Question

find zeroes of x^4 + 2x^3 - 13x^2 - 14x + 24​

Answer 1

Answer:

there are 4 zeroes out of 4 one zero is 1

Answer 2

Solution -

x⁴ + 2x³ - 13x² - 14x + 24

=> x⁴ - 2x³ - 5 x² + 6x + 4x³ - 8x² - 20x + 24

=> x ( x³ - 2x² - 5x + 6 ) + 4 ( x³ - 2x² - 5x + 6 )

=> ( x + 4 )( x³ - 2x² - 5x + 6 )

=> ( x + 4 )( x³ + x² - 2x - 3x² - 3x + 6 )

=> ( x + 4 )[ x ( x² + x - 2 ) - 3 ( x² + x - 2 ) ]

=> ( x + 4 )( x - 3 )( x² + x - 2 )

=> ( x + 4 )( x - 3 )( x² + 2x - x - 2 )

=> ( x + 4 )( x - 3 )[ x ( x + 2 ) - 1 ( x + 2 ) ]

=> ( x + 4 )( x - 3)( x - 1)( x + 2 )

Thus , the required zeroes are -4 , 3, 1 and -2 respectively .

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Additional Information -

( a + b )² = a² + 2ab + b²

( a - b )² = a² - 2ab + b²

( a + b )( a - b ) = a² - b²

( a + b )³ = a³ + 3ab ( a + b ) + b³

( a - b )³ = a³ - 3ab ( a + b ) - b³

( a + b + c )³ = a³ + b³ + c³ + 3 ( a + b )( b + c )( c + a )

a³ + b³ + c³ - 3abc = ( a + b + c )( a² + b² + c² - ab - bc - ca )

When a + b + c = 0 ,

a³ + b³ + c³ = 3abc .

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