Question

2x^3-x^2-3x=210
the answer is 5 but I want to know why.

Answer 1

You want to know why what?

Answer 2

`x=5,(-9+√(255)i )/(4) ,(-9-√(255)i )/(4)`

Explanation:

1) Move all terms to one side.

`2x^(3) -x^(2) -3x-210=0`

2) Factor
`2{x}^(3)-{x}^(2)-3x-210` using Polynomial Division.

1 - Factor the following.

`2x^(3) -x^(2) -3x-210`

2 - First, find all factors of the constant term 210.

`1,2,3,4,5,6,7,10,14,15,21,30,35,42,70,105,210`

3) Try each factor above using the Remainder Theorem.

Substitute 1 into x. Since the result is not 0, x-1 is not a factor..

`2*1^(3) -1^(2) -3*1-210=-212`

Substitute -1 into x. Since the result is not 0, x+1 is not a factor..

`2(-1)^(3) -(-1)^(2) -3*-1-210=-210`

Substitute 2 into x. Since the result is not 0, x-2 is not a factor..

`2*2^(3) -2^(2) -3*2-210=-204`

Substitute -2 into x. Since the result is not 0, x+2 is not a factor..

`2{(-2)}^(3)-{(-2)}^(2)-3* -2-210 = -224`

Substitute 3 into x. Since the result is not 0, x-3 is not a factor..

`2* {3}^(3)-{3}^(2)-3* 3-210 = -174`

Substitute -3 into x. Since the result is not 0, x+3 is not a factor..

`2{(-3)}^(3)-{(-3)}^(2)-3* -3-210 = -264`

Substitute 5 into x. Since the result is 0, x-5 is a factor..

`2* {5}^(3)-{5}^(2)-3* 5-210 =0`

------------------------------------------------------------------------------------------

⇒
`x-5`

4) Polynomial Division: Divide
`2{x}^(3)-{x}^(2)-3x-210` by
`x-5`.

`2x^(2)`
`9x`
`42`

-------------------------------------------------------------------------

`x-5` |
`2x^(3)`
`-x^(2)`
`-3x`
`-210`

`2x^(3)`
`-10x^(2)`

-----------------------------------------------------------------------

`9x^(2)`
`-3x`
`-210`

--------------------------------------------------------------------------

`42x`
`-210`

`42x`
`-210`

-------------------------------------------------------------------------

5) Rewrite the expression using the above.

`2x^2+9x+42`

`(2x^2+9x+42)(x-5)=0`

3) Solve for
`x.`

`x=5`

4) Use the Quadratic Formula.

1 - In general, given
`a{x}^(2)+bx+c=0` , there exists two solutions where:

`x=\frac{-b+\sqrt{b^(2) -4ac} }{2a} ,(-b-√(b^2-4ac) )/(2a)`

2 - In this case,
`a=2,b=9` and
`c = 42.`

`x=(-9+√(9^2*-4*2*42) )/(2*2) ,(-9-√(9^2-4*2*42) )/(2*2)`

3 - Simplify.

`x=(-9+√(255)i )/(4) ,(-9-√(255)i )/(4)`

5) Collect all solutions from the previous steps.

`x=5,(-9+√(255)i )/(4) ,(-9-√(255)i )/(4)`