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Obtain all other zeroes of 3x^4 +6x^3-2x^2-10x-5, if two of it

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Obtain all other zeroes of $3 x^{4} + 6 x^{3} - 2 x^{2} - 10 x - 5$ , if two of its zeroes are $\frac{5}{3}$ and $- \frac{5}{3}$ .

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Obtain all other zeroes of

3 x^{4} + 6 x^{3} - 2 x^{2} - 10 x - 5

, if two of its zeroes are

\frac{5}{3}

and

- \frac{5}{3}

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Text solutionVerified

Two zeroes are

\frac{5}{3}

and

- \frac{5}{3}

So we can write it as, x =

\frac{5}{3}

and x =

- \frac{5}{3}

we get

x - \frac{5}{3} = 0

and

x + \frac{5}{3} = 0

Multiply both the factors we get,

x^{2} - \frac{5}{3} = 0

Multiply by

3

we get

3 x^{2} - 5 = 0

is the factor of

3 x^{4} + 6 x^{3} - 2 x^{2} - 10 x - 5

Now divide,

3 x^{4} + 6 x^{3} - 2 x^{2} - 10 x - 5

3 x^{2} - 5 = 0

we get,
Quotient is

x^{2} + 2 x + 1 = 0

Compare the equation with quadratic formula,

x^{2} - (S u m o f roo t) x + (P ro d u c t o f roo t) = 0

\Rightarrow

Sum of root

= 2

\Rightarrow

Product of the root

= 1

So, we get

\Rightarrow

x^{2} + x + x + 1 = 0

\Rightarrow

x (x + 1) + 1 (x + 1) = 0

\Rightarrow

x + 1 = 0, x + 1 = 0

\Rightarrow

x = - 1, x = - 1

So, our zeroes are

- 1, - 1,

\frac{5}{3}

and

- \frac{5}{3}

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Question Text	Obtain all other zeroes of $3 x^{4} + 6 x^{3} - 2 x^{2} - 10 x - 5$ , if two of its zeroes are $\frac{5}{3}$ and $- \frac{5}{3}$ .
Answer Type	Text solution:1
Upvotes	150

Obtain all other zeroes of 3x4+6x3−2x2−10x−5, if two of its zeroes are 35​​ and −35​​.

Text solutionVerified

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Obtain all other zeroes of $3 x^{4} + 6 x^{3} - 2 x^{2} - 10 x - 5$ , if two of its zeroes are $\frac{5}{3}$ and $- \frac{5}{3}$ .