# Math Help - How can I multiply these radicals?

1. ## How can I multiply these radicals?

At first I was thinking I could just use FOIL, but it's confusing because the first set of parenthesis has THREE numbers, while the second set only has TWO. So I'm not sure if that would work or not?

Here's the problem: (4+26)(3-√6)

2. To begin with, stop relying on FOIL; it only works for binomials, so if you don't know how to multiply general polynomials, then you get into situations like this.

Basically, to multiply two polynomials:

$\left(a_1+a_2+\dots+a_m\right)\left(b_1+b_2+\dots+ b_n\right)$

You first multiply $a_1$ by each $b$ term, then do the same with $a_2$, etc. and then add them all up. So for instance, $(x+2y+z)(-x+y-z)$ would be $[x\cdot (-x) + x\cdot y + x\cdot(-z)] +$ $[2y\cdot(-x) + 2y\cdot y + 2y \cdot (-z)] +$ $[z\cdot(-x) + z\cdot y + z\cdot(-z)] = -x^2 + 2y^2 - z^2 - xy - yz - 2xz$. If you're the kind of guy (like me) who needs to have an explanation of a formula in order to really understand and be able to use it, that follows from distributivity. (the fact that $a(b+c) = ab + ac$)

Back to your question: both parentheticals do have 2 numbers; the term $2\sqrt6$ means 2 times the square root of 6, not plus. So in this case you technically could use FOIL (but make sure you know how to do polynomials with more terms as well!), so the answer is

$(4+2\sqrt6)(3-\sqrt6) = 4\cdot3 - 4\sqrt6 + 2\sqrt6 \cdot 3 - 2\sqrt6\cdot\sqrt6$

which you would then simplify.

3. Originally Posted by error792
To begin with, stop relying on FOIL; it only works for binomials, so if you don't know how to multiply general polynomials, then you get into situations like this.

Basically, to multiply two polynomials:

$\left(a_1+a_2+\dots+a_m\right)\left(b_1+b_2+\dots+ b_n\right)$

You first multiply $a_1$ by each $b$ term, then do the same with $a_2$, etc. and then add them all up. So for instance, $(x+2y+z)(-x+y-z)$ would be $[x\cdot (-x) + x\cdot y + x\cdot(-z)] +$ $[2y\cdot(-x) + 2y\cdot y + 2y \cdot (-z)] +$ $[z\cdot(-x) + z\cdot y + z\cdot(-z)] = -x^2 + 2y^2 - z^2 - xy - yz - 2xz$. If you're the kind of guy (like me) who needs to have an explanation of a formula in order to really understand and be able to use it, that follows from distributivity. (the fact that $a(b+c) = ab + ac$)

Back to your question: both parentheticals do have 2 numbers; the term $2\sqrt6$ means 2 times the square root of 6, not plus. So in this case you technically could use FOIL (but make sure you know how to do polynomials with more terms as well!), so the answer is

$(4+2\sqrt6)(3-\sqrt6) = 4\cdot3 - 4\sqrt6 + 2\sqrt6 \cdot 3 - 2\sqrt6\cdot\sqrt6$

which you would then simplify.

I'm not sure if I simplified this right or not.
Would it be 2√6?

4. You've got it!

5. Originally Posted by deathtolife04
Here's the problem: (4+26)(3-√6)
Don't you like square roots?

Let's define a change of variables according to $u=\sqrt6$

$\implies(4+2\sqrt6)(3-\sqrt6)=(4+2u)(3-u)=12+2u-2u^2$, so we turn back $u$ into $\sqrt6$ and this yields

$12+2u-2u^2=12+2\sqrt6-2\cdot6=2\sqrt6$