Question

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Question :


Find all the zeroes of the polynomial $\sf{ax^4+bx^3+cx^2+dx+e}$, if $\sf{a+c+e=0\:and\:b+d=0}$


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Answer 1

Hi friend here is your answer

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It is given that

$a + c + e = 0 \:$

$b + d = 0$

So

$c = - (e + a) \: and \: d = - b$

$a {x}^{4} + b {x}^{3} + c {x}^{2} + dx + e$

Substitute the values of c and d in above equation

$a{x}^{4} + b{x}^{3} + ( - (a + e)2 + ( - bx) + e$

On solving further

$a {x}^{2} ( {x}^{2} - 1) - e({x}^{2} - 1) + bx({x}^{2} - 1)$

$= ({x}^{2} - 1)(a {x}^{2} - e + bx)$

$= (x + 1)(x - 1)(a {x}^{2} - e + bx)$

As (x+1) and (x-1) are the factors of the above equation

Hence the above equation is divisible by both

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Hope it helps you.................!!

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Day One Ish

Answer 2

$\huge{\mathcal{\pink{Hello}}}$

$\huge\underline\mathfrak{Your answer}$