Question

(i) If the polynomial f(x) = 3x4 + 3x3 - 11x2 - 5x + 10 is completely divisible by 3x2 – 5, find all its zeroes.​

Answer 1

$\pink{3x {}^{4} + 3 {x}^{3} - 11x {}^{2} - 5x + 10 }$

$\frac{ \: \pink{3x {}^{4} + 3 {x}^{3} - 11x {}^{2} - 5x + 10 } \: \: }{3x {}^{2} - 5}$

(look in the attachment )

$\blue{quotient \: = x {}^{2} + x - 2 }$

$\blue{reminder = 0}$

now splitting the term

$x { }^{2} + x - 2$

There are two methods for finding zeros of a quadratic equation

$1 {}^{st} case$

$x {}^{2}$

is alone

let a and b the number the middle term is going to get into

$x {}^{2} + x - 2$

$a \times b = - 2 \\ a + b = 1$

$2 \times - 1 = - 2 \\ 2 - 1 = 1$

$x {}^{2} + 2x - 1x - 2 \\ x(x + 2) - 1(x + 2)$

$(x + 2) \: (x - 1)$

$x = - 2 \: and \: 1$

$d = \sqrt{b {}^{2} - 4ac}$

$x = \frac{ - b \frac{ + }{ - } d}{2a}$

$d = \sqrt{1 {}^{2} - 4 \times 1 \times - 2}$

$d = \sqrt{1 + 8}$

$d = \sqrt{9} \\ d = 3$

$x {}^{} = \frac{- 1 + 3}{1 \times 2}$

$x {}^{} = \frac{ 2}{2}$

$x {}^{} = 1$

$x = \frac{ \: - 1 - 3}{2 \times 1} \\ x = \frac{ - 4}{2} \\ x = - 2$

$3x {}^{2} - 5$

$( \sqrt{3} x ) {}^{2} - ( \sqrt{5 }) {}^{2}$

$a { }^{2} - b {}^{2} = (a - b)(a + b)$

$( \sqrt{3} x+ \sqrt{5} )( \sqrt{3 } x - \sqrt{5} )$

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

$\frac{ - \sqrt{5} }{ \sqrt{3} } \: \: \: \: \: \frac{ \sqrt{5} }{ \sqrt{3} } \: \: \: - 2 \: \: \: \ \: 1$

Hope it helps

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