What are the solutions to the equation 3x^3+6x^2-24x=ax(x+b)(x+c) ?

The solutions to the equation 3x^3+6x^2-24x=ax(x+b)(x+c) are

Find the zeros of 3x^3+6x^2-24x=ax(x+b)(x+c)

The zeros of 3x^3+6x^2-24x=ax(x+b)(x+c) are

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3 x^{3} + 6 x^{2} - 24 x = a x (x + b) (x + c)

x = 0, x = \frac{- 6 + a c + ab + a ^{2} b ^{2} + a ^{2} c ^{2} - 2 a ^{2} b c + 12 ab c - 12 ab - 12 a c - 96 a + 324}{2 ( 3 - a )}, x = \frac{- 6 + a c + ab - a ^{2} b ^{2} + a ^{2} c ^{2} - 2 a ^{2} b c + 12 ab c - 12 ab - 12 a c - 96 a + 324}{2 ( 3 - a )}; a \neq = 3

Solution steps

3 x^{3} + 6 x^{2} - 24 x = a x (x + b) (x + c)

Move

a x (x + b) (x + c)

to the left side

3 x^{3} + 6 x^{2} - 24 x - a x (x + b) (x + c) = 0

Factor

3 x^{3} + 6 x^{2} - 24 x - a x (x + b) (x + c) : x (3 x^{2} + 6 x - a x^{2} - a c x - ab x - ab c - 24)

x (3 x^{2} + 6 x - a x^{2} - a c x - ab x - ab c - 24) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 3 x^{2} + 6 x - a x^{2} - a c x - ab x - ab c - 24 = 0

Solve

3 x^{2} + 6 x - a x^{2} - a c x - ab x - ab c - 24 = 0 : x = \frac{- 6 + a c + ab + a ^{2} b ^{2} + a ^{2} c ^{2} - 2 a ^{2} b c + 12 ab c - 12 ab - 12 a c - 96 a + 324}{2 ( 3 - a )}, x = \frac{- 6 + a c + ab - a ^{2} b ^{2} + a ^{2} c ^{2} - 2 a ^{2} b c + 12 ab c - 12 ab - 12 a c - 96 a + 324}{2 ( 3 - a )}; a \neq = 3

The solutions are

x = 0, x = \frac{- 6 + a c + ab + a ^{2} b ^{2} + a ^{2} c ^{2} - 2 a ^{2} b c + 12 ab c - 12 ab - 12 a c - 96 a + 324}{2 ( 3 - a )}, x = \frac{- 6 + a c + ab - a ^{2} b ^{2} + a ^{2} c ^{2} - 2 a ^{2} b c + 12 ab c - 12 ab - 12 a c - 96 a + 324}{2 ( 3 - a )}; a \neq = 3

a, a^{3} - b^{3} = 0

(2 x^{2} - 7 x)^{2} + 2 x^{2} - 7 x = 20

(x)^{3} + (3 x)^{2} - x - 3 = 0

(x^{2} + x + 1)^{2} = x^{4} + 8 x^{2} + 1

x^{3} + x^{3} = 2

What are the solutions to the equation 3x^3+6x^2-24x=ax(x+b)(x+c) ?
The solutions to the equation 3x^3+6x^2-24x=ax(x+b)(x+c) are
Find the zeros of 3x^3+6x^2-24x=ax(x+b)(x+c)
The zeros of 3x^3+6x^2-24x=ax(x+b)(x+c) are