# Solve Alegebraically: `x^3-7x-6=0`

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`x^3-7x-6=0`

To solve algebraically, we may apply factor by grouping.

To do so, group the first two terms.

`(x^3-7x)-6=0`

Factor out its GCF.

`x(x^2-7)-6`

In order that x^2-7 becomes a difference of two squares, subtract by 2.

`->`    `x^2-7-2=x^2-9`

Since outside the parenthesis of x^2-7 has an x, it means that when we subtract it by 2, we add 2x. So that when the polynomial is simplified, it returns to the original equation.

`x(x^2-7-2)+2x-6=0`

`x(x^2-9)+2x-6=0`

Then, group the last two terms and factor out its GCF.

`x(x^2-9)+(2x-6)=0`

`x(x^2-9)+2(x-3)=0`

Also, factor x^2-9.

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

Factor out the GCF of the two groups.

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

Simplify the expression inside the nested parenthesis.

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

Factor x^2+3x+2.

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

Then, set each factor equal to zero and solve for x.

For the first factor:

`x-3=0`

`x=3`

For the second factor:

`x+2=0`

`x=-2`

And for the third factor:

`x+1=0`

`x=-1`

Hence, the solution to the given equation is `x={-1,-2,3}` .

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You also should approach the problem in the following way, such that:

`x^3 - 7x - 6 = x^3 - (1 + 6)x - 6`

`x^3 - 7x - 6 = x^3 - x - 6x - 6`

You need to group the terms such that:

`(x^3 - x) - (6x + 6) = 0`

Factoring out x in the first group yields:

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

Factoring out 6 in the second group yields:

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

You need to convert the difference of squares `x^2 - 1` into a prodcut, such that:

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

Factoring out x + 1 yields:

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

You need to use the factored form of quadratic equation `x^2 -x - 6` , such that:

`x^2 - x - 6 = (x - x_1)(x - x_2)`

`x_(1,2) = (1 +- sqrt(1 + 24))/2 =>x_(1,2) = (1 +- 5)/2 `

`x_1 = 3 ; x_2 = -2`

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

Replacing `(x - 3)(x + 2)` for `x^2 -x - 6` yields:

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

You may solve for x the equation `(x + 1)(x - 3)(x + 2) = 0` , using zero prodcut rule, yields:

`(x + 1)(x - 3)(x + 2) = 0 => {(x +1 = 0),(x - 3 = 0),(x + 2 = 0):}`

`{(x= -1),(x = 30),(x = -2):}`

Hence, evaluating the solutions to the given equation, using factorization, yields `{(x= -1),(x = 3),(x = -2):}.`

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Another method that sometimes works (for "nice" textbook problems with small roots, especially) is to find a zero of the corresponding polynomial by trial and error. Using the Rational Root Test gives candidates for the rational roots, and in this problem it is easily seen that `-1` is a zero of `x^3-7x-6.` It's actually an obvious number to try because 7 and 6 differ by 1.

Anyway, since `-1` is a zero of `x^3-7x-6,` ``then `x-(-1)=x+1` must be a factor of `x^3-7x-6.` By long division, we see that

`x^3-7x-6=(x+1)(x^2-x-6).`

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