# binomial

• May 7th 2008, 04:02 PM
lax600
binomial
I need to factor this completely:

\$\displaystyle 16x^2 - (y + 3)^2\$

\$\displaystyle
16x^2 - (y + 3)^2
= [16x - (y+3)]^2
= [4x - (y+3)]^2
= (2x - y - 3)(2x + y + 3)\$

but the answer in the packet says:

\$\displaystyle (4x + y + 3)(4x - y - 3)\$

My answer differs from the one in the packet at the last step. Is it because I cannot turn 16x into 4x?
• May 7th 2008, 04:10 PM
icemanfan
Quote:

Originally Posted by lax600
I need to factor this completely:

\$\displaystyle 16x^2 - (y + 3)^2\$

\$\displaystyle
16x^2 - (y + 3)^2
= [16x - (y+3)]^2
= [4x - (y+3)]^2
= (2x - y - 3)(2x + y + 3)\$

but the answer in the packet says:

\$\displaystyle (4x + y + 3)(4x - y - 3)\$

My answer differs from the one in the packet at the last step. Is it because I cannot turn 16x into 4x?

You're just not correctly converting the difference of squares. We have:

\$\displaystyle 16x^2 = a^2\$

\$\displaystyle (y+3)^2 = b^2\$

Hence, \$\displaystyle a = 4x\$ and \$\displaystyle b = y + 3\$.

So, using the formula \$\displaystyle a^2 - b^2 = (a + b)(a - b)\$ we have:

\$\displaystyle (4x + y + 3)(4x - y - 3)\$.
• May 7th 2008, 04:17 PM
lax600
Oh! Alright thanks ^^