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...

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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}

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