How would you factor...?

Question

9x^2- 30xy + 25y^2


&&

4m^2 + 40m + 100


thanks so much. im soo confused. soo i would greatly appreciate anyones help. =]

Answer 1

9x^2 is a perfect square, as is 25y^2. So you will factor the first one thusly:

(3x - 5y) * (3x - 5y) = (3x - 5y)^2

Note that you have 2 * (3x)(-5y) = -30xy in the middle term.

For the second one, again note that 4m^2 and 100 are perfect squares, so take their square root.

(2m + 10) * (2m + 10) = (2m + 10)^2

If set equal to a number, these two equations would describe circles.

edit: It looks like David reduced and I didn't. *shrug*

Answer 2

9x^2- 30xy + 25y^2
= ( 3 x - 5 y )^2

4m^2 + 40m + 100
= 4 ( m^2 + 10 m + 25 )
= 4( m + 5 )^2

Answer 3

Notice in both cases that you have perfect squares in the first and last terms, so test out....

(3x - 5y)^2 (I get the - from the - in the middle term)...

Then you multiply the first and last terms of the factor by 2 to see if you get the middle term of your original answer....

2(3x)(-5y) = -30xy ... Bingo! That's your factor.

To show you again, I'll do the other one...

4m^2 + 40m + 100

= (2m + 10)^2... (Check... 2(2m)(10) = 40m... right)

Note that you'd probably want to factor that further....
2(m+5)^2

EDIT: Whoops, I forgot a 2 in the answer above (because that answer factored out the 4 first. You should use that one.)

Answer 4

You are going to find the greates common factor among the group of numbers. Also you will use a letter if it is in the equation more than once. Use the quadratic formula:
(3x+5) (3x-5)
You cannot factor it because 25 does not have a common greatest common factor.

4(m^2+10m+25) = m^2+10m+25
Put it in order from greates number to least number but it is already that way.

Answer 5

9x^2- 30xy + 25y^2
(3x-5y)(3x-5y)

4m^2 + 40m + 100
(2m + 10)(2m+10) = 4(m+5)(m+5)

Answer 6

question #1.
The question is, 9x^2-30xy+25y^2
It's look like as a^2-2ab-b^2.
The question can be rewritten as,
(3x)^2-30xy+(5y)^2
here, a=3x, b= 5y.
so, (3x)^2 - 2*(3x)*(5y) + (5y)^2
we know that (a-b)^2 = a^2-2ab+b^2
so, we get,
9x^2-30xy+25y^2 = (3x-5y)^2, which is a required factor.

question #2:
4m^2+40m+100
divide each term by 2 and get,
2m^2+20m+50
the factors of 100 is 10*10, so that we will get additon as 20.
then,
2m^2+20m+50 = 2m^2+10m+10m+50
= 2m(m+5) + 10(m+5)
= (2m+10)(m+5)
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Answer 7

Okay. 1st one is in the pattern a^2 + 2ab + b^2 which equals (a+b)^2. a and b in this case are 3x and 5y, respectively, so (3x+5y)^2

2nd one: factor out a 4: 4(m^2 + 10m + 25). Also in a^2 + 2ab + b^2 format: 4(m+5)^2

Answer 8

The second one will have complex roots. You'll need the quadratic formula to find them. You can factor out a 4 first to make the calculation easier.

The first one is tricky. All we can say at first is that it will probably have the form (ax - by)(cx - dy). Since 9 and 25 are perfect squares, I tried using their square roots and it worked: (3x-5y)(3x-5y).

Answer 9

The quadratic formula can be used for both of these; if the roots are r1 and r2, then the factorization is (x-r1)(x-r2). The second one has a factor of 4 that can be taken out first.

Answer 10

The first one is...

(3x+5y)(3x+5y)

Then second is...

4(m+5)(m+5)