Question

find the zeros of the polynomial xsqare +x-12 and verify the relation between the zeros and Co efficient of the polynomial

Answer 1

x=0

x-12=0
x=12
now,
coefficient is=0

Answer 2

$hey \: mate \: here \: is \: your \: answer...$
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${x}^{2} + x - 12 \\ {x}^{2} + (4 - 3)x - 12 \\ {x}^{2} + 4x - 3x - 12 \\ x(x + 4) - 3(x + 4) \\ (x - 3)(x + 4) \\ \\ x - 3 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x + 4 = 0 \\ \: \: \: \: \: \: \: \: x = 3 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = - 4$


$\: \: \: \: \: \: \: \: \: \: verification$


$sum \: of \: zeros = \frac{ - b}{a} \\ \\ \: \: \: \: \: \: \: \: \: 3 + ( - 4) = \frac{ - ( 1)}{1} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 3 - 4 = - 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 1 = - 1$


$product \: of \: zeros = \frac{c}{a} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 3 \times - 4 = \frac{ - 12}{1} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 12 = - 12$


$hence \: verified...$

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$hope \: it \: helps.....$