Question

find the zeros of quadratic polynomial FX = x square - 3x - 28 and verify the relation between zeros and its coefficient

Answer 1

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Answer 2

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✔Solution HERE ↔↔↔↔↔↔↔↔↔↔↔
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Let the given Polynomial be denoted by f (x) , Than

$f(x) = {x}^{2} - 3x - 28 \\ = > {x}^{2} - 7x + 4x - 28 \\ = > x(x - 7) + 4(x - 7) \\ = > (x + 4)(x - 7) \\ \\ f(x) = 0 \\ \\ = > (x + 4)(x - 7) = 0 \\ = > x + 4 = 0 \: \: \: or \: \: \: x - 7 = 0 \\ = > x = - 4 \: \: or \: \: x = 7$
So , the zeros of f(x) are -4 and 7 .

we have to verify .
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$sum \: \: of \: \: zeros \: ( - 4 + 7) = - 3 = \frac{ - ( coefficient \: of \: \: x) \: }{coefficient \: \: of \: {x}^{2} }$


$product \: \: of \: \: zeros = - 4 \times 7 = - 28 = \frac{constant \: \: term}{coefficient \: \: of \: {x}^{2} }$

Hence verified .

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Hope it's helps you.
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