Help Factorising this equation
[From: ] [author: ] [Date: 12-07-11] [Hit: ]
where a does not = 0, drop the a and multiply it to c. So we have x^2+bx+ac=0. Simplify as normal then, but when you factored it, multiply a to the first bit with x.......
Hence
(3x - 1)(2x + 5) = 0
Hope that helps!!!!
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I know a method which I use all the time when the lead coefficient doesn't = 1.
If you have ax^2+bx+c=0, where a does not = 0, drop the a and multiply it to c. '
So we have x^2+bx+ac=0. Simplify as normal then, but when you factored it, multiply a to the first bit with x. If you want to prove it, the prove goes along the lines that the sum and product of the roots is the same. I'll get back to you on that if you wish.
Applying it to your problem,
6x^2+13x-5=0
x^2+13x-30=0 (see, 6*5=30)
(x+15)(x-2)=0
(6x+15)(6x-2)=0 (I multiplied 6 to both x's)
That will get you the same roots that your original trinomial had.
6x+15= 0, x=-15/6 | 6x-2=0, x=1/3
EDIT: I just noticed you wanted a factorization for it. My method does not give a legit factoring but it does give the roots.
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6x^2 +13x -5 = 0
product of coefficient of first term and 3rd term = 6 × (–5) = – 30
factors of 30 such that sum or difference is 13
15 × 2 = 30
+15 – 2 = 13 and 15 ×(– 2) = – 30
split middle term 13 x as + 15x – 2x
6x² + 15x – 2x – 5 = 0
3x(2x + 5) – 1(2x + 5) = 0
(3x – 1)(2x + 5) = 0
x = 1/3 or – 5/2
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6x² + 13x - 5 = 0
(3x - 1)(2x + 5) = 0
x = 1/3 , x = - 5/2
I just use trial and error to factorise.
