# College Factoring Simple yes or no will help alot!

• Dec 8th 2009, 11:55 PM
cpuboye11
College Factoring Simple yes or no will help alot!
Hey guys and Gals.

I have two factoring problems here and they don't seem to be factor-able. I might be wrong. A simple yes or no will do just fine. ^_^

1. \$\displaystyle 4x^2+20x+25-y^2\$

2. \$\displaystyle 36x^2+12xy+y^2-16w^2\$

• Dec 9th 2009, 12:14 AM
11rdc11
Quote:

Originally Posted by cpuboye11
Hey guys and Gals.

I have two factoring problems here and they don't seem to be factor-able. I might be wrong. A simple yes or no will do just fine. ^_^

1. \$\displaystyle 4x^2+20x+25-y^2\$

2. \$\displaystyle 36x^2+12xy+y^2-16w^2\$

yes both are factorable

1. \$\displaystyle (2x+y+5)(2x-y+5)\$

2 \$\displaystyle (6x+y-4w)(6x+y+4w)\$
• Dec 9th 2009, 12:21 AM
cpuboye11
Quote:

Originally Posted by 11rdc11
yes both are factorable

1. \$\displaystyle (2x+y+5)(2x-y+5)\$

2 \$\displaystyle (6x+y-4w)(6x+y+4w)\$

hey thanks a lot. I didn't need you to do that !! (Rofl)
• Dec 9th 2009, 02:03 AM
jgv115
Quote:

Originally Posted by 11rdc11
yes both are factorable

1. \$\displaystyle (2x+y+5)(2x-y+5)\$

2 \$\displaystyle (6x+y-4w)(6x+y+4w)\$

may I ask how you did the second one?
• Dec 9th 2009, 03:48 AM
HallsofIvy
Quote:

Originally Posted by cpuboye11
Hey guys and Gals.

I have two factoring problems here and they don't seem to be factor-able. I might be wrong. A simple yes or no will do just fine. ^_^

1. \$\displaystyle 4x^2+20x+25-y^2\$

\$\displaystyle 4x^2+ 20x+ 25\$ is a "perfect square"- it is \$\displaystyle (2x+ 5)^2\$.
So \$\displaystyle 4x^2+ 20x+ 25- y^2\$ is a "difference of squares"- it is \$\displaystyle (2x+5)^2- y^2\$ and so can be factored as a "sum and difference": \$\displaystyle (2x+ 5+ y)(2x+ 5- y)\$

Quote:

2. \$\displaystyle 36x^2+12xy+y^2-16w^2\$
Same idea: \$\displaystyle (6x+ y)^2- 16w^2= (6x+y- 4w)(6x+y+ 4w)\$

Quote: