Factor out completly: 6x^2y − 3xy − 24xy^2 + 12y^2

The factor of 6x²y − 3xy − 24xy² + 12y² is 3y(2x - 1)(x - 4y)

Given:

6x²y − 3xy − 24xy² + 12y²

find the highest common factor of all the coefficient of each term

6 = 1, 2, 3, 6

6 = 1, 2, 3, 63 = 1, 3

6 = 1, 2, 3, 63 = 1, 324 = 1, 2, 3, 4, 6, 8, 12, 24

6 = 1, 2, 3, 63 = 1, 324 = 1, 2, 3, 4, 6, 8, 12, 2412 = 1, 2, 3, 4, 6, 8, 12

The highest common factor of all the coefficient of each term is 3

So,

6x²y − 3xy − 24xy² + 12y²

= 3y(2x² - x - 8xy + 4y)

(2x² - x - 8xy + 4y) = (2x - 1)(x - 4y)

Therefore,

6x²y − 3xy − 24xy² + 12y = 3y(2x - 1)(x - 4y)

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geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

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