Math Help Forum: FACTORING TRINOMIALs

25a2 +9b2 + 30ab

2x2 - 16x + 32

5c2 +30c + 45

128x2 +96xy + 18y2

450a2 +242c2 + 660ac

y2 - xy - 56x2


Can you show me how to do these promblems. I realy what to now how to learn to do theses problems.

Originally Posted by frow4

25a2 +9b2 + 30ab

2x2 - 16x + 32

5c2 +30c + 45

128x2 +96xy + 18y2

450a2 +242c2 + 660ac

y2 - xy - 56x2


Can you show me how to do these promblems. I realy what to now how to learn to do theses problems.

We did a bunch for you this afternoon. Can you please show us some of your work now? What have you tried?

i can do simple ones like

x2n + 10xn + 16

(x^n + 8)(X^n+2)

y2x - yx - 20

(y^x-5) (y^x-4)

what i am having troubles on

128x2 +96xy + 18y2 ; 450a2 +242c2 + 660ac

because of the x^2 and Y^2 and xy I have no idea what to do with those powers

Originally Posted by frow4

y2 - xy - 56x2

I'll do this one for you.

The coefficient of is 1.

So we can say this:



Now take a look at 56. We want factors of 56 so that the difference of the two factors gives us -1

7 and 8 clearly satisfies this condition.

So the final answer is:

Originally Posted by frow4

128x2 +96xy + 18y2

So You times 128 by 18 to get 2304. factors of 2304 That equals 96 is 48.

So it (128x+48y)^2

then you reduces it to
(8x+3y)^2

Is that right

Originally Posted by frow4

So You times 128 by 18 to get 2304. factors of 2304

Why are you doing this?

128x2 +96xy + 18y2

First factor out a 2.

Originally Posted by Soroban

.

Example:

Write the two pairs of parentheses and "split the x's": .

Use the first coefficient twice: .


Multiply the first coefficient by the last coefficient: .

We will factor into two parts.
Note the sign of the last term.
. . If "+", think sum.
. . If "-", think difference.

i look this up and it tells me to do it