The trick is that to assume (a+b) as x for example.

So the expression becomes:

6x^2 + 11x + 3

This could be factored as:

k ( x - x1 ) ( x - x2 )

where

k here = 6 (of x^2)

x1 and x2 are the roots (values that let the expression equals zero)

But do you know how to find the roots of the equation:

ax^2 + bx + c = 0 ?

if not, then other steps should be followed.

Let me assume you know the root formula for the general form:

x1 = [ - b + SQRT(b^2 - 4ac) ] / 2a

x2 = [ - b - SQRT(b^2 - 4ac) ] / 2a

here

a= 6

b = 11

c = 3

So

x1 = [ - 11 + SQRT( 11^2 - 4*6*3) ] / (2*6)

= - 4 / 2*6 = -1 / 3

and

x2 = [ - 11 - SQRT( 11^2 - 4*6*3) ] / (2*6)

= -18 / 2*6 = - 3 / 2

Therefore

6x^2 + 11x + 3

= 6 * ( x - x1 )( x - x2 )

= 3 * 2 * ( x - (-1/3 )( x - (-3/2) )

= 3(x+1/3)*2(x+3/2)

= (3x + 1) ( 2x +3)

What remains is to substitute x with a+b

= [ 3(a+b) + 1] [ 2(a+b) + 3 ]

= ( 3a + 3b + 1 ) ( 2a +2b + 3)

You factor it out completely, using the FOIL method for the 6(a+b) part and then distribution for the rest. You should get..

6(a^2+b^2+2ab) +11a + 11b + 3

6a^2 + 6b^2 + 12ab +11a + 11b +3

Then group all your like terms..

6a^2 + 11a + 6b^2 + 11b +12ab + 3

You simplify it like you'd simplify a trinomial but theres more terms.

(3a + 3b + 1)(2a + 2b + 3)

3a and 2a = 6a^2

3b and 2b = 6b^2

3a and 2b / 3b and 3a= 6ab + 6ab = 12ab

3 x 3a/ 3 x 3b/ 1 x 2a/ 1 x 2b/ = 9a + 9b + 2a + 2b = 11a and 11b

3 x 1= +3

If you were to factor out that finished explination, you would get the first equation aswell.

There you go. Its tricky but do-able if you really look at it and remember to do your distributive property. practice is needed for these questions. good luck.