Question #50ed4

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Question #50ed4

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Answer 1 · 2018-02-22T09:58:38

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

Explanation:

$given the zeros of a polynomial say$ $"given the zeros of a polynomial say"$

$x=a and x = b then the factors are$ $"x=a and "x=b" then the factors are"$

$(x - a) and (x - b)$ $(x-a)" and "(x-b)$

$and the polynomial can be written as$ $"and the polynomial can be written as "$

$f (x) = k (x - a) (x - b) \leftarrow k is a multiplier$ $f(x)=k(x-a)(x-b)larrcolor(blue)"k is a multiplier"$

$here x = 1 \pm \sqrt{5}$ $"here "x=1+-sqrt5$

$\Rightarrow (x - (1 + \sqrt{5})) (x - (1 - \sqrt{5})) are the factors$ $rArr(x-(1+sqrt5))(x-(1-sqrt5))" are the factors"$

$\Rightarrow f (x) = k (x - 1 - \sqrt{5}) (x - 1 + \sqrt{5})$ $rArrf(x)=k(x-1-sqrt5)(x-1+sqrt5)$

$let k = 1$ $"let "k=1$

$\Rightarrow f (x) = {(x - 1)}^{2} - {(\sqrt{5})}^{2}$ $rArrf(x)=(x-1)^2-(sqrt5)^2$

$\Rightarrow f (x) = x^{2} - 2 x + 1 - 5$ $color(white)(rArrf(x))=x^2-2x+1-5$

$\Rightarrow f (x) = x^{2} - 2 x - 4$ $color(white)(rArrf(x))=x^2-2x-4$

$to find the zeros solve x^{2} - 2 x - 4 = 0$ $"to find the zeros solve "x^2-2x-4=0$

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Question #50ed4

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