Factorising quadratics... again

RickyOswaldIOW · 2005-12-17 06:52:07

I have problems when trying to factorise a quadratic where the co-efficient of x^2 is < -1 i.e.

y = -2x^2 - 3x + 2

What are the steps I need to take to get an answer in the form (ax + b)(cx + d)?

krassi_holmz · 2005-12-17 07:02:55

You can (what is the word for taking divisor out of brackets),so
-2x^2 - 3x + 2 =
=-(2x^2 + 3x - 2)=
=-2(x^2+3/2x-2/2)=
=-2(x^2+2.(3/4)x+-(3/4)^2-1)=
=-2((x+3/4)^2-(1+9/16))=
=-2((x+3/4)^2-25/16)=
=-2((x+3/4)^2-(5/4)^2)=
=-2(x+3/4-5/4)(x+3/4+5/4)=
=-2(x-1/2)(x+2)=
=-(2x-1)(x+2)=
= (x+2)(1-2x).

Last edited by krassi_holmz (2005-12-17 07:05:43)

RickyOswaldIOW · 2005-12-17 07:06:57

so you're saying I should divide the whole equation through by the co-efficient of x^2 and then factorise normally, multiplying the co-efficient back in at the end?

krassi_holmz · 2005-12-17 07:25:30

Yes. It must work.

mathsyperson · 2005-12-17 07:26:55

(what is the word for taking divisor out of brackets) --> factorise

krazzi_holmz is right, but I think there's an easier way than his. His way involves fractions, and you should always avoid that if you can.

-2x² - 3x + 2 = 0

Factorise out -1 to make the x² term positive:
-(2x² + 3x - 2)

Then use the method that I described here:

-(2x² + 3x - 2)
-(2x² + 4x - x - 2)
-(2x(x+2) - (x+2))
-(x+2)(2x-1)

The just add the - coefficient to one of the terms. It would look nicer if we added it to the second one.

y = (x+2)(1-2x)

So, I get the same as krazzi_holmz, but without complicating it with fractions.

Of course, it would be easier to just use the quadratic equation.

krassi_holmz · 2005-12-17 07:35:58

Yes, your method is truely better in a ways.

RickyOswaldIOW · 2005-12-17 07:36:01

Merry Christmas Krassi and Mathsy

krassi_holmz · 2005-12-17 07:44:28

Merry Christmas Ricky and Mathsy!

I have a problem:
What picture is better than e^Pii?
This is e^Pii with christmas hat on e!
But when I try to change my picture, I get the message that the picture has been refreshed and then I go back. But then I see my previous picture. Has anybody had the same problem as mine?

Last edited by krassi_holmz (2005-12-17 07:46:30)

RickyOswaldIOW · 2005-12-17 07:55:42

The picture of yours that I see has the hat on! Are you trying to change it back to your normal one?

RickyOswaldIOW · 2005-12-17 08:06:37

I'm still having a little trouble with this.

y = ax^2 + bx + c
and our answer needs to be in the format
(dx + e)(fx + g) where:

e * g = a * c
and
e + g = b

in a question like y = -x^2 + 3 + 4 I would take the negative co-efficient from x^2 and reverse the other signs, factorise and then put the - back in front of d making d * f = a??

If I have a question like y = -2x^2 - 3x + 2
I'll change it to -(2x^2 + 3x - 2)
like this I can see that e = 4, g = -1, d = 2 and f = 1.
This is where I go wrong, into my answer I would now write
-[(2x + 4)(x - 1)]
Which of cours, is wrong. Do I need to divide e by d?

Last edited by rickyoswaldiow (2005-12-17 08:08:01)

mathsyperson · 2005-12-17 08:16:56

krassi_holmz wrote:

Merry Christmas Ricky and Mathsy!
I have a problem:
What picture is better than e^Pii?
This is e^Pii with christmas hat on e!
But when I try to change my picture, I get the message that the picture has been refreshed and then I go back. But then I see my previous picture. Has anybody had the same problem as mine?

Merry Christmas Ricky and krazzi!

I had that problem a while ago, back in my days of frequent avatar changing. I think it's because the forum has something that only lets you change avatars once every [insert amount of time here]. In my experience, the forum saves the picture you uploaded but doesn't actually put it as your avatar until the time limit is up, so stop the forum from slowing down.

To Ricky, you want the format (dx + e)(fx + g), where:

e * g = c
d * g + e * f = b.

krassi_holmz · 2005-12-17 08:20:53

I'll try something...

Last edited by krassi_holmz (2005-12-17 08:22:30)

krassi_holmz · 2005-12-17 08:41:32

Here is it:
Let
F[x]=ax^2+bx+c and F[x] is not factorisible.
Proof that in the rank
F[1], F[2], ...
exist infinitely many prime numbers.

krassi_holmz · 2005-12-17 08:48:36

Here's another question-when the
ax^2 + by^2 + cxy + dx + ey + f
is factorisible?
Here's an example:
3x^2+5y^2+2xy+7x+9y+10(not factorisible)

Is there a method such for this task?

kempos · 2006-01-07 21:43:47

In my opinion the best way is to write the coefficient of x^2 in front of the brackets and then factorise expression in the brackets using completing the square or other method). I prefer completting the square as it always work very well and you do not have to wast your time for finding e * f = c etc. Work well with any question of factorizing quadratics, especially when you have to deal with questions such as 6a^2 - 23ab + 20b^2

Regards

kempos79 · 2006-01-07 22:10:05

Another easy way is to use determinant(det) b^2 - 4ac

Remember that ax^2 + bx + c = a(x-x1)(x-x2) where x1 and x2 are x1 = (-b - sqroot det)/2a
and x2 = (-b + sqroot det)/2a

once you have your equation in the form a(x-x1)(x-x2) you can leave it like this or multiply and bracket by a or two brackets by the factors of a - this is in case you are solving a multiple choice test and you need to get the answer they expect.

Example:

5m^2 + 3m - 14

answers: a) (5m+7)(m-5) b) (5m+8)(m-5) c) (5m-7)(m+2)

find determinant:

Det = b^2 - 4ac = 9 - 4*5*(-14) = 9 + 280 = 289 Square root of det = 17

m1 = (-b - sqrootdet)/2a = (-3 - 17) / 10 = -20/10 = -2
m2 = (-b + sqrootdet)/2a = (-3 + 17) / 10 = 14/10 = 7/5

5m^2 + 3m - 14 = 5(m+2)(m-7/5) to get the answer they expect multiply the second bracket by 5 and you get (m+2)(5m-7)

Another example where you will have to multiply both brackets to get the answer:

6a^2 - 23ab + 20b^2 in this case our a = 6 our b = 23b and our c = 20b^2

answers: a)(6a-b)(a-20b) b)(5a-4b)(a-5b) c)3a-3b)(2a-5b) d)(3a-5b)(2a-4b)

Find determinant

det = b^2-4ac = 529b^2 - 480b^2 = 49b^2 square root = 7b

a1 = (23b-7b)/12 = 16b/12 = 4b/3
a2 = (23b+7b)/12 = 30b/12 = 5b/2

6(a-4b/3)(a-5b/2) multiply the first bracket by 3 and the second by 2 to remove fractions
(3a-4b)(2a-5b)

mathsyperson · 2006-01-07 22:37:52

That's the easiest way in general, because it's guaranteed to work, you're just plugging into a formula every time. But this question asked you to factorise, so that method wouldn't be allowed.

kempos · 2006-01-08 02:02:49

what do you mean not allowed!?

you get the final answer, any way, as long as it is mathematically correct, can be used.

Math Is Fun Forum

#1 2005-12-17 06:52:07

Factorising quadratics... again

#2 2005-12-17 07:02:55

Re: Factorising quadratics... again

#3 2005-12-17 07:06:57

Re: Factorising quadratics... again

#4 2005-12-17 07:25:30

Re: Factorising quadratics... again

#5 2005-12-17 07:26:55

Re: Factorising quadratics... again

#6 2005-12-17 07:35:58

Re: Factorising quadratics... again

#7 2005-12-17 07:36:01

Re: Factorising quadratics... again

#8 2005-12-17 07:44:28

Re: Factorising quadratics... again

#9 2005-12-17 07:55:42

Re: Factorising quadratics... again

#10 2005-12-17 08:06:37

Re: Factorising quadratics... again

#11 2005-12-17 08:16:56

Re: Factorising quadratics... again

#12 2005-12-17 08:20:53

Re: Factorising quadratics... again

#13 2005-12-17 08:41:32

Re: Factorising quadratics... again

#14 2005-12-17 08:48:36

Re: Factorising quadratics... again

#15 2006-01-07 21:43:47

Re: Factorising quadratics... again

#16 2006-01-07 22:10:05

Re: Factorising quadratics... again

#17 2006-01-07 22:37:52

Re: Factorising quadratics... again

#18 2006-01-08 02:02:49

Re: Factorising quadratics... again

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