How can I solve the equation : 120=x(x-1)(x-2)(x-3)

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The equation to be solved is : 120=x(x-1)(x-2)(x-3)

120 = x(x - 1)(x - 2)(x - 3)

=> 120 = x^4 - 6*x^3 + 11*x^2 - 6*x

=> x^4 - 6*x^3 + 11*x^2 - 6*x - 120 = 0

=> x^4 - 5x^3 - x^3 + 5x^2 + 6x^2 - 30x + 24x -120 = 0

=> x^3(x - 5) - x^2(x - 5) + 6x(x - 5) + 24(x - 5) = 0

=> (x - 5)(x^3 - x^2 + 6x + 24) = 0

=> (x - 5)(x^3 + 2x^2 - 3x^2 - 6x + 12x + 24) = 0

=> (x - 5)(x^2(x + 2) - 3x(x + 2) + 12(x + 2)) = 0

=> (x - 5)(x + 2)(x^2 - 3x + 12) = 0

This gives the roots x1 = 5 and x2 = -2

From x^2 - 3x + 12 = 0, we get the roots

x3 = 3/2 + sqrt (9 - 48)/2

=> 3/2 + i*sqrt 39/2

x4 = 3/2 - i*sqrt 39/2

**The solutions of the given equation are (5, -2, 3/2 + i*sqrt 39/2, 3/2 - i*sqrt 39/2)**

To solve 120 = x(x-1)(x-2)(x-3).

Let f(x) = x(x-1)(x-2)(x-3) = x^4-6x^3+11x^2-6x-120

We know that 5 = 120 = 5!

=> 120 = 5*4*3*2. This implies x1 = 5 is a solution of f(x) = 0 . So x-5 is a factor of f(x).

Also 120 = (-2)(-3)(-4)(-5). So x2 = -2 is a solution of f(x) = 0. Or x+2 is a factor.

Therefore f(x) has the factors (x+2)(x-5) = x^2-3x-10.

So f(x)/(x^2-3x-10) = (x^4-6x^3+11x^2-6x-120)/(x^2-3x-10) = x^2-3x-12) . Actual division is not shown please.

Therefore x^2-3x-12 = 0 gives other two solutions of x^4-6x^3+11x^2-6x-120 = 0 or 120 = x(x-1)(x-2)(x-3).

The quadratic formula gives x= {3 + or - sqrt(3^2-4*12)}/2 as the solution of x^2-3x-12 = 0

**Therefore the solutions of 120 = x(x-1)(x-2)(x-3) are x1=5, x2 =-2, x3 = {3+sqrt(-39)}/2 and x4 = {3-sqrt(-39)}/2.**

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