Hey, I was just wondering if there is an easy method of factorising quadratics when the coefficient is greater than 1. For example, 1 question is 6x^2-11x-7=0. In school, we are just taught to do it in our heads but on questions like this it gets hard so I want to know a written method of solving this sort of questions
Method of factorising quadratics when coefficient is greater than 1? Watch
- Thread Starter
- 02-10-2016 11:02
- 02-10-2016 11:10
AC= 6*-7= -42
something that multiplys to get -42 but adds to get -11
= -14 +3 = -11
=-14 x 3 =-42
new equation = 6x^2-14x+3x-7
factorise (both brackets must be equal)
- 02-10-2016 11:11
2x(3x-7) + (3x-7)
(3x-7) (2x+1) ?
not sure if its plus one in last bracket though
- 02-10-2016 16:18
You can always plug the numbers into the quadratic solution to get the roots directly, then back out the brackets that way.
x = 7/3 and -1/2, from the general quadratic equation solution
so the factors must be (3x - 7) and (2x + 1)
i.e. when these factors are zero, x equals the answers we calculated.