geno3141

This problem has no easy factors.

You can either solve by completing the square or by the quadratic formula.

If you want to use the quadratic formula:  a = 1     b = 3     c = -2

x  =  [ -b ± √( b² - 4·a·c ) ] / ( 2·a )

   =  [ -3 ± √( 3² - 4·1·-2 ) ] / ( 2·1 )

   =  [ -3 ± √( 9 + 8 ) ] / ( 2 )

   =  [ -3 ± √17 ] / 2

There are many different types of problems that can be factored.

Could you give us an example of the type of problem that you want factored?

5/12  =  0.416666...

1/3  doubled is  1/3 + 1/3  =  2/3

This can also be calculated as  2 x 1/3  =  2/3

Because she's moving upward ...

Her work must be expressed upward because she is moving upward against the force of gravity which is pulling her downward.

To simplify these, multiply both the numerator and denominator of the fraction by the conjugate of the denominator.

The conjugate of   x + √y  is x - √y.

The conjugate of   x - √y  is x + √y.

The conjugate of  x + a√y  is x - a√y

As an example:

To simplify  (2 + √3) / (4 - √5)  multiply both the numerator and denominator by 4 + √5:

     (2 + √3) / (4 - √5) x (4 + √5) / (4 + √5)

Numerator:  (2 + √3)(4 + √5)  =  8 + 4√3 +2√5 + √15

Denominator:  (4 - √5)(4 + √5)  =  16 - 4√5 + 4√5 - √25  =  16 - 5  = 11

Answer:  ( 8 + 4√3 +2√5 + √15 ) / 11

Any questions about this example?

Percentage is Part ÷ Whole x 100:     £4 / 16£ x 100  =  0.25 x 100  =  25% 

3x² + 7x + 4 + ( ??? )  =  11x² - 15

Take these one term at a time:

The x² term:  3x² + ? = 11x²   --->  must be 8x²

The x term:  7x + ? = 0x (no x-term on the right side)   -->  must be -7x

The number:  4 + ? = -15   --->  must be -19

So the missing part must be:  8x² - 7x - 19

I'm going to assume that the choices are random choices, otherwise, Bill could always choose to win.

I'm also going to assume that the game is played "with replacement", meaning that both players choose from all four posibilities, not that one player gets two of the numbers, while the other player gets the two that are left. 

So, what are the possibilities for Darla?

1 + 2 = 3     1 + 3 = 4     1 + 4 = 5       2 + 3 = 5     2 + 4 = 6     3 + 4 = 7

There are 6 possibilities for Darla (we must count the two ways for Darla to get 5 as separate possibilities).

Now, what are the possibilities for Bill?

1 x 2 = 2     1 x 3 = 3     1 x 4 = 4     2 x 3 = 6     2 x 4 = 8     3 x 4 = 12

Each of the possibilities for Bill has a 1/6th chance of occuring.

In how many different ways can Darla beat Bill when

Bill gets 2?  6 ways  So Carla has a 6/6 chance to win.

Bill gets 3?  5 ways  So Carla has a 5/6 chance to win.

Bill gets 4? 4 ways  So Carla has a 4/6 chance to win.

Bill gets 6?  1 way  So Carla has a 1/6 chance to win.

Bill gets 8?  0 ways  So Carla has a 0/6 chance to win.

Bill ets 12?  0 ways  So Carla has a 0/6 chance to win.

So now we need to multiply the probability that Bill gets a certain result times the probability that Darla beats him and add all those results together:

(1/6)(6/6) + (1/6)(5/6) + (1/6)(4/6) + (1/6)(1/6) + (1/6)(0/6) + (1/6)(0/6)  =  16/36  =  4/9

The probability that Darla beats bill is  4/9.

If the choice is "without replacement" (one of them gets two of the numbers, the other gets the other two numbers) needs a different analysis. Re-post if that is the problem.