i needed to edit this question :p

Question

anonymous · Answer

i am lousy at factoring how about  you?

anonymous · Answer

you kind have a choice for the first part 
could be $$(2x+a)(2x+b)$$ or $$(4x+a)(x+b)$$

anonymous · Answer

there are people who tell you there is a method for doing it, but i have never seen one that did not involve knowing the answer to begin with
i usually grind it till you find it 
we can try $$(4x+2)(x+3)$$ and see if it works, or at least see why it does not work

jim_thompson5910 · Answer

4x^2+11x+6


The first coefficient is 4
The last term is 6


Multiply them: 4*6 = 24


Now find two numbers that multiply to 24 and add to 11 (middle coefficient) at the same time


These two numbers are 3 and 8


so we can break up the 11x into 8x+3x


----------------------------


4x^2+11x+6


turns into


4x^2+8x+3x+6


Now factor by grouping

jim_thompson5910 · Answer

Alternatively, you can use the quadratic formula to find the roots. Then you use the roots to find the factorization

anonymous · Answer

you will definitely get $4x^2$ and also get the $6$ but you will not get $11x$ because $2x+12x\neq 11x$

anonymous · Answer

try $$(4x+3)(x+2)$$

anonymous · Answer

@jim_thompson5910  yes, if we know this 

Now find two numbers that multiply to 24 and add to 11 (middle coefficient) at the same time 

then we know it, but that seems to me the crux of the matter

jim_thompson5910 · Answer

yeah the guess and check can be a bit tedious sometimes. Which is why the quadratic formula method is probably more efficient

anonymous · Answer

true. but i have NEVER seen it taught that way.  always factoring comes first, then solving by factoring, then completing the square, then the quadratic formula

anonymous · Answer

with the third and most useful step usually skipped

anonymous · Answer

@G33k  did you try multiplying  $$(4x+3)(x+2)$$ to see if it works?