Question

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

Answer 1

`(x-3)(x^2+2x-1)`

Explanation:

`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)`.

Answer 2

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

`color(white)(x^3-x^2-7x+3) = (x-3)(x+1-sqrt(2))(x+1+sqrt(2))`

Explanation:

`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))`