Is Distributing or Using FOIL Better for Expanding (5t-1)(25t^2+5t+1)?

[thomas576]

(5t-1)(25t^2+5t+1)
i just distributed 5t-1 thought the 2nd part and got?

this right way to do it?



125t^3-1 + 25t^2-1 + 5t

 

[arildno]

No; it is totally wrong.
Do you understand what you are supposed to be doing?

If a,b,c are numbers, then we have the rule:
a*(b+c)=a*b+a*c
Does this rule seem weird to you?
Or can you think of examples in the real world where you use multiplication and where that rule ought naturally to hold?

 

[Data]

You can't just do that. Let's examine it a more efficient way than you might have learned. We want to expand

(5t-1)(25t^2+5t+1).

Examine the coefficients of each power of t when you multiply it out: to get t^3, we have to multiply t^2 with t. The only way we can do that is (5t)(25t^2), so the coefficient of t^3 is 5(25)=125. Now consider the coefficient on t^2. To get t^2, we either have to multiply t^2 with a constant, or t by itself. The only ways we can do that here are 25t^2(-1) and (5t)(5t), so the coefficient on t^2 is 25(-1)+(5)(5)=0.

Using a similar process for the coefficients of t and t^0, you find that the expansion is just

125t^3 - 1.

 

[BobG]

(5t-1)(25t^2+5t+1)
i just distributed 5t-1 thought the 2nd part and got?

this right way to do it?



125t^3-1 + 25t^2-1 + 5t

No.

If you're distributing it, then you really have:

25t^2(5t-1) + 5t (5t-1) + 1 (5t - 1)
(25t^2*5t - 25t^2*1) + (5t*5t - 5t*1) + (1*5t - 1*1)
125t^3 - 25t^2 + 25t^2 - 5t + 5t - 1
125t^3 -1

FOIL only works as a memory aid for multiplying polynomials with two terms each:

(a + b)(c + d) = ac + ad + bc + bd

You can kind of do the same thing with what you have. You're multiplying the first term of the first polynomial (a) by each of the terms in the second polynomial, then the second term by each of the second polynomial terms.

5t*25t^2 + 5t*5t + 5t*1 - 1*25t^2 - 1*5t - 1*1

 

[OlderDan]

(5t-1)(25t^2+5t+1)
i just distributed 5t-1 thought the 2nd part and got?

this right way to do it?

125t^3-1 + 25t^2-1 + 5t

There are several ways to do this correctly, as others have shown. I just want to make it clear that FOIL could be used here if you applied it correctly, but I recommend you start thinking in terms of distribution the way BobG did it. As he said, to use FOIL you have to have two groups of two, but each one part of those twos can contain multiple terms.

(5t-1)(25t^2 + 5t + 1)

(5t-1)[(25t^2 + 5t) + 1]

------F---------O--------I---------L
5t(25t^2 + 5t) + 5t*1 - 1*(25t^2 + 5t) -1

125t^3 + 25t^2 + 5t - 25t^2 - 5t -1

125t^3 -1

There is no advantage to doing it this way, but it illustrates that FOIL is not "broken" when you have multiple terms, it just gets more complicated in its application.

FOIL is just an extension of the distributive law to cases where the single term in a simple distribution is replaced by a split term.

A(c + d) = Ac + Ad

If A = (a + b) this is

(a + b)(c + d) = (a + b)c + (a + b)d

(a + b)(c + d) = ac + bc + ad + bd

This result is in the order FIOL, but who can remember a silly thing like FIOL, so somebody decided to reverse the order of the inner two terms and make it FOIL. You will always be able to keep things straight if you develop an orderly distribution approach of multiplying the potentially complicated first factor (i.e., first thing in parentheses) times each term in the potentially complicated second factor, and then distribute a second time across the first factor for each term created. This is exactly what BobG did. It is an excellent model to follow. When you become more proficient, you might want to adopt Data's method because that keeps like-powered terms arranged together which makes simplifying easier.