There are two ways to find it using long division and synthetic division.
using Long Division:
*For ( x + 1 ) to be a factor you must have a zero remainder.
since there is no remainder the (x + 1) is a factor of ( x² – 9x – 10 )
using Synthetic Division:
x + 1 = x – a *we first need to find a 1 = -a a = -1
using a we do the synthetic division:
Another example:
Divide 3x³ – 2x ² + 3x – 4 by x – 3 using synthetic division x – 3 = x + a a = 3
REMAINDER THEOREM
If P(x) is polynomial, then P(a) is equal to the remainder when P(x) is divided by ( x – a )
Find the remainder 6x³ – 5x² + 4x – 17 with (x + 3) P(3) = 6x³ – 5x²+ 4x – 17 = 6(3)³ – 5(3)² + 4(3) – 17 = 6(27) – 5(9) + 12 – 17 = 6(27) – 5(9) + 12 – 17 = 162 – 45 + 12 – 17 = 112 * (x + 3) is not a factor of 6x³ – 5x² + 4x – 17 because the remainder is not zero
Find the remainder when x^6 + 5x^5 + 5x^4 + 5x³ + 2x² – 10x – 8 with ( x + 1 ) P(1) = x^6 + 5x^5 + 5x^4 + 5x³ + 2x² – 10x – 8 = (1)^6 + 5(1)^5 + 5(1)^4 + 5(1)³ + 2(1)² – 10(1) – 8 = 1 + 5 + 5 + 5 + 2 – 10 – 8 = 0 * ( x + 1 )is a remainder of x^6 + 5x^5 + 5x^4 + 5x³ + 2x² – 10x – 8 because the remainder is zero. COMPLETELY FACTOR * using both factor and remainder theorem f(x) = x³ - 3x² - 13x + 15 steps: 1) find the factors of the constant. Constant = 15 factors of 15 : ± 1, ± 15, ± 3, ± 5 2) pick one of the factors then evaluate P(1) = x³ - 3x² - 13x + 15 = (1)³ - 3(1)² - 13(1) + 15 = 1 - 3 - 13 + 15 = 0 *because the remainder is zero ( x - 1 ) is one of the remainders 3) now use synthetic division (because its easier) to reduce and find the other factor.
x³ - 3x² - 13x + 15 = ( x - 1 ) ( x - 5 ) ( x + 3 )
*** can only use if dividing by (x ± a ) coefficient must be 1.
ok.. i think that's all for today.. our homework is exercise 53 #'s 1 - 7.. and the next blogger is Youna. thats all.. bye!! im going to change the next blogger because i found out that youna isnt gonna be at school tommorow soo here it comes!...... THE NEXT BLOGGER IS ELVIN!
Today we defined the word “inverse”, were taught how to inverse a function and how to figure out if two given functions are the inverse of each other. Well, let’s get started…
Inverse: the set of ordered pairs obtained by interchanging the coordinates of each ordered pair in relation.
Example:
Original set of pairs --> [ ... (-3,3), (1,4), (4,5) ... ]
Inverse of set of pairs --> [ ... (3,-3), (4,1), (5,4) ... ]
* notice that all i did was switch the x and y values around in each pair
When graphing the original and inverse sets of pairs, each point must reflect the other across the y=x line on the graph. So, it will look like this:
Now, the next question is: How do you find the inverse of a function? Well, according to Mrs. Ingram, if you follow these four steps, you'll be fine:
given function: f(x) = 3x - 2 find: f^-1(x)
STEP 1: Replace f(x) with y y = 3x - 2
STEP 2: Interchange the x and y x = 3y -2
STEP 3: Solve for y x + 2 / 3 = y
STEP 4: Replace y with f^-1(x) f^-1(x) = x+2 / 3
And there we have it. The inverse of the function f(x)=3x - 2 is f^-1(x)=x+2 / 3
But, what about if you are given: f(x)=3x-2 & m(x) =x+2 / 3 and it asks if those two functions are inverses of each other? To find out we must do the whole composition of functions thing ... lol. Here we go:
f(m(x)) & m(f(x))
f(x+2 / 3) = 3(x+2 / 3) - 2 ** the threes cancel out = x + 2 - 2 = x
Both compositions equal x which means the two functions are inverses of each other.
And... that's about it. We've come to the conclusion of the blog and let's make this quick. Next blogger shall be ... well remyshire asked for it so yeah.. it's her. Ex. 52 # 1-5,8,9,11 are the homework for tonight. And... well that's all folks. vamoose.
Now for the final part, composition of functions. Composition of functions looks something like this: (f ○ g)(x) or f(g(x)) In these equations, 'g' replaces 'x' in the f(x). Confused? Thought so. Well here's some examples.
Try these questions. Given: f(x) = 3x + 2 and g(x) = x²
a) f (g (x)) b) f (f (x)) c) g (f (x))
Take you time, no rush. Done? Good! Here are the answers.
a) f (g (x)) → f (x) = 3x + 2 ............................ = 3x² + 2
b) f (f (x)) = f (3x + 2) ...............= 3 (3x + 2) + 2 ...............= 9x + 6 + 2 ...............= 9x + 8
c) g (f (x)) = (3x + 2)² ................= 9x² + 12x + 11
Lastly, we have Exercise 51, Questions 1 - 7 for homework. Have fun! =D
That's it for me. Finally done. Sorry it's kinda of late. Well anywho... it's time fo the best part of this whole, who shall be the next scribe! Well, since this person didn't do it before and sits across the room, I chose Chrycel! WAIT! Can I chose Mrs. Ingram as the next scribe?! If I can, I chose her! If not then... I guess it's Chrycel once again.
Anyways, circles. Like richard said they started out pretty easy, and gradually increased in difficulty but hows that different from any other unit? They weren't ridiculously hard for me, but I kept forgetting stuff, so there was always that one easy question that I could't do, because I forgot the tangent thing, or the inscribed angle thing. Most of the time it was simple logic thatsolved the questions (with the help of alittle existential knowledge on the subject.)
Yeah so Circles, = 7 out of 10 on the ridiculously hard scale.
Good luck on thetest everyone, I know I'll need it -_-;
Sorry I wasn't there today. I wasn't feeling very well. For all those who blogged, yes you will still get the mark even though it wasn't asked for on the test. I hope the test went well for all, I'll have them marked in a couple days. Tomorrow we start functions
Circles.. I found this unit very difficult. it started out easy and when it started to get into the unit it became harder and harder maybe it was because i have miss so many classes in this unit or maybe it is because this unit hates me.
What can I say about this unit... hmmmm? I'd say that it was the easiest unit yet. Even easier than the parabola unit because all you have to do is memorize the silly rules like the inscribed angle is half of the central angle. I'm glad we didn't have to do a dictionary on it too. Anyways, nothing outrageously tricky, just like I said before, memorizing those rules.
Yes now I get my one mark! Victory Fanfare plays in the background* nyuk nyuk nyuk nyuk nyuk
I'm really not in the mood. But whatever. It's one mark. My day hasn't been so great. I'm sleepy its 1 in the morning and I haven't had time to do homework. I've just realized while doing that homework that I had to blog. Well here it is. HMM circles. Not fun. Mrs. Ingram made it LOOK understandable and simple and it probably is but I'm probably just thinking too hard. Wasn't a fun unit but in any case, which isnt? To the point. It was an alright unit and I understood most of it, but my siblings warned me about circles, and how NOT fun they are. Correct they were. I found the inscribed angle was alright and all but having to prove an angle or arc was very hard because you'd have to put it into words. I THOUGHT THIS WAS MATH?! It's really simple if you study it. I studied but in actuality, I probably won't remember much because I'll have been so sleep deprived from this night. Well it was an alright unit all in all So yeah. My hands are starting to cramp up. I need water. So I say good day for now. Tired I am
