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

RickyOswaldIOW
Member
Registered: 2005-11-18
Posts: 212

Factorising quadratics... again

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)?


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

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Factorising quadratics... again

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)


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

RickyOswaldIOW
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Registered: 2005-11-18
Posts: 212

Re: Factorising quadratics... again

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?


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

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Factorising quadratics... again

Yes. It must work.


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

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

Re: Factorising quadratics... again

(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.


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

krassi_holmz
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Registered: 2005-12-02
Posts: 1,908

Re: Factorising quadratics... again

Yes, your method is truely better in a ways.


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

RickyOswaldIOW
Member
Registered: 2005-11-18
Posts: 212

Re: Factorising quadratics... again

Merry  Christmas  Krassi  and  Mathsy smile


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

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Factorising quadratics... again

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! smile
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)


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

RickyOswaldIOW
Member
Registered: 2005-11-18
Posts: 212

Re: Factorising quadratics... again

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


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

RickyOswaldIOW
Member
Registered: 2005-11-18
Posts: 212

Re: Factorising quadratics... again

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)


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

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Factorising quadratics... again

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! smile
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.


Why did the vector cross the road?
It wanted to be normal.

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

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Factorising quadratics... again

I'll try something...

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


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

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Factorising quadratics... again

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.


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

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Factorising quadratics... again

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?


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

kempos
Guest

Re: Factorising quadratics... again

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

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

kempos79
Guest

Re: Factorising quadratics... again

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)

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

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Factorising quadratics... again

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.


Why did the vector cross the road?
It wanted to be normal.

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

kempos
Member
Registered: 2006-01-07
Posts: 77

Re: Factorising quadratics... again

what do you mean not allowed!?

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

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