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

#color(blue)1x^3-x^2-7x+color(red)3#

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

The possible rational zeros are the factors of the constant #color(red)3# divided by the factors of the leading coefficient #color(blue)1#.

#x=(+-3)/(+-1)= +-1,+-3#

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 #(x-c)#.

I will test #x=3# by synthetic division, which represents #(x^3-x^2-7x+3)-:(x-3)#. If the remainder is zero, the factor is #(x-3)#.

#3|1color(white)a-1color(white)(a)-7color(white)color(white)(a)-3#
#color(white)(aa)darrcolor(white)(aaa)3color(white)(aa^1)6color(white)(a^1)-3#
#color(white)(a^1a)color(magenta)1color(white)(aa^1a)color(magenta)2color(white)(a)color(magenta)(-1)color(white)(aaaa)color(orange)0#

The remainder is #color(orange)(zero)#, which means #x=3# is a zero.

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

#color(magenta)1x^2+color(magenta)2xcolor(magenta)(-1)#

This polynomial is not factorable. So #x^3-x^2-7x+3# factors to
#(x-3)(x^2+2x-1)#.

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

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #3# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-3#

We find:

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

So #x=3# is a zero and #(x-3)# a factor:

#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 #a=(x+1)# and #b=sqrt(2)# as follows:

#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))#