac{ - (Coefficient \: of \: x) }{Coefficient \: of \: {x}^{2} }$

$\sf Product \: of \: zeroes = 0 \times -2 = 0$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{c}{a} = \dfrac{0}{4} = 0$

$\sf \therefore Product \: of \: zeroes = \dfrac{ Constant\: term}{Coefficient \: of \: {x}^{2} }$

Hence Verified