Question

Find zeros of polynomial 2x4-3x3-5x2+9x-3 if 2 of its zeroes are root 3 and-root 3

Answer 1

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Answer 2

Answer:

The zeros of the polynomial are $\frac{1}{2},1,\sqrt3,-\sqrt3$

Step-by-step explanation:

The given polynomial is $2x^4-3x^3-5x^2+9x-3$

Two zeros are $\sqrt3,-\sqrt3$

Factor Theorem: If a and b be the zeros of a function f, then from factor theorem its factors are (x-a)(x-b).

Hence, using this theorem, the factors are

$(x-\sqrt3)(x+\sqrt3)\\\\x^2-(\sqrt3)^2\\\\x^2-3$

Therefore, for other zeros, we divide x^2-3 to the given polynomial.

$\frac{2x^4-3x^3-5x^2+9x-3}{x^2-3}=2x^2-3x+1$

Hence,

$2x^2-3x+1=0\\\\(2x-1)(x-1)=0\\\\x=\frac{1}{2},1$

Therefore,  zeros are $\frac{1}{2},1,\sqrt3,-\sqrt3$