Answer:
(x+5)(x-1)(x+1)
Step-by-step 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).
Answer:
[tex]\large\boxed{x^3+5x^2-x-5=(x+5)(x-1)(x+1)}[/tex]
Step-by-step explanation:
[tex]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)[/tex]
