How would i factor completly the following. 10 points for best answer!?

Question

1. 18x squared-3x

2. 2x(x-3) +4(x-3)

3.25 x squared -20x +4

4. 3x squared -27 y squared

Answer 1

18x² - 3x = 3x(6x - 1)

2x(x - 3) + 4(x - 3) = (2x + 4)(x - 3) = 2(x + 2)(x - 3)

25x² - 20x + 4 = (5x - 2)(5x - 2) = (5x - 2)²

3x² - 27y² = 3(x² - 9y²) = 3(x + 3y)(x - 3y)

And I posted my answer before Pradyumna N posted his.

Answer 2

1 (3x)(6x-1)
2 (2x + 4)(x - 3) = 2(x + 2)(x - 3)

For these first two, you're looking for a factor that is common and can be pulled out. For the first one, you can pull out a 3x. For the second, you see that both 2x and 4 are both multiplying the same thing so you can add them together.

3 (5x-2)(5x-2) = (5x + 2) squared
4 3(x squared - 9 y squared) = 3(x - 3y)(x + 3y)

For number 3, this is just factoring. You want to find multiples of 25 and multiples of 4 so that the sum of their products = -20.
For number 4, once you pull out the 3, you have a difference of two squares(since x squared = (x) squared and 9y squared = (3y) squared). The difference of two squares and be written (a - b)(a + b). In this case a = x and b = 3y

Answer 3

*Have you give these a try?

1. 18x^2 - 3x
First: find the least common factor > find a number divisible by the coefficients & find a common variable which is, 3x ...

Sec: factor out the 3 and multiply it by the remaining terms to get the original expression...

3x(6x - 1)

2. First: its already been factored for you
Sec: since you have two sets of "x - 3" > you just write it once & combine it with the outer terms...

(x - 3)(2x+4)

3. 25x^2 - 20x + 4
First: multiply the 1st and 3rd coefficient to get 100. find two numbers that give you 100 when multiplied and - 20 (2nd-middle coefficient) when added/subtracted. The numbers are -10 & - 10

Sec: rewrite the expression with the new middle coefficients...

25x^2 - 10x - 10x + 4

Third: when you have 4 terms - group "like" terms & factor...

(25x^2 - 10x) - (10x + 4)
5x(5x - 2) - 2(5x - 2)

*Combine one of the innter terms with the outer term...

(5x - 2)(5x - 2)
(5x - 2)^2

4. give this a try-its similar to #1

Answer 4

I just felt that providing just the factors was too easy so , i just solved them!


1.
18x^2-3x = 0
=> 18x^2 = 3x
We remove one x from the equation by x=0.
=> 18x = 3
=> x = 1/6
-------------------------------------

2.
2x(x-3) +4(x-3) = 0
=> (2x + 4) ( x-3) = 0
=> 2x = -4 or x- 3 = 0
=> x = -1/2 or x = 3
-------------------------------------

3.
25 x^2 -20x +4 = 0
=> (5x)^2 + 2(-2)(5x) + 2^2 = 0
=> (5x-2)^2 = 0 [ as (a-b)^2 = a^2 -2ab + b^2]
=>5x = 2
=> x = 2/5
-------------------------------------

4.
3x^2 -27 y^2 = 0
=> 3(x^2 - 9y^2) = 0
=> x^2 - (3y)^2 = 0
=> (x+3y)(x-3y) = 0 [ as (a+b)(a-b) = a^2 - b^2]
=> x = -3y or x = 3y


The last one, is in terms of y. Cant help it.

Thats pretty much it.

Peace out.

Answer 5

He was born in room 1995, I added this answer to explain the following: 2010 is the current year, so he was born in the year 2005 leap year is not possible, since 2010 is not a leap year, and neither would 1995 have been! for those of you who answered leap year: a leap year occurs every 4 years, except every 100 years it doesn't, and except to that every 400 years it does (eg: 1996 was a leap year, 1900 not, but 2000 was)

Answer 6

1) 18x^2-3x

3x(6x-1)

2) 2x(x-3) + 4(x-3)

(x-3)(2x+4)

3) 25x^2-20x+4

(5x-2)(5x-2)

4) 3x^2-27y^2

3(x^2-9y^2)
3(x+3y)(x-3y)

Answer 7

1. 3*x*(6x-1)
2. (x-3)(x+2)*2
3) 25*(x- 2/5)`^2
4. (3x -27y)*(3x+27y)

Answer 8

You first..................

Answer 9

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