Factorising Quadratics

RickyOswaldIOW · 2005-11-26 06:09:14

I was revising some chapters of my A level maths, I was wondering;
When factorising a quadratic where the coefficient of the x^2 is more than 1, is there a way of figuring out 'a' and 'b' where (ax + y) (bx + z) without using trial and error?
i.e. 6x^2 - 11x + 3
= (2x - 3) (3x - 1)

I start of by guessing with (6x + _) (x + _) and try to fill in the blanks, if it dosent work I move on to (2x + _) (3x + _) and so on. If there is a quicker way to determin the coefficients of x, it would be tres handy

Last edited by rickyoswaldiow (2005-11-26 06:09:58)

mathsyperson · 2005-11-26 07:31:17

The best way to describe the method is just to give an example.
Let's do it with your one.

6x² - 11x + 3 = 0

The way to solve this is similar to solving ones where the x² coefficient is 1, but instead of finding 2 numbers that add to give -11 and multiply to give 3, we need numbers that add to give -11 and multiply to give 6x3 = 18, because the x² coefficient is 6.

Two such numbers are -9 and -2.
We can use these numbers to rewrite the original equation in a different form:
6x² - 9x - 2x + 3 = 0

Factorising the first two terms gives: 3x(2x - 3) - 2x + 3 = 0
Factorising the last two terms gives: 3x(2x - 3) -1(2x - 3) = 0 [1 included for clarity]

These two terms will combine to give (2x-3)(3x-1) = 0
The two x coefficients will always factorise with the other 2 terms to make the overall factorisation easier, no matter what coefficients you start off with. (Provided, of course, that they factorise in the first place)

If you're doing A level, you probably know this already, but you can check if a quadratic will factorise by looking at the discriminant: b² - 4ac, where a, b and c come from ax² + bx + c = 0.

For the quadratic to factorise, the discriminant has to be a square number.

RickyOswaldIOW · 2005-11-26 07:59:23

so if we have ax^2 + bx + c we need to find 2 numbers that add to give b and multiply to give a*c?

Last edited by rickyoswaldiow (2005-11-26 08:00:51)

mathsyperson · 2005-11-26 08:59:54

Exactly. You can then use those two numbers to split up the b and that helps you factorise the quadratic easier.

RickyOswaldIOW · 2006-01-23 02:53:55

I am stuck on the question y=-2x^2 - 7x + 15 where I need to sketch the curve.
x = 0 => y = 15

-2x^2 - 7x + 15
-[2x^2 + 7x - 15]
Here, I can see that the two numbers I need are +10 and -3 since:
10 * -3 = 2 * -15
and
10 - 3 = 7
so would the answer be...

RickyOswaldIOW · 2006-01-23 02:58:13

(ax + b) (cx + d)
so
-2x^2 - 7x + 15
-[2x^2 + 7x - 15]
-[(2x - 3)(x + 5)]
my problem is removing the square set of brackets here.

Last edited by rickyoswaldiow (2006-01-23 03:09:42)

RickyOswaldIOW · 2006-01-23 05:54:29

solved

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

irspow · 2006-01-23 08:55:49

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

I wouldn't bother trying to get that negative sign out of this problem, as you can see, it only served to confuse you. Mathsyperson's detailed description is all that you need.

-2x² - 7x + 15 (you realized that -10 and +3 were the factors needed)

-2x² - 10x + 3x + 15

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

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

I am actually thrilled to be able to do this myself. I hadn't seen, or did not remember, this method until Mathsyperson layed the steps out perfectly.

RickyOswaldIOW · 2006-02-12 07:00:35

I'm still having quite a bit of confusion over finding the co-efficients of the x in (dx + e) (fx + g). I can follow your examples and do other similar sums myself but I want to know an exact way of getting those co-efficients for my C++ based computer program.

for(i=0; i<255; i++){
for(j=0; j<255; j++){
while(a*c==i*j & b==i+j){ cout << "(x + " << i << ")(x + " << j) };
}
}

a, b and c are doubles entered by the user, how do I expand this program to take in values of a (ax^2 + bx + c) that are not 1 and then display the co-efficients of x in the answer?

irspow · 2006-02-12 07:49:32

In finding the factors of ax² + bx + c = 0

ac = n

d + e = b

d = n / e

d = ac / e

b = ac / e + e

b = (ac + e²) / e

be = ac + e²

e² - be + ac = 0

e = (b ± √(b² - 4ac)) / 2a

The two solutions for e actually represent both factors i.e., d and e

This definition will lead you to;

ax² + dx + ex + c

But now you would need to factorize;

ax² + dx and also ex + c

I'll have to think about this one some more.

edit*

It seems that after you had a, d, e, and c, you would then need an expression which would find the greatest common factor of a and d and then the gcf of e and c.

If f1 were the greatest factor of a and d and f2 were the gfc of e and c;

Your solution would be;

(f1 + f2) (a/f1 + d/f1)

If (a/f1 + d/f1) ≠ (e/f2 + c/f2), then it is not factorizable.... I think.

I have to leave this for someone smarter than me from this point. Sorry I couldn't be more helpful.

Last edited by irspow (2006-02-12 08:47:36)

irspow · 2006-02-12 10:05:18

You could try something like this to find f1 or f2.

Declare a variable n and then make an iterative loop where n starts at the smaller value of the two numbers in question. For each iteration is would decrease by 1. Then declare a variable f1 (greatest common denominator)

In the loop make a statement so that if;

a%n = a/n and d%n = d/n then f1 = n

RickyOswaldIOW · 2006-02-14 03:34:18

what does the % operator do?

irspow · 2006-02-14 08:55:29

It's been a while, but I believe that in C++ it is the modulus(?) function. It only returns an integer value for a fraction (or division). What the statement up there would do is check whether it is true that n goes evenly into a and d. If it didn't go in evenly then the integer answer wouldn't match the float answer. Being false, it would not assign f1 = n.

Ricky · 2006-02-14 09:19:40

% returns the remainder of division:

5 % 2 = 1
10 % 3 = 1
100 % 10 = 0
5 % 5 = 0
22 % 4 = 2

If you wish to see whether n goes into a (n divides a), you would do:

if (a % n == 0)

irspow · 2006-02-14 09:45:00

Thanks Ricky, it's been a while.

Ricky · 2006-02-14 11:04:41

Yep. The syntax is a bit weird becaue there is no mathimatical equivalent (that I know of). The only thing close is:

a = x mod n

Where x would be the result from a % n.

Math Is Fun Forum

#1 2005-11-26 06:09:14

Factorising Quadratics

#2 2005-11-26 07:31:17

Re: Factorising Quadratics

#3 2005-11-26 07:59:23

Re: Factorising Quadratics

#4 2005-11-26 08:59:54

Re: Factorising Quadratics

#5 2006-01-23 02:53:55

Re: Factorising Quadratics

#6 2006-01-23 02:58:13

Re: Factorising Quadratics

#7 2006-01-23 05:54:29

Re: Factorising Quadratics

#8 2006-01-23 08:55:49

Re: Factorising Quadratics

#9 2006-02-12 07:00:35

Re: Factorising Quadratics

#10 2006-02-12 07:49:32

Re: Factorising Quadratics

#11 2006-02-12 10:05:18

Re: Factorising Quadratics

#12 2006-02-14 03:34:18

Re: Factorising Quadratics

#13 2006-02-14 08:55:29

Re: Factorising Quadratics

#14 2006-02-14 09:19:40

Re: Factorising Quadratics

#15 2006-02-14 09:45:00

Re: Factorising Quadratics

#16 2006-02-14 11:04:41

Re: Factorising Quadratics

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