How do you simplify this equation?

3 x ² + 9 x
---------------
x ² + 4 x + 3

( 3 x ) (x + 3)
----------------
(x + 3) (x + 1)

3x
-------
(x + 1)

Im guessing break up the x[2] + 4x + 3 to (x+3)(x+1) and break up 3x[2] + 9x to (3x)(x+3) and then cancel out (x+3). Then you're left with 3x divide by x+1. Do

numerator is 3x^2 + 9x
taking 3x common
3x (x +3 )

denominator is x^2+4x + 3
i.e.
(x+3) (x + 1)

now [3x (x + 3 )] / [(x + 3) (x + 1)]
cancelling (x + 3 )

final answer is 3x / (x +1)

3x^2 + 9x / x^2 + 4x + 3

Factorise top and the bottom.....

3x (x + 3) / (x + 3) (x + 1)

Simplify......i.e. the (x + 3) term cancels....

3x / (x + 1) - Solution

(3x^2 + 9x)/(x^2 + 4x + 3)
= 3x(x + 3)/(x + 3)(x + 1)
= 3x/(x + 1)

[3x^2+9x]/[x^2+4x+3]

3x[x+3]/([x+3][x+1])

3x/[x+1]

3x[2]+9x = 3x(x+3 )
x[2]+4x+3 = (x+1)(x+3)
divid both by (x+3), then the simpification will be = 3x/(x+1)

(3x^2 + 9x)/(x^2 + 4x + 3)
= 3x(x + 3)/(x + 3)(x + 1)
= 3x/(x + 1)

hi!

(3x^2 + 9x) / (x^2 + 4x + 3)

factor both the numerator and the denominator
[3x(x + 3)] / [(x + 3) (x + 1)]

then, cancel same terms that are present both in the numerator and denominator
3x / (x + 1)

so, the simplified term is
3x / (x + 1)

i hope this helps

=)