RE:
What? His text book says I was correct!
i'm afraid that method of expansion is incorrect. let's prove it.
try you method of expansion to calculate:
(1 + 2 + 3)(4 + 5)
you will get 120
the correct answer is 54
or even look at the line where you have the expression for (e + f)^3 as e^3f + 2e^2f^2 + ef^3, you kow that's wrong
yeah, but after you expanded with respect to the first term, you then expanded with respect to the second term using the new terms you expanded using the first term, as opposed to the original terms. ok, that didn't make any sense. but look at your work again with e and f, and replace e and f with two numbers (preferably above 1), you'll realize that you don't get the right answer in the end
well, i never went that far...i don't even know how to use a graphing calculator. drawing graphs by hand has always been kind of enjoyable for me (i must sound like a real geek saying that, but i'm not) so i didn't let technology do it for me. but routine addition and subtraction and all that other stuff is just monotonous and painful, so i just whip out my calculator. i want to be able to use my head again and do this stuff, but i'm too far gone it seems, the habit sticks with me
Hello, bobchiba!
First of all, we know what the terms look like:Does anyone know of an easy to remember way of expanding binomials?
An example would be: .(x + y)^5
. . . x^5 . . x^4·y . . x³·y² . . x²·y³ . . x·y^4 . . y^5 . . [1]
It begins with x^5 and ends with y^5
. . In between, the x-exponents decrease and the y-exponents increase.
. . Note that, in each term, the exponents add up to 5.
Now look at Pascal's Triangle and find the row that begins 1, 5.
. . We find the numbers: .1, 5, 10, 10, 5, 1
And those are the coefficients of the six terms in [1].
Therefore: .(x + y)^5 .= .x^5 + 5x^4·y + 10x³·y² + 10x²·y³ + 5x·y^4 + y^5
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Another example: .(2a + 3b)³
The terms are: . (2a)³ . . (2a)²(3b) . . (2a)(3b)² . . (3b)³
From Pascal's Triangle, we find: .1, 3, 3, 1
So we have: .1(2a)³ + 3(2a)²(3b) + 3(2a)(3b)² + 1(3b)³
. . = .1(8a³) + 3(4a²)(3b) + 3(2a)(9b²) + 1(27b³)
. . = .8a³ + 36a²b + 54ab² + 27b³