How do I use FOIL method with a binomial cubed ? Our teacher said we need to use FOIL method.?

Question

How do I use FOIL method with a binomial cubed ? Our teacher said we need to use FOIL method.
EX. (mn - 5)(mn - 5)(mn - 5) And I have to use FOIL method with that.

^_^ OOH!!! THANK YOU SO MUCH ALL OF YOU ARE VERY HELPFUL TO ME !! tank you thank you.... I finally got it....thank you....

Answer 1

first multiply the first two factors and multiply the product by the third factor

To multiply something complicated like (4x + 6)(5x - 3)(15 - x), just do FOIL on two of the binomials and then distribute the answer onto the remaining binomial

Answer 2

Multiply the first two quantities together and then multiply that answer by the third quantity. The first pair gives you m squared n squared minus 10 mn + 25. This answer times the last quantity gives you m cubed n cubed minus 15 m squared n squared plus 75 mn minus 125. Foil stands for first - outer - inner - last.

Answer 3

Start with one pair of terms:
(mn - 5)(mn - 5)

First --> (mn)²
Outer --> -5mn
Inner --> -5mn
Last --> 5² = 25

This simplifies to:
(mn)² - 10mn + 25

But you still have one more pair:
[ (mn)² - 10mn + 25 ](mn - 5)

This takes a little more work... first multiply mn through:
[(mn)^3 - 10(mn)² + 25mn ] ...

Then multiply -5 through:
+[ -5(mn)² +50mn -125 ]

And put it all together:
(mn)^3 - 10(mn)² + 25mn -5(mn)² +50mn -125

Now group like terms:
(mn)^3 - 15(mn)² + 75mn - 125

There's your expanded form of (mn - 5)^3

Note: in general if you have:
(a + b)^3

It expands to:
a^3 + 3a²b + 3ab² + b^3

(1, 3, 3, 1 is row 3 of Pascal's triangle)

So in your case:
a = mn
b = -5

Substituting:
(mn)^3 + 3(mn)²(-5) + 3(mn)(-5)² + (-5)^3

This simplifies to:
(mn)^3 - 15(mn)² + 75(mn) - 125

This matches with the prior result...

Answer 4

Good Evening,

Foil the first two terms like usual, then distribute...

(mn-5)(mn-5)(mn-5) =
(mn-5)(m^2n^2 - 10mn + 25)
mn(m^2n^2 - 10mn + 25) -5(m^2n^2 -10mn + 25)
m^3n^3 - 10m^2n^2 + 25mn - 5m^2n^2 +50mn - 125
m^3n^3 - 15m^2n^2 + 75mn - 125

Hope this helps,
~i~

Answer 5

i in my opinion do no longer think of of FOIL whilst i'm multiplying binomials, yet greater often than not only as a results of fact that is much less stressful to do all polynomials the comparable way rather of thinking "FOIL" for binomials and "Distribution" for all different polynomials. even though, having been an Algebra I/Algebra II instructor for 3 years, it fairly is an invaluable coaching device and helps scholars discover ways to multiply polynomials in an prepared way (the main mandatory area). it is likewise useful for after a protracted smash whilst young ones are forgetful and puzzled. they are able to bear in mind FOIL and dive suitable back into binomial multiplication.

Answer 6

FOIL the first two quantities
m^2n^2 - 10mn + 25
now multiply that by the last quantity
m^3n^3 - 10m^2n^2 +25mn - 5m^2n^2 + 50mn - 125
simplify
m^3n^3 - 15m^2n^2 + 75mn - 125

Answer 7

foil the first two
((mn)^2-10mn+25)(mn-5)
then "foil" the rest by grouping if you want
(((mn)^2-10mn)+25)(mn-5)
mn((mn)^2-10mn)+-5((mn)^2-10mn)+25((mn)^2-10mn)-125
(mn)^3-10(mn)^2+25mn-5(mn)^2+50mn-125
then simplify
(mn)^3-15(mn)^2+75mn-125

Answer 8

I think the answer is m2n2-5mn-5mn+25. You have to break it up into pieces