Integrated math 3
- Dividing polynomials by linear expressions
- Dividing polynomials by linear expressions: missing term
- Divide polynomials by linear expressions
- Factoring using polynomial division
- Factoring using polynomial division: missing term
- Factor using polynomial division
This lesson guides you through the process of dividing polynomials by linear factors, showing you how to neatly organize your work and avoid common errors. You'll learn how to handle missing degree terms and ensure correct subtraction. The result? You'll confidently simplify and rewrite complex rational expressions.
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- What is "The Price is Right"?(4 votes)
- “The Price is Right” is a game show with different pricing games that rewards contestants for knowledge of prices of goods.(22 votes)
- How do you constrain the domain? Do you need to know how to?(5 votes)
- you need to, yes, and to do so you just need to know what the domain of functions are.
For instance if a function has a square root, you know the things under the square root cannot be less than 0. Logs cannot be 0 or less, and so on.
So a kinda complex example that will show what to do is something like sqrt(ln(1/(x-1)))
So you have the square root of the natural log of 1/(x-1) so let's take it one step at a time.
square root has to be 0 or greater, so the domain is 0 to infinity.
natural log has to just be greater than 0, so not the domain changes to be greater just greater than 0, this is because this has both square root and a log.
the last part is 1/(x-1) for functions with a variable int he denominator of a fraction, the denominator cannot be 0. so when does x-1=0? you just use some algebra, xo it is 0 when x = 1.
Now the domain changes again to be greater than 0 except for the value 1.
Let me know if that doesn't help.(14 votes)
- what if your trying to divide a polynomial has a cube and than a square and than a regular number. what do u insert for the missing term?(2 votes)
- A third-degree polynomial would look like this:
𝑎𝑥³ + 𝑏𝑥² + 𝑐𝑥 + 𝑑
However, in our case the 𝑥-term is missing, so we would replace 𝑐 with 0.(12 votes)
- At1:05, why did Sal add "0x^2" between "2x^3" and "-47x" in the expression "2x^3 - 47x - 15"?(3 votes)
- It leaves the spot needed to divide the number. It would be comparable to regular division such as 305/5, without the 0 it would get an incorrect answer.(5 votes)
- I dont understand this format p(x)+k/x+1 that is asked in practices exercises(3 votes)
The main thing to worry about is the denominator, as x=-1 is the vertical asymptote. Formatting wise,
p(x) is any polynomial
k is any constant term
hopefully that helps !(3 votes)
- I'm struggling with how I would write the integers of the final polynomial. For example, when I did this question, I thought it was -10x not +10x in the answer. It would help if i had a rule for the integers..?(0 votes)
- There is a rule :D
First, let me explain how negatives and positives work. Note that I use multiplication and division in the example because it may make more sense, but this also applies to addition and subtraction.
+ = positive
- = negative
★ A + number multiplied or divided by a + number is positive (you probably knew that).
★ A - number multiplied or divided by a + number is negative.
★ A - number multiplied or divided by a - number is +.
This may seem confusing or hard to remember at first, but there is a cool way of remembering this!
+ = friend
- = enemy
★ A friend of a friend is your friend.
Translates to: + times + = +.
★ An enemy of your friend is your enemy.
Translates to: - times + = -.
★ An enemy of your enemy is your friend.
Translates to: - times - = +.
Kind of makes sense if you think about it, right? That's how positives and negatives work!
- - -
In your example from the video, Sal got +10x^2 because he was subtracting a negative number from 0x^2. Subtracting a negative number translates to: 10x^2-(-10x^2)= 10x^2.
Hopefully that helps :). I'm not the best at explaining things, so if you would like me to clarify anything, please ask!(3 votes)
- if a divisor was x^2-4, would we put missing terms for it? like x^2+0x-4? If no, Why?
why does putting missing terms work for dividens and not divisors?(1 vote)
- Do you have to write 0x^2 or keep it blank?(0 votes)
- I would strongly suggest adding 0x^2, as doing the division without it can be messy and easier to make a mistake on.(2 votes)
- can i just use the box method(0 votes)
In front of us, we have another screenshot from Khan Academy and I've modified it a little bit so I have a little bit of extra space. And it says, divide the polynomials. The form of your answer should either be a straight up polynomial or a polynomial plus the remainder over x minus five, which we have here in the denominator, where p of x is a polynomial and k is an integer. So, we've done stuff like this, but like always, I encourage you to pause this video and work on this on your own. And if you were doing this on Khan Academy, there's a little bit of an input box here where you'd have to type in the answer, but let's just do it on paper for now. All right. So, we're trying to figure out what x minus five divided into two x to the third power. Actually, I want to be careful here because I want to be very, very organized about my different degree columns. So, this is my third degree column. And then I want my second degree column. But there is no second degree term here. There's a first degree term. So, I'll write it out here. So, minus 47x. And actually, to be even more careful I'll write plus zero x squared. And then I have minus 15. By putting that plus zero x squared, that's making sure I'm doing good, I guess degree place column hygiene. All right, so now we can work through this. And first we could say hey, how many times does x go into the highest degree term here? Well, x goes into two x to the third power two x squared times. We'd want to put that in the second degree column. Two x squared. You can see how it would've gotten messy if I'd put negative 47x here. I'd be like where do I put that two x squared and you might confused yourself, which none of us would want to happen. All right, two x squared times negative five is negative 10x squared. Two x squared times x is 2x to the third power. Now we want to subtract what we have in red from what we have in blue. So, I'll multiply them both by negative one. So that becomes a negative and then that one becomes a positive. And that's actually one of the biggest areas for careless errors. If you had negative here and you just want to subtract it because you know you have to subtract it. Be like no, I'm subtracting a negative, it needs to be a positive now. All right. So, zero x squared plus 10x squared is 10x squared. And then the two x to the third minus two x to the third is just zero. And then we can bring down that negative 47x. And once again, we look at the highest degree terms. X goes into 10x squared 10x times. So, plus 10x. 10x times negative five is negative 50x. 10x times x is 10x squared. And once again, we want to subtract what we have in teal from what we have in red. So we could multiply both of these times negative one. That becomes a negative. This one becomes a positive. Now, negative 47x plus 50x is positive three x. And then 10x squared minus 10x squared gets canceled out. Bring down that 15. Come on down. Used to watch a lot of Price is Right growing up. Never quite made it to the show. All right. X goes into three x how many times? It goes three times. Three times negative five is negative 15. Three times x is three x. We want to subtract the orange from the teal. And so, this becomes a negative, this becomes a positive. Negative 15 plus 15 is zero. And three x minus three x is zero. So, you're just left with zero. So, no remainder. So this whole thing, you could re-express or simplify as two x squared plus 10x plus three. And once again, if this was on Khan Academy, there would be a little bit of an input box that looks something like this and you would have to type this in. Now, if you wanted these to be exactly the same expression you would also need to constrain the domain. You would say, okay four x does not equal positive five. And the reason why you have to constrain that is the whole reason why we can even divide by x minus five is we're assuming that x minus five is no equal to zero. And it's generally not equal to zero as long as x does not equal to positive five. But for the sake of that exercise, you don't need to constrain the domain like this.