Question

$if \: \alpha \: and \: \beta are \: zeroes \: of \: the \: quadratic \: polynomial \: f(x) = 3 {x}^{2} - 5x - 2 \\ then \: evaluate \: { \alpha }^{3} + { \beta }^{3}$
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Answer 1

$\huge{\mathcal{Hi\: there!}}$

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Given :

$\mathsf{f(x) = 3x {}^{2} - 5x - 2}$

Solution :

$\mathsf{3x {}^{2} - 5x - 2 = 0 } \\ \\ \mathsf{3x {}^{2} - 6x + x - 2 = 0} \\ \\ \mathsf{3x(x - 2) + 1(x - 2) = 0} \\ \\ \mathsf{(3x + 1)(x - 2) = 0} \\ \\ \mathsf{(3x + 1 = 0) \: and \: (x - 2 = 0)} \\ \\ \mathsf{x = \frac{ - 1}{3} \: and \: x = 2 }$

Now,

$\mathsf{let \: \alpha \: = \frac{ - 1}{3} \: and \: \beta \: = 2}$

So,

$\mathsf{( \alpha ) {}^{3} + ( \beta ) {}^{3} } \\ \\ \mathsf{( \frac{ - 1}{3}) {}^{3} + (2) {}^{3} } \\ \\ \mathsf{ \frac{ - 1}{27} + 8} \\ \\ \mathsf{ \frac{ - 1 + 216}{27} } \\ \\ \therefore\mathsf{ \frac{215}{27} }$

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Thanks for the question !

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Answer 2

$<b>Hello Friend$

▶ If α and β are roots of the quadratic polynomial, say, f(x) = ax^2 + bx + c then,
we have
α + β = - b / a and αβ = c / a

▶ Given :

f (x) = 3x^2-5x-2

a = 3, b = - 5, c = - 2

1. α + β = - ( - 5 ) / 3

α + β = 5 / 3

2. αβ = - 2 / 3

▶ To Find : α^3 + β^3

α^3 + β^3 = (α + β)^3 - 3αβ (α + β)

α^3 + β^3 = ( 5 / 3 )^3 - 3 × ( - 2 / 3 ) ( 5 / 3 )

α^3 + β^3 = 125 / 27 + 2 ( 5 / 3 )

α^3 + β^3 = 125 / 27 + 10 / 3

α^3 + β^3 = 125 + 90 / 27

α^3 + β^3 = 215 / 27

Hope It Helps

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