Factorise: 6x^2-3x-3=0

Since the equation shows a 6x^2, we know that in the brackets we must have either: (2x+a)(3x+b) or (6x+a)(x+b) However we know the last two numbers must multiply to make -3, hence a and b must be +1 and -3 respectively and so we know the second solution is the one to use. This leads us to the final step --> the inner number must be the added product of the x terms and a and b. This means that to get -3x we need to place the numbers as follows: (6x+3)(x-1) This is because (6x)(-1)=-6x and (x)(3)=3x and so the sum of these numbers gives us the middle term.

WP
Answered by William P. Maths tutor

3661 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

x^2 - 10x + 21 = 0


Solve 6x^2 > 3 - 7x


n is an integer greater than 1. Prove algebraically that n^2-2-(n-2)^2 is always an even number


What if my equation doesn't factorise?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences