We're solving for X.
A. Apply special formula (ax^2+2bx+c^2)
3 (x^2+2[2x]+2^2) = 48
3(x^2+4x+4)=48
B.Multiply the coefficient with the trinomial
3x^2+12x+12=48
C. Set the equation to zero by subtracting 48 from both sides.
3x^2+12x-36=0
D. Factor using either trial and error or ac method of cross method
I'll choose trial and error.
a. Right away we can see 3(1) will be the beginning of your factors.
Why? Because nothing else multiplies to make 3.
(3x+?) (x+?)
b. Now check that -36 out. What multiplies to -36?
Remember, when using trial and error you're building off standard form, ax^2+bx+c^2 so you also have to apply whatever numbers you choose to multiply into the c^2 = -36 to add or subtract into bx=12x.
So for this problem I chose -2 and 18 because 3(-2)=-6 and 18-6=12.
Thus...
3x^2+12x-36=0
(3x+18)(x-2)
E. Set both binomials to zero and solve for X. I'll leave this to you.
