Wednesday

April 16, 2014

April 16, 2014

Posted by **Sarah** on Sunday, January 21, 2007 at 9:20pm.

1.solve: 3x+5=x√7 (that's a radical sign)

2.factor: (2k+3)2-(2x+3)(y-2)-20(y-2)2 (the "2" after (2k+3) ans (y-2) is suppose to be a square sign.

thanks if you can help, or even if you can't

For number 2:

First, FOIL out (2k+3)^2. You should get:

4k^2 + 12k + 9.

Then, FOIL out (2x+3)(y-2). You should get:

2xy - 4x + 3y - 6. You wanted to subtract the first part by the second:

4k^2 + 12k + 9 -(2xy - 4x + 3y - 6).

Distribute the negative:

4k^2 + 12k + 9 - 2xy + 4x - 3y + 6.

Combine like-terms:

4k^2 + 12k - 2xy + 4x - 3y + 15.

Before multiplying 20 by (y-2)^2, FOIL out the (y-2)^2. You should get:

y^2 - 4y + 4.

Now multiply by 20:

20y^2 - 80y + 80.

Now, take the equation from 6 lines up and subtract by this part:

4k^2 + 12k - 2xy + 4x - 3y + 15 -(20y^2 - 80y + 80).

Distribute the negative:

4k^2 + 12k - 2xy + 4x - 3y + 15 - 20y^2 + 80y - 80.

Now, combine like-terms to finish:

4k^2 + 12k - 2xy + 4x + 77y - 20y^2 - 65.

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