Question

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

Answer 1

`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 tha 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)`

We can factor the remaining quadratic 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`

`=(x-1)^2-(sqrt(2))^2`

`=(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))`