Welcome to our free-to-use Q&A hub, where students post questions and get help from other students and tutors.

Follow the trail of responses and if you have anything to add please sign up or sign in.

You can ask your own question or look at similar Mathematics questions.

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


If you need further help, feel free to contact me for tuition services


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

Footer Graphic