Question

find one of the zeroes of the polynomial 2x^2+7x-4​

Answer 1

Answer:

x= 1/2 & -4

Step-by-step explanation:

${2x}^{2} + 7x - 4 = 0$

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

$2x(x + 4) - 1(x + 4) = 0$

$(2x - 1)(x + 4) = 0$

$2x - 1 = 0$

$x = \frac{1}{2}$

$x + 4 = 0$

$x = - 4$

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Answer 2

$\huge\underline\bold\red{AnswEr}$

X = -4 or X = 1/2

SOLUTION:-

${2x}^{2} + 7x - 4 \\.......but \: 4 \times 2 = 8 \: and \: 8 \times 1 = 8 \: therefore \: 8 + ( - 1) = 7 \\ = {2x}^{2} + 8x - x - 4x \\ = 2x(x + 4) - 1(x + 4) \\ hence.... \\ (x + 4)(2x - 1) \\ x = - 4 \: and \: x = \frac{1}{2}$