Question

obtain all zeroes of 3x4-15x3+13x2+25x-30 if two zeroes are root5/3 and root -5/3

Answer 1

Heya dear .

Solution Given here .
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The given Polynomial is
$f(x) = (3 {x}^{4} - 15 {x}^{3} + 13 {x}^{2} + 25x - 30).$
$since \sqrt{ \frac{5}{3} } \: \: and \: \: - \sqrt{ \frac{5}{3} }$
are the zeros of f (x) , it follows that each one of
(x- √5/3 ) and (x + √5/3) is a factor of f(x).

$(x - \sqrt{ \frac{5}{3} } )(x + \sqrt{ \frac{5}{3} }) = ( {x}^{2} - \frac{5}{3} ) = \frac{(3 {x}^{2} - 5) }{3} \\ is \: \: a \: \: factor \: \: of \: f(x)$
consequently , ( 3x^2 - 5) is a factor of f (x)

on dividing f(x) by (3x^2 - 5) , we get
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【SEE in the above attachment 】
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$= > f(x) = 3 {x}^{4} - 15 {x}^{3} + 13 {x}^{2} + 25x - 30 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = ( 3 {x}^{2} - 5)( {x}^{2} - 5x + 6) \\ \: \ = ( \sqrt{3}x + \sqrt{5} )( \sqrt{3} x - \sqrt{5} )(x - 2)(x - 3). \\ \\ = > f(x) = 0 \\ = > ( \sqrt{3} x + \sqrt{5}) = 0 \: \: \: or \: ( \sqrt{3} x - \sqrt{5} ) = 0 \\ or \: (x - 2) = 0 \: \: or \: \: (x - 3) =0 \\ \\ \\ = > x = - \sqrt{ \frac{5}{3} } \: \: or \: x = \sqrt{ \frac{5}{ 3} } \: \: or \: x = 2 \: \: or \: x = 3. \\ \\ hence \: \: all \: zeros \: \: of \: f(x) \: \: are \: \\ \sqrt{ \frac{5}{3} } \: \: or \: - \sqrt{ \frac{5}{3} } \: \: or \: \: 2 \: \: and \: \: 3$
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Hope it's helps you.
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Answer 2

hi mate

other two zeroes are 2 and 3

Step-by-step explanation:

for step by step explaination refer the attachment above.

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