Question

X^{3} -x^{2} -14x +24=0

Answer 1

Hi,

Explanation:

If you problem is to find roots of the equations:

`x^3-x^2-14x+24\\\\=x^3-2x^2+x^2-2x-12x+24\\\\=x^2(x-2)+x(x-2)-12(x-2)\\\\=(x-2)(x^2+x-12)\\\\=(x-2)(x^2-3x+4x-12)\\\\=(x-2)(x(x-3)+4(x-3))\\\\=(x-2)(x-3)(x+4)\\\\\boxed{x=2\ or\ x=3\ or \ x=-4}\\`

Answer 2

Hello!

Answer:

`\Large \boxed{\sf x =-4~~~x =2~~~x =3}`

Explanation:

→ We want to solve this equation:

`\sf x^(3) -x^(2) -14x +24=0`

→ It's a 3rd degree equation because it's on the form ax³ + bx² + cx + d = 0.

→ To solve a 3rd degree equation, let's factorize this equation:

`\sf x^3 + 2x^2 - 8x - 3x^2 - 6x +24 = 0`

`\sf x(x^2+ 2x - 8) - 3(x^2 + 2x - 8) = 0`

`\sf (x-3)(x^2+2x-8) = 0`

`\sf (x-3)(x^2-2x+4x-8) = 0`

`\sf (x-3)(x(x-2)+4(x-2)) = 0`

`\sf (x-3)(x-2)(x+4) = 0`

→ So now he have 3 equations to find the solutions:

`\sf x - 3 = 0`

`\sf x - 2 = 0`

`\sf x +4 = 0`

→ So the solutions are:

-4, 2, 3

Conclusion:

The solutions of the equation x³ - x² - 14x + 24 = 0 are -4, 2 and 3.