# `3x^3 -3x^2-7x+5= x^3-2x^2-1` : Solve for x3x^3-3x^2-7x+5=x^3-2x^2-1 How do you reach :x=1, 3/2 and -2

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Reduce your equation by placing like terms together:

`3x^3 -x^3 -3x^2 +2x^2 -7x +5 +1 =0`

Simplify:

`therefore 2x^3-x^2 -7x +6 =0`

Try substituting values of x such as x=1, x=2, x=-1, x=-2 to find the first value of x:

`therefore 2(1)^3 - (1)^2 -7(1) + 6 = 0`

If the LHS (left hand side) reduces to zero then you know that x = 1 and therefore (x-1) is a factor:

`2 - 1 - 7 + 6 = 0`

`therefore 0=0`

`therefore x=1 ` and (x-1) is a factor

Now complete a division sum to find the other 2 factors. Synthetic division is the easiest:

Take the factor 1 and take the co-efficients (the numbers in front of the xs and the constant and add them according to the rule in order. Note that the answer to one is then added to the next, except the first number (in this case the 2 which is simply brought down as follows:

(from x=1) 1 ..... / 2 :-1 : -7: 6

now bring down the 2 and transfer that same 2 to match up with the -1 and add together thus:

2 (first factor)....

(2+(-1))= 1. (2nd factor) Now take that 1 and add to -7 thus

2....1.....1+(-7) = -6 (3rd factor). Now take that -6 and add to the 6 thus

2.....1.....-6....(-6) +6=0

Now you have the makings of the trinomial to accompany (x-1)

2....1....-6....0

These are the co-efficients you will use as follows:

`(x-1)(2x^2+x-6) =0`

Factorize (factors of 2 are 1 x 2 thus (x+...)(2x +...) and

factors of 6 that work in this case are 2x3.

To render the correct miiddle term (1x) we must use the correct combination: (x+2)(2x-3)

`therefore` (x-1)(x+2)(2x-3) =0

now make each equation equal to zero and solve separately. Follow the link below for guidelines on synthetic division

So **x=1, x=-2** and 2x=3 so **x=3/2**