So.. In class we learnt how to expand and simplify using the binomial expansion:
Example 1:
(3x-y)^5
Okay, first off, we start off by listing our values.
Our a = 3xb = -y -> don't forget the negative sign!n value = 5 because our expression is to the power of 5.
Using the formula, we're going to expand the expression now. I wrote it out because it was really hard typing out all the power signs.
In the first step, we expanded out the equation (3x-y)^5 following the formula. We did this by plugging in our values into the equation.
If you can't remember the formula, you can always remember that all the powers should equal to whatever the value of n is. In this case n is equal to 5. If you notice in the first term 5C0(3x)^5(-y)^0 the powers are 5 and 0. 5+0 = 5. Your terms 5C0, 5C1... etc should keep adding up until you reach 5. Which is what you should get in the end -> 5C5.
In step 2, what we have to do next is multiply the brackets through. I'm just going to do the first term just because it's really difficult typing everything out. First term: 5C0(3x)^5(-y)^0
First we start off with 5C0. 5C0 is simply just 1. You can either plug it into your calculator to find out the answer or you can use the formula: nCr = n!
----
r! (n-r)!
5C0, our 5 = our n value, and 0 is our r value. Then we will get
5! 5!
------- = ------ = ------------------
0!(5-0)! 0!5! (1)
So our top is cancelled, and we're left with 1 at the bottom. 1/1 = 1. So then 1 is our answer for the 5C0. Still don't believe me? You can double check by plugging 5C0 into your calculator.
Next, we're trying to simplify the (3x)^5
(3x)^5 if we multiply that through (3)(3)(3)(3)(3) = 3^5 and (x)(x)(x)(x)(x) = x^5 |
Then we simplify (-y)^0
(-y)^0 = 1 because anything to the power of 0 is equal to 1.
Following that, you will continue to simplify the rest of the equation.
You should be left with what I have in step 2.
Then in your last step, you simplify what's remaining. Again, I'm only going to do the first term. (1)(3x)^5(1) = (1)(3)^5(x)^5(1) = 243x5. After you're done simplifying and multiplying everything, you should be left with the answer that I have boxed in.
Get it? Got it? Good. Next example.
Example 2) Find the 3rd term of (2x3+y2)^3
-> In this example we're looking for a specific term, and when we want to know a specific term we can use the formula:
tk+1 = nCka^n-k(b)^k
Again, we start off by listing our values.
a = 2x3
b = y2
n = 3 because our exponent is to the power of 3.
k = ? -> our k value is always 1 less then the term we're looking for. In this case, we're looking for the 3rd term, so our k value should be 3-1 = 2.
Since we have all the values we can simply just plug it into the formula now and solve!
tk+1 = nCka^n-k(b^k)
t2+1 = 3C2(2x3)^3-2(y2)^2
--> Now after we plug everything in we can just simplify the expression.
= (3)(2x3)^1(y4)
Now you just multiply and simplify the expression. You should get 6x3y4 as your answer.
Example 3) Find the coefficient of x2 in expansion of (x-3)^4
Notice how in this example it says find the COEFFICIENT, not find the TERM? Which means we have to do things a little differently as our first step this time.
It says find the coefficient of x2 in expansion of (x-3)^4.
Before we do anything, we have to find a pattern with x2 in the expression. We can do this by expanding out some of the terms.
First always list your terms!
a = x
b = -3
n = 4
(x-3)^4 = 4C0(x^4)(-3)^0 + 4C1(x)^3(-3)^1 + 4C2(x^2)(-3)^2
-> x = x4 in the 1st term, -> x3 -> x2. Is this what we're looking for? Yes! = 3rd term.
After we find out the pattern, and we find that the 3rd term of the expression is = x2 we can plug it into this formula again:
tk+1 = nCka^n-k(b)^k
Values:
a = x
b = -3
n = 4
k = 2, this is because our 3rd term is = x2, so one less than 3 = 2.
t2+1 = 4C2(x)^4-2(-3)^2
=6(x^2)(9)
= 54x2
-> In this case our answer is going to be 54, because it's only asking for the COEFFICIENT, not the TERM.
Still having trouble? Check out the YouTube video above! ^^
Well.. Our time is up everyone. Hope I helped and have a fabulous day everyone!!! :)
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