how do you factorise 14x^2 - 29x + 12?

In some of the answers already posted the numbers -21 and -8 appear as if out of the blue. I wonder whether you know where they came from?

Here is a way of obtaining them in all case where the factors are rational numbers.

A quadratic expression is of the form ax² + bx + c (where a, b and c are constants)

You need two numbers that ADD to 'b' and MUTIPLY to 'a*c'.

In your case a = 14, b = -29 and c = 12

So you need two numbers to add to -29 mutiply to 168

Having seen previous answers the numbers are
'obviously!!!!!!!' -8 and -21

Not easy to 'see' these so if you are stuck then focus on the a*c value, 168, and write down all pairs which when mutiplied give 168
such as 1*168, 2*84, 4*42,....until you get there (Be careful with any negative values that may occur in other examples)

Some other answers posted take it from there.

Apologies for the 'sermon' but this should give you a good grounding for all quadratic factorisation and hopefully in future you will be able to 'see' the pair you require.
Good luck

14x^2-21x-8x+12

7x(2x-3)-4(2x-3)

(7x-4)(2x-3)
Hope that helps.
Please choose me as the best answer!!

14x^2 - 29x + 12
= 14x^2 - 21x - 8x + 12
= (14x^2 - 21x) - (8x - 12)
= 7x(2x - 3) - 4(2x - 3)
= (2x - 3)(7x - 4)

(7x - 4)(2x - 3)

look at factors

14 == 2 & 7

12 == 3 & 4

29 = 21 + 8 = 3*7 + 2* 4

so
14x^2 - 29x + 12 = (7x - 4)(2x -3)

its an art :)

Go to school and they will teach you all you need to know.