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given the equation, we can say that we have (x+4) in common.
So, we take that out, (x+4), we will have another bracket, (3x+3).

We can say that (x+4)(3x+3)=0

now we can equate, we can say either (x+4)=0 or (3x+3)=0                                                                                                x=0 -4  or 3x=0-3                                                                                                  x= -4 or x=-3/3                                                                                                      x=-4   or x=-1

expand the brackets

3x^2 + 12x + 3x + 12 = 0

Gather like terms

3x^2 + 15x + 12=0

simplify by dividing the whole equation by 3

x^2 + 5x + 4 = 0

Factorise (two numbers that added give you 5 and multiplied give you 4)

(x+1)(x+4) = 0

Thus, the two solutions are x=-1 and x=-4

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First, expand the brackets

So 3x^2+12x + 3x+12 =0

Then simplify it all out - ax^2+bx+c=0 is the format you want the question in to solve

3x^2 + 15x +12 =0

This is the same as 3(x^2+5x+4)=0

Now you need to find 2 numbers to multiply to make 4 (c in the equation) that also add to make 5 - 1 and 4.

Therefore x=-1 and x=-4

Check it in the equation and see if it equals 0, if it does it's right

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