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 (2005-12-18 06:05:43)

Yes. It must work.

Yes, your method is truely better in a ways.

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 (2005-12-18 06:46:30)

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 (2005-12-18 07:08:01)

I'll try something...

Last edited by krassi_holmz (2005-12-18 07:22:30)

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?

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)

what do you mean not allowed!?

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