Factorising harder quadratics

paulb203 · 2025-06-18 10:48:27

Maths Genie method

Example;

5x^2+2x-3

1. Set out brackets thus; (5x )(5x )
2. Thinking in terms of ax^2+bx+c, multiply a by c, i.e., multiply 5 by -3, giving us -15
(We do this because we’re going to eventually divide our answer by 5)
3. Find factors of -15, i.e.,
1,-15
3,-5;
-1,15
-3,5
4. Choose pair that adds to give b term, i.e., 2, which is -3,5
4. plug those into prepared brackets re step 1;
(5x-3)(5x+5)
5. Divide by 5;
(5x-3)(5x+5)/5
=(5x-3)(x+1) which is your answer
Q. Is this not a convoluted way of doing this?
Q. Why not find the factor pair that gives the b term (2) when you multiply one of them by 5, the other by 1?

Bob · 2025-06-19 02:15:49

I've not ever seen this method before.

Is it a convoluted way? Yes!

I'd advise stick to the way you've been using.

Bob

Is it easier to program using the method?

paulb203 · 2025-06-20 08:37:53

Thanks, Bob, although I'm having second thoughts;

Two examples later and my method didn't seem to work, and theirs did;

6x^2 + 11x -10

P.S. Is there a BBCode for superscript that works here?

Bob · 2025-06-20 20:19:07

Here's what I do:

6 and 10 both have more than two factors so there are several alteratives to try.

6 10 6 5 3 10 3 5

1 1 1 2 2 1 2 2

By trial I found that 3 times 5 - 2 times 2 gives 11

(3x )(2x )

The 3x must multiply the 5 so

(3x 2)(2x 5)

I want + 11 so that fixes the signs

(3x -2)(2x + 5)

If you want to check first if an integer solution is lurking there you can check B^2 - 4AC to see if it has an integer root:

B^2 - 4AC = 121 + 240 = 361. This has root 19 so if I used the quadratic formula a simple 'solution' is available. That check saves wasting time looking for a factorisation if there isn't one.

Bob

ktesla39 · 2025-06-20 23:48:02

Hi guys!

In Nepal, we are mainly taught 3 ways of solving quadratic equations: product sum method, completing square method and the formula method. You can read the whole chapter here: https://lms.neemaacademy.com/course/2095/School-Mathematics-Grade-10/component/ynPP0L

In the product sum method, which only works for integers, we do the steps:

For any equation like:

, we do:
- Multiply a with c to make ac
- Check whether it's sum or difference, if c > 0, it's sum else it's difference
- Find two numbers whose sum or difference is b and product is ac
- Once you have the numbers, keep them instead of b in the equation.

For example:

Here, ac = 6*(-10) = -60; use difference
The numbers whose product is 60 and difference is 11 are: 15 and 4.

Hence (2x+5) and (3x-2) are the solutions.

ktesla39 · 2025-06-20 23:50:22

Bob wrote:

B^2 - 4AC = 121 + 240 = 361. This has root 19 so if I used the quadratic formula a simple 'solution' is available. That check saves wasting time looking for a factorisation if there isn't one.
Bob

True to Bob. The quadratic formula is better than any other methods as it works for all.

paulb203 · 2025-06-21 03:08:52

Thanks, Bob; impressive, as ever.

I did Google the Maths Genie method; seems to be common, and called, The AC Method.

Recommended as a fail safe method for exams etc

I gave it a go with a few examples and found it less convoluted than I did a first.

paulb203 · 2025-06-21 03:10:56

@ktestla39

Cheers

Math Is Fun Forum

#1 2025-06-18 10:48:27

Factorising harder quadratics

#2 2025-06-19 02:15:49

Re: Factorising harder quadratics

#3 2025-06-20 08:37:53

Re: Factorising harder quadratics

#4 2025-06-20 20:19:07

Re: Factorising harder quadratics

#5 2025-06-20 23:48:02

Re: Factorising harder quadratics

#6 2025-06-20 23:50:22

Re: Factorising harder quadratics

#7 2025-06-21 03:08:52

Re: Factorising harder quadratics

#8 2025-06-21 03:10:56

Re: Factorising harder quadratics

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