Here are some questions that I can't solve...

factorization:

1.a^2x+2+2a^2+x

2.-(3+a)(5-b)+3c(b-5)

3.-a-b+3ab+3a^2

4.x^2-y^2+x-y

5.x^2-y^2+3x+3y

6.2c^2-8d^2+c-2d

That's all, hope you can help me :)

Thanks

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- Mar 17th 2006, 06:11 AMdgolverkFactorization
Here are some questions that I can't solve...

factorization:

1.a^2x+2+2a^2+x

2.-(3+a)(5-b)+3c(b-5)

3.-a-b+3ab+3a^2

4.x^2-y^2+x-y

5.x^2-y^2+3x+3y

6.2c^2-8d^2+c-2d

That's all, hope you can help me :)

Thanks - Mar 17th 2006, 07:24 AMearbothQuote:

Originally Posted by**dgolverk**

with your problems, you always have to seek equal factors at two summands:

to 1: $\displaystyle a^2x+2+2a^2+x=x(a^2+1)+2(a^2+1)=(a^2+1)(x+2)$

to 2: $\displaystyle -(3+a)(5-b)+3c(b-5)=(b-5)(3+a+3c)$

to 3: $\displaystyle -a-b+3ab+3a^2=-(a+b)+3a(a+b)=(a+b)(3a-1)$

to 4: $\displaystyle x^2-y^2+x-y=(x+y)(x-y)+(x-y)=$

$\displaystyle (x-y)(x+y+1)$

to 5: $\displaystyle x^2-y^2+3x+3y=(x+y)(x-y)+3(x+y)=$

$\displaystyle (x+y)(x-y+3)$

to 6: $\displaystyle 2c^2-8d^2+c-2d=2(c+2d)(c-2d)+(c+2d)=$

$\displaystyle (c-2d(2c+4d+1)$

Hope this was of some help.

Greetings

EB - Mar 17th 2006, 08:10 AMdgolverkThanks,
That was exactly what I was looking for :)

- Mar 17th 2006, 08:32 AMdgolverkHmm..
While doing my homework I found 3 more problems...

1.3x(a-b)+4y(b-a) I don't understand how can I make two (b-a) or (a-b)

2.24X^2(d-c)+3x^2(-d+c)

3.3(a+b-c)-d(a+b-c)

Those are the 3 last questions in my homework...

Thanks again for helping! - Mar 17th 2006, 11:28 AMearbothQuote:

Originally Posted by**dgolverk**

it's me again:

to 1.: $\displaystyle (b-a)=(-1)(a-b)$

$\displaystyle 3x(a-b)+4y(b-a) =(a-b)(3x-4y)$

to 2.: $\displaystyle 24x^2(d-c)+3x^2(-d+c)=(d-c)(24x^2-3x^2)=$

$\displaystyle 21x^2 \cdot (d-c)$

to 3.: $\displaystyle 3(a+b-c)-d(a+b-c)= (a+b-c)(3-d)$

Nice Weekend to you.

Greetings

EB - Mar 17th 2006, 10:11 PMdgolverkThanks
I made a mistake in the last question it supposed to be like that:

3(a+b+c)-d(a+b-c)

What is the answer for that one?

and those:

x^3-xy^2+y^2+xy

x^4-y^4+y-x - Mar 18th 2006, 09:44 AMdgolverk
guys it's really urgent, that's for tomorrow...

Thanks