Question

Verify the relationship between zeroes and coefficients of quadratic and cubic polynomials formed by you.

Answer 1

Step-by-step explanation:

hope this will help you

Answer 2

$i ) Let \: a \: Quadratic \: polynomial : \\p(x) = x^{2} - 2x - 8$

$We \: get \: zeroes \: of \: the \: polynomial \\p(x)\: we \: will \:take \:p(x) = 0$

$\implies x^{2} - 2x - 8 = 0$

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

$\implies x( x - 4 ) + 2( x - 4 ) = 0$

$\implies (x-4)(x+2) = 0$

$\implies x - 4 = 0 \: Or \: x + 2= 0$

$\implies x = 4 \: Or \: x = -2$

$\therefore The \: zeroes \: of \: x^{2} - 2x - 8\:are \: -2 , \: 4$

$\blue { Sum \:of \: the \: zeroes } \\= - 2 + 4 \\=2\\= \frac{- Coefficient \: of \: x }{Coefficient \: of \:x^{2}}$

$\blue { Product \:of \: the \: zeroes } \\= ( - 2 )\times 4 \\ = -8= \frac{Constant \:term }{Coefficient \: of \:x^{2}}$

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$ii ) Let \: a \: cubic \: polynomial :\\ g(x) = 2x^{3} - 5x^{2} - 14x + 8$

$We \: see \: that \: g(x) = 0 \\for \: x = 4 , -2 \: \frac{1}{2}$

$Since, g(x)\: can \: have \: at \: most \: three \\zeroes ,these \: are \: the \: zeroes \: of \\2x^{3} - 5x^{2} - 14x + 8$

$\blue { Sum \:of \: the \: Zeroes } \\= 4 + (-2) + \frac{1}{2} \\= 2 + \frac{1}{2} \\= \frac{4 + 1}{2} \\= \frac{5}{2} \\= \frac{ - Coefficient \:of \:x^{2}}{Coefficient \: of \: x^{3}}$

$\blue { Sum \: of \:the \: products \: of \: the }\\\blue { zeroes \: taken \: two \:at \:a \:time } \\= 4 \times (-2) + (-2) \times \frac{1}{2} + \frac{1}{2} \times 4 \\= -8 - 1 + 2 \\= -7 \\= \frac{-14}{2} \\= \frac{Constant \: of \: x}{Coefficient \:of \:x^{3}}$

$\blue { Product \: of \: zeroes } \\= 4 \times (-2) \times \frac{1}{2} \\= - 4 \\= \frac{-8}{2} \\= \frac{ - (constant\:term )}{coefficient \:of \:x^{3}}$

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