Here are a few examples, including some degenerate cases:

*x*^{2}(*x*^{3}+ 5) ≡*x*^{2}·*x*^{3}+*x*^{2}· 5 ≡*x*^{5}+ 5*x*^{2};*x*^{2}(*x*+ 5) ≡*x*^{2}·*x*+*x*^{2}· 5 ≡*x*^{3}+ 5*x*^{2};*x*(*x*^{3}+ 5) ≡*x*·*x*^{3}+*x*· 5 ≡*x*^{4}+ 5*x*;*x*(*x*+ 5) ≡*x*·*x*+*x*· 5 ≡*x*^{2}+ 5*x*;

Here is a slightly more complicated family of examples, where I have to multiply some constants as well:

- 3
*x*^{2}(2*x*^{3}+ 5) ≡ 3*x*^{2}· 2*x*^{3}+ 3*x*^{2}· 5 ≡ 6*x*^{5}+ 15*x*^{2}; - 3
*x*^{2}(2*x*^{3}− 5) ≡ 3*x*^{2}· 2*x*^{3}+ 3*x*^{2}· (−5) ≡ 6*x*^{5}− 15*x*^{2}; - −3
*x*^{2}(2*x*^{3}+ 5) ≡ −3*x*^{2}· 2*x*^{3}− 3*x*^{2}· 5 ≡ −6*x*^{5}− 15*x*^{2}; - −3
*x*^{2}(2*x*^{3}− 5) ≡ −3*x*^{2}· 2*x*^{3}− 3*x*^{2}· (−5) ≡ −6*x*^{5}+ 15*x*^{2};

First, treat the factor *t* + 3 as a single expression,
ignoring that it can be broken down as a sum.
Then you can use the distributive law on the factor *t* + 2
to get *t*(*t* + 3) + 2(*t* + 3).
That done, now you can use that *t* + 3 is a sum
and apply the distributive law in the other direction
(twice, in fact, once for each term)
to get *t*^{2} + 3*t* + 2*t* + 6,
which simplifies (by combining terms)
to *t*^{2} + 5*t* + 6.

Let's see that again, slowly:

- (
*t*+ 2)(*t*+ 3) — original expression *t*(*t*+ 3) + 2(*t*+ 3) — use the distributive law (on the right) to distribute*t*+ 3 to both*t*and 2;- (
*t*·*t*+*t*· 3) + (2 ·*t*+ 2 · 3) — use the distributive law (on the left) to distribute*t*to both*t*and 3, and again to distribute 2 to both*t*and 3; - (
*t*^{2}+ 3*t*) + (2*t*+ 6) — simplify each monomial term; *t*^{2}+ (3*t*+ 2*t*) + 6 — rearrange order of addition;*t*^{2}+ (3 + 2)*t*+ 6 — use the distributive law backwards to combine like terms;*t*^{2}+ 5*t*+ 6 — since 3 + 2 is 5.

Probably the most important step to understand in the list above
is the long one:
(*t* · *t* + *t* · 3) +
(2 · *t* + 2 · 3).
The parentheses here aren't important
—they just tell you to add the terms in a specific order,
which doesn't make any difference—
but they're helpful to keep track of where each term comes from.
The first pair comes from multiplying *t* by both *t* and 3,
while the second pair comes from multiplying 2 by both *t* and 3.
In other words, to simplify (*t* + 2)(*t* + 3),
both *t* and 2 are multiplied by both *t* and 3.

This is just one example of the general rule:
*Multiply each term by each term.*
That is, when multiplying one polynomial (a sum of monomial terms)
by another polynomial (another sum of monomial terms),
multiply each term of the first polynomial
by each term of the second polynomial.
These multiplications, of course,
use the monomial techniques from the last lecture;
afterwards, you combine like terms
using the techniques from the lecture before.

Here are some more examples. In each case, the first step involves multiplying each term by each term, the second step uses monomial techniques to simplify each product term, the next step rearranges the order of addition of the terms, and the last step combines like terms to get the final answer. Try to get good enough at this that you only need to write down every other formula (the original expression, the middle expression, and the final result).

- (
*t*+ 2) (*t*+ 3) ≡ (*t*·*t*+*t*· 3) + (2 ·*t*+ 2 · 3) ≡ (*t*^{2}+ 3*t*) + (2*t*+ 6) ≡*t*^{2}+ (3*t*+ 2*t*) + 6 ≡*t*^{2}+ 5*t*+ 6; - (4
*t*+ 2) (5*t*+ 3) ≡ (4*t*· 5*t*+ 4*t*· 3) + (2 · 5*t*+ 2 · 3) ≡ (20*t*^{2}+ 12*t*) + (10*t*+ 6) ≡ 20*t*^{2}+ (12*t*+ 10*t*) + 6 ≡ 20*t*^{2}+ 22*t*+ 6; - (4
*t*− 2) (−5*t*+ 3) ≡ (4*t*· −5*t*+ 4*t*· 3) + (−2 · −5*t*+ −2 · 3) ≡ (−20*t*^{2}+ 12*t*) + (10*t*− 6) ≡ −20*t*^{2}+ (12*t*+ 10*t*) − 6 ≡ −20*t*^{2}+ 22*t*− 6; - (
*x*^{2}+ 4*x*− 2) (3 − 5*x*) ≡ (*x*^{2}· 3 +*x*^{2}· −5*x*) + (4*x*· 3 + 4*x*· −5*x*) + (−2 · 3 + −2 · −5*x*) ≡ (3*x*^{2}− 5*x*^{3}) + (12*x*− 20*x*^{2}) + (−6 + 10*x*) ≡ −5*x*^{3}+ (3*x*^{2}− 20*x*^{2}) + (12*x*+ 10*x*) − 6 ≡ −5*x*^{3}− 17*x*^{2}+ 22*x*− 6; - (
*x*+*x**y*) (*y*+ 3) ≡ (*x*·*y*+*x*· 3) + (*x**y*·*y*+*x**y*· 3) ≡ (*x**y*+ 3*x*) + (*x**y*^{2}+ 3*x**y*) ≡*x**y*^{2}+ (*x**y*+ 3*x**y*) + 3*x*≡*x**y*^{2}+ 4*x**y*+ 3*x*.

How then
should you simplify an expression like (*x* + 2)^{2}?
The *wrong* answer would be *x*^{2} + 4;
instead you should turn the exponentiation into multiplication
and use the techniques above:

- (
*x*+ 2)^{2}— original expression - (
*x*+ 2)(*x*+ 2) — rewritten using multplication; - (
*x*^{2}+ 2*x*) + (2*x*+ 4) — multiply each term by each term; *x*^{2}+ 4*x*+ 16 — combine like terms.

- (
*x*+ 2)^{3}— original expression - (
*x*+ 2)(*x*+ 2)(*x*+ 2) — rewritten using multplication; - (
*x*^{2}+ 4*x*+ 16)(*x*+ 2) — using previous calculation of (*x*+ 2)(*x*+ 2); - (
*x*^{3}+ 4*x*^{2}+ 16*x*) + (2*x*^{2}+ 8*x*+ 32) — multiply each term by each term; *x*^{3}+ 6*x*^{2}+ 24*x*+ 32) — combine like terms.

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