Dimensión 7: Dar un Rótulo apropiado al texto
II. Marco metodológico
3.2 Resultados inferenciales
3.2.6 Inteligencia emocional de estado de ánimo
One important consequence of the distributive law is the FOIL rule of algebra,which says that for any numbers or variables a, b, c, d,
(a+b)(c+d) =ac+ad+bc+bd
FOIL gets its name from First-Outer-Inner-Last. Here, ac is the product of the first terms in (a+b)(c+d). Then ad is the product of the outer terms. Then bc is the product of the inner terms. Then bd is the product of the last terms.
To illustrate, let’s multiply two numbers using FOIL:
23×45= (20+3)(40+5)
= (20×40) + (20×5) + (3×40) + (3×5)
=800+100+120+15
=1035
Aside
Why does FOIL work? By the distributive law (with the sum part writ-ten first) we have
(a+b)e=ae+be Now if we replace e with c+d, this gives us
(a+b)(c+d) =a(c+d) +b(c+d) =ac+ad+bc+bd
where the last equality comes from applying the distributive law again.
Or if you prefer a more geometric argument (when a, b, c, d are positive), find the area of the rectangle below in two different ways.
a
c d
bc bd
b
ac ad
On the one hand, the rectangle has area (a+b)(c+d). On the other hand, we can decompose the big rectangle into four smaller rectangles, with areas ac, ad, bc, and bd. Hence the area is also equal to ac+ad+ bc+bd. Equating these two areas gives us FOIL.
Here is a magical application of FOIL. Roll two dice and follow the instructions in the table below. As an example, let’s suppose when you roll the dice, the first die has 6 on top and the second die has 3 on top.
Their bottom numbers are 1 and 4, respectively.
Roll two dice (say we get 6 and 3):
Multiply top numbers: 6×3 = 18
Multiply bottom numbers: 1×4 = 4 Multiply first top by second bottom number: 6×4 = 24 Multiply first bottom by second top number: 1×3 = 3
Total: 49
In our example, we arrived at a total of 49. And if you try it yourself with any normal six-sided dice you will arrive at the same total. It’s based on the fact that on every normal six-sided die, the opposite sides add to 7. So if the dice show x and y on the top, then they must have 7−x and 7−y on the bottom. So using algebra, our table looks like this.
Roll 2 dice (xandy):
Multiply top numbers: xy = xy
Multiply bottom numbers: (7− x)(7− y) = 49−7y−7x+xy First top times second bottom: x(7− y) = 7x− xy First bottom times second top: (7− x)y = 7y− xy
Total: = 49
Notice how we use FOIL in the third row (and that−x times−y is positive xy). We could also arrive at 49 using less algebra by looking at the second column of our table and noticing that these are precisely the four terms we would get by FOILing(x+ (7−x))(y+ (7−y)) = 7×7=49.
In most algebra classes, FOIL is mainly used for multiplying expres-sions like the ones below.
(x+3)(x+4) =x2+4x+3x+12=x2+7x+12
Notice that in our final expression, 7 (called the coefficient of the x term) is just the sum of the two numbers 3+4. And the last number, 12 (called
the constant term) is the product of the numbers 3×4. With practice, you can immediately write down the product. For example, since 5+7=12 and 5×7=35, we instantly get
(x+5)(x+7) =x2+12x+35
It works with negative numbers, too. Here are some examples. In the first example, we exploit the fact that 6+ (−2) =4 and 6× (−2) = −12.
(x+6)(x−2) =x2+4x−12 (x+1)(x−8) =x2−7x−8 (x−5)(x−7) =x2−12x+35 Here are examples where the numbers are the same.
(x+5)2 = (x+5)(x+5) =x2+10x+25 (x−5)2 = (x−5)(x−5) =x2−10x+25
Notice that, in particular, (x+5)2 = x2+25, a common mistake made by beginning algebra students. On the other hand, something interesting does happen when the numbers have opposite signs. For example, since 5+ (−5) =0,
(x+5)(x−5) =x2+5x−5x−25=x2−25
In general, it’s worth remembering the difference of squares formula:
(x+y)(x− y) =x2− y2
We applied this formula in Chapter 1, where we learned a shortcut for squaring numbers quickly. The method was based on the following algebra:
A2 = (A+d)(A− d) +d2
Let’s first verify the formula. By the difference of squares formula, we see that [(A+d)(A−d)] +d2 = [A2−d2] +d2 = A2. Hence the for-mula works for all values of A and d. In practice, A is the number that is being squared, and d is the distance to the nearest easy number. For example, to square 97, we choose d =3 so that
972 = (97+3)(97−3) +32
= (100×94) +9
=9409
Aside
Here’s a proof of the difference of square law using pictures. It shows how a geometric object with area x2−y2can be cut and rearranged to form a rectangle with area(x+y)(x−y).
We also learned in Chapter 1 a method for multiplying numbers that were close together. We focused on numbers that were near 100 or began with the same digit, but once we understand the algebra behind it, we can apply it to more situations. Here is the algebra behind the close-together method.
(z+a)(z+b) =z(z+a+b) +ab
The formula works because(z+a)(z+b) =z2+zb+za+ab, and then we can factor out z from the first three terms. The formula works for any numbers, but we typically choose z to end in zero (which is why I chose the letter z). For example, to do the problem 43×48, we let z=40, a=3, b =8. Then our formula tells us that
Notice that the original numbers being multiplied have a sum of 43+48=91, and the easier numbers being multiplied also have a sum of 40+51 =91. This is not a coincidence, since the algebra tells us that the original numbers being multiplied have a sum of(z+a) + (z+b) = 2z+a+b, and this is also the sum of the easier numbers z and z+a+b.
With this algebra, we see that we could also round up to easy numbers.
For instance, the last calculation could also have been performed with z = 50, a = −7, and b = −2, so our initial multiplication will be 50× 41. (An easy way to get 41 is to notice that 43+48 = 91 = 50+41.) Consequently,
43×48= (50−7)(50−2)
= (50×41) + (−7× −2)
=2050+14
=2064
Aside
In Chapter 1, we used this method for multiplying numbers that were just above 100, but it also works magically with numbers that are just below. For example,
96× 97= (100− 4)(100− 3)
= (100× 93) + (−4 × −3)
=9300+12
=9312
Note that 96+97=193=100+93. (In practice, I just add the last digits, 6+7, to know that 100 will be multiplied by a number that ends in 3, so it must be 93.) Also, once you get the hang of it, you don’t have to multiply two negative numbers together, but just multiply their positive values. For example,
97× 87= (100− 3)(100− 13)
= (100× 84) + (3× 13)
=8400+39
=8439
(continues on the following page)
Aside(continued)
The method can also be applied to numbers that are just below and above 100, but now you have to do a subtraction at the end.
For instance,
109×93= (100+9)(100−7)
= (100×102)−(9×7)
=10,200−63
=10,137
Again, the number 102 can be obtained through 109−7 or 93+9 or 109+93−100 (or just by summing the last digits of the original numbers; 9+3 tells you that the number will end in 2, which may be enough information). With practice, you can use this to mul-tiply any numbers that are relatively close together. I’ll illustrate using three-digit numbers of moderate difficulty. Notice here, the numbers a and b are not one-digit numbers.
218×211= (200+18)(200+11)
= (200×229) + (18×11)
=45,800+198
=45,998
985×978= (1000−15)(1000−22)
= (1000×963) + (15×22)
=963,000+330
=963,330