Factorise fully 12x^2-20x+3

By factorising we're attempting to simplify this expression. When we factorise quadratics, typically this will result in two brackets. The content of these brackets, can be simply worked out with trial and error. I like to start by writing a list of pairs of factors for 12 and also 3. (12 and 1, -12 and -1, 6 and 2, -6 and -2, 3 and 4, -3 and -4) (3 and 1, -3 and -1). Now use trial and error to work out which two pairs multiplied with each other, and then added together make -20. In this case it is ( -6 and -2) as well as (3 and 1) since 3x-6=-18 and 1x-2=-2, and then -18+-2= -20. Now you just place these in the form (ax+b)(cx+d), ensuring that the placements are correct so that the right numbers are multiplying. i.e. with the same numbers your two brackets could expand to make a completely different quadratic. Eg. (-6x+3)(-2x+1) expands to make 12x^2-12x+3. Therefore the correct factorisation is (-6x+1)(-2x+3)