# Find all rational zeros of P(x) = x3 - 7x + 6

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P(x) = x^3 - 7x + 6

First we will factor 6 to find possible roots for the function f9x):

6 = 1, -1, 2, -1, 3, -3, 6, -6

Let us try and substitute x = 1:

==> P (1) = 1 - 7 + 6 = 0

Then x= 1 is one of the roots for P(x):

==> ( x -1) is a factor for P(x):

Then we could wrtie:

P (x) = (x-1) * R(x)

Now we will divide P(x) by (x-1) to determine the other factors:

==> P(x) = (x-1)(x^2 +x - 6)

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

**Then roots for P(x) are:**

**x = { 1, 2, -3}**

We'll write 7x = 6x + x

We'll re-write the equation:

x^3 - 6x - x + 6 = 0

We'll find the roots of the equation between the divisor of the last terms divided by the divisors of the coefficient of the first term.

D6 = +/-1 ; +/-2 ; +/-3 ; +/-6

D1 = +/-1

We'll put x = 1:

1^3 - 6 - 1 + 6 = 0

0 = 0

So x = 1 is the first root of the equation.

Now, we'll use Viete's relations:

x1 + x2 + x3 = 0

1 + x2 + x3 = 0

x2 + x3 = -1

x1*x2 + x1*x3 + x2*x3 = -7

x2 + x3 + x2*x3 = -7

But

x2 + x3 = -1

-1 + x2*x3 = -7

x2*x3 = -7+1

x2*x3 = -6

x^2 + x - 6 = 0

We'll apply the quadratic formula:

x2 = [-1+sqrt(1+24)]/2

x2 = (-1+5)/6

x2 = 2/3

x3 = -6/6

x3 = -1

**The roots of the equation are: {-1 ; 2/3 ; 1}.**