Question

plz solve it in ur own handwriting

Answer 1

24 ) f ( x ) = x³+2x²-5ax-7

=> (x+1) = 0 => x = -1

we have , f ( -1 ) = R1


=> f ( -1 ) = ( -1 )³ + 2 ( -1 )² - 5a ( -1 ) - 7

=> R1 = -1 + 2 + 5a - 7

=> R1 = 5a - 6 ( eqn 1 )

g ( x ) = x³ + ax² - 12x + 6

=> x-2 = 0 => x = 2

we have , g ( 2 ) = R2

=> g ( 2 ) = ( 2 )³ + a ( 2 )² - 12 ( 2 ) + 6

=> R2 = 8 + 4a - 24 + 6

=> R2 = 4a - 10 (eqn 2)


given , 2R1 + R2 = 6

=> 2( 5a-6 ) + 4a - 10 = 6
=> 10a - 12 + 4a - 10 = 6

=> 14a = 6+22 = 28

=> a = 28/14 = 2


5 ) factorize => y³ - 2y² - 29y - 42

=> this is to be done by trial and error method

lets suppose y = 1

=> f(y) = y³ - 2y²- 29y - 42

=> f(1) = 1³ - 2(1)² - 29(1) - 42

=> f(1) = 1-2-58-42 ≠ 0

hence , ( y - 1 ) is not a factor


=> in the same way

=> let y = -2

=> f (-2) = (-2)³ - 2(-2)² - 29(-2) - 42

=> f (-2) = -8 - 8 + 58 - 42 => -16-42+58 => -58+58 = 0

therefore , ( y+2 ) is a factor of f (y)


now lets divide f(y) by (y+2) to get other factors


For further solution see the pic


hence , answer is ( y+2 ) ( y+3 ) ( y-7 )


hope this helps

if any confusion u know where to find me