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# Method of factorising quadratics when coefficient is greater than 1? watch

1. 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

Thanks.
2. 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
(6x^2-14x)+(3x-7)
factorise (both brackets must be equal)
(6x^2-14x)+(3x-7)
3. 2x(3x-7) + (3x-7)
(3x-7) (2x+1) ?
not sure if its plus one in last bracket though
4. You can always plug the numbers into the quadratic solution to get the roots directly, then back out the brackets that way.

E.g.

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.

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