Answer
$x^3+5x^2+4x = (x)~(x+4)~(x+1)$
Work Step by Step
Let's consider a trinomial in this form: $~x^2+bx+c$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = c$
Then we can factor the trinomial as follows:
$~x^2+bx+c = (x+r)~(x+s)$
Let's consider this trinomial: $~x^3+5x^2+4x$
First we can factor by using the GCF:
$x^3+5x^2+4x = (x)~(x^2+5x+4)$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = 5$ and $r\times s = 4$. We can see that $4+1 = 5~$ and $4\times 1 = 4$
We can factor the trinomial as follows:
$x^3+5x^2+4x = (x)~(x^2+5x+4)$
$x^3+5x^2+4x = (x)~(x+4)~(x+1)$
