(2x+a)(x+3) = 2x^2 + 4ax+ b?

(2x+a)(x+3) = 2x^2 + 4ax+ b

2x² + 6x + ax + 3a = 2x² + 4ax+b

2x² + (6+a)x + 3a = 2x² + 4ax + b

For this to be true for all x, the coefficients of x^n on the LHS must equal the coefficients of x^n on the RSH.

2 = 2 : trivial

6+a = 4a → 3a = 6 → a = 2

3a = b = 3*2 = 6

a = 2

b = 6

Check:

(2x+2)(x+3) = 2x² + 8x + 6

2x² + 8x + 6 = 2x² + 8x + 6

0 = 0

Edit:

"why doesn't line 2 become a= -2 could you please expand on line two , thanks"

Because in line 2, 3a = 6 and a = 6/3 = 2. If 3a = -6, then a would equal -6/3 = -2.

If you mean the LHS expanded gives the RHS - no, it doesn't - it gives 2x^2 + x(a + 6) + 3a; If you are trying to factorise the RHS to get the LHS, you can't - you could get 2x(x + 2a) + b; if you are asking us to work out the values of a & b assuming the LHS = RHS, then look at the expansion I gave you at the start, to see that 4a = a + 6, so 3a = 6, so a = 2; also 3a = b, so b = 3 x 2 = 6