Question

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10. Obtain all other zeroes of 3x4+ 6x3 – 2x2 – 10x – 5, if two of its zeroes
are √(5/3) and-√(5/3).






➳ REQUIRED GOOD ANSWER​

Answer 1

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Step-by-step explanation:

Two zeros are

$\sqrt{ \frac{5}{3} } and -\sqrt{ \frac{5}{3} }$

so we can write as x =

$\sqrt{ \frac{5}{3} }and \: \times = \: - \sqrt{ \frac{5}{3} }$

we get

$x - \sqrt{ \frac{5}{3} } = 0 \: and \: x +\sqrt{ \frac{5}{3} } \: = 0$

Multiply both the factor we get,

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

Multiply by 3 we get

${3x}^{2} - 5 = 0 \: is \: the \: factor \: of \: {3}^{4} + \: {6x}^{3} - {2x}^{2} - 10x - 5$

Now divide

${3x}^{4} + {6x}^{3} - {2x}^{2} - 10x - 5 \: by \: {3x}^{2} - 5 = 0 \: we \: get$

Quotient is

${x}^{2} + \: 2x + 1 = 0$

Compare the equation with quadratic formula

${x }^{2} - \: (sum \: of \: root \: )x + (product \: of \: root) = 0$

☞ sum of root = 2

☞ Product of the root = 1

So, we get

$→ {x}^{2} + x + x + 1 = 0 \\ →x(x + 1) + 1(x + 1) = 0 \\ →x + 1 = 0. \: x + 1 = 0 \: →x = - 1. \\ x = - 1$

So, our zeros are

, $- 1, - 1, \sqrt{ \frac{5}{3} } and - \sqrt{ \frac{5}{3} }$

Answer 2

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