Question

Obtain all other zeroes of the polynomial x^4+6x^3+x^2-24x-20,if two of its zeroes are 2 and -5

Answer 1

it is right please follow me ok

Answer 2

$<b>$

$\boxed{\sf{P(x) =x^{4}+6x^{3}+x^{2}-24x-20}}$

Two of zeroes are 2 and (-5)

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According to FACTOR THEOREM

(x-2) and (x+5) are factors are p(x)

Also,

$(x - 2)(x + 5) \\ \\ = {x}^{2} + 5x - 2x - 10 \\ \\ = {x}^{2} + 3x - 10$

x²+3x-10 is also a factor of p(x)

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Now,

Diving p(x) by x²+3x-10

We get :-

Q(x) = x² + 3x +2

r(x) = 0

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We have :-

$p(x) = ({x}^{2} + 3x -10)( {x}^{2} + 3x + 2) \\ \\ \\ = (x - 2)(x + 5)( {x}^{2} + x + 2x + 2) \\ \\ \\ = (x - 2)(x + 5)(x + 1)(x + 2)$

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Thus

Zeroes are as follows :-

$x - 2 = 0 \\ \\ x = 2$

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$x + 5 = 0 \\ \\ x = ( - 5)$

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$x + 1 = 0 \\ \\ x = ( - 1)$

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

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