maths

Method 1 : x^3+3x^2-18x-40= x^3 - 4x^2 + 7x^2 - 18x - 40= (x^2)(x - 4) + (7x + 10)(x - 4)= (x - 4)(x^2 + 7x + 10)= (x - 4)(x + 2)(x + 5)Method 2 : Putting x as some factors of - 40 to x^3+3x^2-18x-40 , When you trying x = 4 , x = - 2 or x = - 5 , x^3+3x^2-18x-40 = 0 ,By remainder theorem (x - 4) , (x + 2) and (x + 5) are factors of
x^3+3x^2-18x-40.;Method 1 : 2x^3 + ax^2 + bx - 90 = (x + 5)(2x^2 + 5x - 18)= 2x^3 + 5x^2 - 18x + 10x^2 + 25x - 90= 2x^3 + 15x^2 + 7x - 90a = 15
b = 7Method 2 : 2x^3 + ax^2 + bx - 90 is divisible by 2x^2 + 5x - 18 = (2x + 9)(x - 2)Let f(x) be 2x^3 + ax^2 + bx - 90 , by factor theorem ,f(2) = 2(8) + a(4) + b(2) - 90 = 0==> 2a + b = 37 .....(1)f( -9/2) = 2(- 729/8) + a(81/4) + b(- 9/2) - 90 = 0==> - 729 + 81a - 18b = 360==> 9a - 2b = 121 .....(2)(1)*2 + (2) :4a + 9a = 37*2 + 12113a = 195a = 15
b = 37 - 2*15 = 7

x^3+3x^2-18x-40 = (x + 5)(x + 2)(x - 4)