# x(x+1)(x+2)(x+3)=120find x

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You need to open the brackets such that:

`(x^2+x)(x^2+3x+2x``+6` `)=120`

`(x^2+x)(x^2+5x+6)=120`

`x^4 + 5x^3 + 6x^2 + x^3 + 5x^2 + 6x - 120 = 0`

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

You should perform a root test, hence you need to form a set consisting of divisors of 120 such that:

`D_120:{+-1,+-2,+-3,+-4,+-5,+-6,+-8,+-10,+-12,+-15,+-20,+-24,...,+-120` }

You need to substitute each divisor from this set in equation to check if it cancels the equation such that:

`x=-5`

`625 - 750 + 275 - 30 - 120 = 0`

Notice that `x=-5` cancels the equation, hence `x=-5` is a root of polynomial such that:

`x^4 + 6x^3 + 11x^2 + 6x - 120 = (x+5)(x^3+x^2+6x-24)`

Notice that x=2 cancels the equation `x^3+x^2+6x-24 = 0` , hence x=2 is a root of polynomial such that:

`x^4 + 6x^3 + 11x^2 + 6x - 120 = (x+5)(x-2)(x^2+3x+12)`

You should use quadratic formula to find the next two roots `x_3` and `x_4` such that:

`x_(3,4)=(-3+-sqrt(9-48))/2 `

`x_(3,4)=(-3+-i*sqrt39)/2`

**Hence, evaluating the roots of polynomial yields two real roots `x_1=-5, x_2=2` and two complex roots `x_(3,4)=(-3+-i*sqrt39)/2.` **

X=2