Question

What is the complete factorization of the polynomial below x^3+5x^2-x-5

Answer 1

`\large\boxed{x^3+5x^2-x-5=(x+5)(x-1)(x+1)}`

Explanation:

`x^3+5x^2-x-5\qquad\text{distributive}\\\\=x^2(x+5)-1(x+5)\\\\=(x+5)(x^2-1)\\\\=(x+5)(x^2-1^2)\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=(x+5)(x-1)(x+1)`

Answer 2

(x+5)(x-1)(x+1)

Explanation:

Let's attempt factoring by grouping:

So what this means we first want to group the first two terms together and second two terms together, like so:

(x^3+5x^2)+(-x-5)

Now we factor what we can from each pair:

x^2(x+5)+1(-x-5)

Notice x+5 doesn't appear to be the same as -x-5 so we should factor out -1 instead of 1 in the second pair of terms:

x^2(x+5)-1(x+5)

You have two terms: x^2(x+5) and -1(x+5); they have a common factor of (x+5) so we can factor it out:

(x+5)(x^2-1)

You can actually factor this more because x^2-1 is a difference of squares.

The formula for factoring a difference of squares is u^2-v^2=(u-v)(u+v).

So the factored form of x^2-1 is (x-1)(x+1).

So the complete factored form of our expression we had initially is

(x+5)(x-1)(x+1).