Question

if alpha and beta are the zeros of the polynomial 5x^2+5x+1 find the value of alpha^3+beta^3

Answer 1

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$p(x) = 5 {x}^{2} + 5x + 1$

FINDING ZEROES OF POLYNIMIAL.

$p(x) = 0$

$5 {x}^{2} + 5x + 1 = 0$

$(5 {x}^{2} + 5x) + 1 = 0$

$5x(x + 1) + 1 = 0$

$(x + 1)(5x + 1) = 0$

NOW EITHER (x+1) IS ZERO OR (5x+1).

$x = - 1$

$x = - 1 \div 5$

HENCE,THE ZEROES ARE :

$\alpha = - 1$

$\beta = - 1 \div 5$

NIW LET US FING OUT THE VALUE OF ALPHA CUBE+ BETA CUBE.

$\alpha {}^{3} + \beta {}^{3} = ( \alpha + \beta ) {}^{2} - 3 \alpha \beta ( \alpha + \beta )$

$= ( (- 1) + ( - 1 \div 5)) - 3( - 1)( - 1 \div 5)(( - 1) + ( - 1 \div 5))$

$= ( - 6 \div 5) -( 3 \div 5)( - 6 \div 5)$

$= ( - 6 \div 5) - (18 \div 5)$

$= 24 \div 5$

HENCE,

$\alpha {}^{3} + { \beta }^{3} = 24 \div 5$

HOPE IT HELPS YOU...

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