what is the expression equivalent for

[tara78]

(6n-2)(2n+1)

 

[Mrspi]

(6n-2)(2n+1)

You could do the multiplication....

When you multiply two binomials of the form

(a + b)(c + d)

you are going to use the distributive property three times. First, multiply each term inside the second set of parentheses by the (a + b) which is outside:

(a + b)*c + (a + b)*d

Now, you have two more places to use the distributive property. Multiply each term inside the first set of parentheses by the "c" which is outside, and each term inside the second set of parentheses by the "d" which is outside:

a*c + b*c + a*d + b*d

And the multiplication is complete, except for any simplification that might be possible. Do you see that each term inside the first set of parentheses in the original expression gets multiplied by each term inside the second set of parentheses. This, basically, is how you multiply two polynomials together.

Let's see if we can apply this to your problem:

(6n - 2)*(2n + 1)

Multiply each term in the second set of parentheses by 6n, and by -2:

6n*2n + 6n*1 + (-2)*2n + (-2)*1

You can do some simplifying....I'll leave that part up to you.

 

[Chaim]

(6n-2)(2n+1)

What my teacher has always taught me was FOIL
FOIL stands for First, Outer, Inner, Last

So basically this is how FOIL works
First: Take the first numbers in the paranthesis then multiply them together: 6n * 2n
Outer: Take the first number in the first paranthesis, and the last number in the second paranthesis: 6n * 1
Inner: Take the last number in the first paranthesis, and the first number in the second paranthesis: -2 * 2n
Last: Take the last numbers in the paranthesis then multiply them together: -2 * 1

After you get the results for First, Outer, Inner, and Last, you would be able to simplify it.