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)

Answered by Santiago O. Maths tutor

3585 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Solve the folllowing quadratic equation: y=x^2+x-6


Sketch the inequality x^2 - x - 12 > y on a set of axes.


Divide 711 in the ratio 4:5


Anna has 4 cakes. Three of them are squares with sides of length x, and one is rectangular and measures 2 by (3x+2). The total area of all the cakes is 13. What is the length of x?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences