Question

How do you multiply (4x-√3)(4x+√3)?

Answer 1

`16x^2 -3`

Explanation:

`(4x-sqrt3)(4x+sqrt3)`
Use FOIL - firsts, outers, inners, lasts
first: `4x*4x = 16x^2`

outer: `4x*sqrt3 = 4sqrt3x`

inner: `-sqrt3*4x = -4sqrt3x`

lasts : `-sqrt3*sqrt3 = -3`

Then add all the results
`16x^2+4sqrt3 -4sqrt3 -3`

The `4sqrt3 and -4sqrt3` cancel each other out so you are left with `16x^2 -3`

Answer 2

`color(green)(16x^2-3`

Explanation:

`(4x-sqrt3)(4x+sqrt3)`
`color(white)(aaaaaaaaaaaaa)` `4x-sqrt3`
`color(white)(aaaaaaaaaaa)` `xx underline(4x+sqrt3)`
`color(white)(aaaaaaaaaaaaa)` `16x^2-4xsqrt3`
`color(white)(aaaaaaaaaaaaaaaaaa)` `ul(+4xsqrt3-3)`
`color(white)(aaaaaaaaaaaaa)` `16x^2` `color(white)(aaaaaaaa)-3`

`color(white)(aaaaaaaaaaaaa)` `color(green)(16x^2-3`

Answer 3

One can use the F.O.I.L method. This method is well known but it has the disadvantage of working only with the product of two binomials.

Given: `(4x-sqrt3)(4x+sqrt3) = ?`

Step 1. Multiply the first terms:

`(4x)(4x) = 16x^2`

Step 2. Multiply the outside terms:

`(4x)(sqrt3) = 4sqrt3x`

Step 3. Multiply the inside terms:

`(4x)(-sqrt3) = -4sqrt3x`

Step 4. Multiply the last terms:

`-sqrt3sqrt3= -3`

Step 5. Add all of the terms (please observe that this is a special case where the product of the inside terms and outside terms sum to zero):

`16x^2 color(red)(cancel(+ 4sqrt3x-4sqrt3x)) - 3 = 16x^2-3`

I prefer the distributive method because it has the advantage of working with polynomials of any size.

Given: `(4x-sqrt3)(4x+sqrt3) = ?`

Step 1. Distribute the second factor over the terms of the first factor:

`4x(4x+sqrt3)-sqrt3(4x+sqrt3)`

Please observe that the above step can be done with any number of terms in either factor and which factor you choose to distribute does not matter.

Step 2. Use the distributive property `a(b+c) = ab + ac` to eliminate the parenthesis:

`16x^2+4sqrt3x-4sqrt3x-sqrt3sqrt3`

Please observe that we have the same terms that we obtained using the F.O.I.L. method.

Step 3. Combine like terms:

`16x^2-3`