Twenty points: quadratic factorisation (two questions)

6x^2 + 11x + 4 

a = 6 , b =11 , c= 4 

Start by multiplying a * c = 6 * 4 = 24 

Now find two numbers that multiply to 24 and add up to 11

We could you 8 and 3 because...

     8 + 3 = 11 
      8 * 3 = 24

Now we rewrite ...

6x^2 + 8x + 3x + 4

Pull out like terms...

2x ( 3x + 4 ) + 1 ( 3x + 4) 

The factors of this expression are ( 2x + 1 ) ( 3x + 4) 

There are a few ways to evaluate a quadratic equation when the lead term's coefficient is a number besides a 1. In this case it's a 6. I think this is by far, the easiest approach:

[Step 1: multiply the lead term to the last term] *don't forget we did this, as we'll have to reverse it at the end:
6x²+11x+4: 

x²+11x+24   [Step 2: factor normally]
Factors of 24 which add to 11 are 8 and 3
 (x+8) (x+3)

However, we have to address the 6 that we multiplied at the beginning...[step 3]: divide the 6 from the numbers in each of the binomials:
(x+8/6) (x+3/6)
[Step 4]: Reduce each fraction. We can divide out a 2 from the first fraction, and a 3 from the second fraction:

(x+4/3)(x+1/2).

[Step 5]: Move any remaining denominator value in front of the x-terms:
(3x+4)(2x+1)

And, we're done!

Wow. that is a really cool question.