# How do you factor completely #x^3-x^2-7x+3# ?

##### 2 Answers

#### Explanation:

To factor this expression, first list all the possible rational zeros.

The possible rational zeros are the factors of the constant

Next, pick one of the possible rational zeros and test it with synthetic division. If the remainder is zero, it is a zero of the equation. If the possible rational zero is c, the factor is

I will test

The remainder is

The results of synthetic division are the coefficients of the polynomial quotient of the division of polynomials listed above.

This polynomial is not factorable. So

#### Explanation:

#f(x) = x^3-x^2-7x+3#

By the rational roots theorem, any *rational* zeros of

That means that the only possible *rational* zeros are:

#+-1, +-3#

We find:

#f(3) = 27-9-21+3 = 0#

So

#x^3-x^2-7x+3 = (x-3)(x^2+2x-1)#

The remaining quadratic can be factored using irrational coefficients by completing the square and using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with

#x^2+2x-1 = x^2+2x+1-2#

#color(white)(x^2+2x-1) = (x+1)^2-(sqrt(2))^2#

#color(white)(x^2+2x-1) = ((x+1)-sqrt(2))((x+1)+sqrt(2))#

#color(white)(x^2+2x-1) = (x+1-sqrt(2))(x+1+sqrt(2))#

Putting it all together:

#x^3-x^2-7x+3 = (x-3)(x+1-sqrt(2))(x+1+sqrt(2))#