Question

Compute the zeroes of the polynomial 4x^2 - 4x - 8. Also, establish a relationship between the
xeroes and coefficients.​

Answer 1

Answer:

$4 {x}^{2} - 4x - 8 = 0 \\ = > 4 {x}^{2} - (8 - 4)x - 8 = 0\\ = > 4 {x}^{2} - 8x + 4x - 8 = 0 \\ = > 4x(x - 2) + 4(x - 2) = 0\\ = > (x - 2)(4x + 4) = 0 \\ = > (x - 2) = 0 \: or \: (4x + 4) = 0 \\ = > x = 2 \: or \: 4x = - 4 \\ = > x = 2 \: or \: x = \frac{ - 4}{4} \\ = > x = 2 \: or \: x = - 1$

zeroes of polynomial are 2 and -1

$sum \: of \: zeroes = 2 + ( - 1) = 2 - 1 = 1 \\ = - ( \frac{ - 4}{4} ) = - \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$product \: of \: zeroes = 2 \times ( - 1) \\ = - 2 = \frac{ - 8}{4} = \frac{constant \: term}{coefficient \: of \: {x}^{2} }$