Guys, I'm so sorry, the first checking comprehension should have a + 14 rather than - 14 in the second fraction, that makes it work out nicely. And the second problem is also looking strange and you may have to disregard it. I can't believe I let these errors slide, rest assured the other math comprehensions are not like this!
Thank Christ, I was stuck on the 9x+14 part for ages and couldn't figure out what I was doing wrong. Do you have other checking comprehension examples for this section?
SOLUTION TO PROBLEM 2: I hope you find this comment, because I'm here to say that if you weren't able to solve the second problem is cause it's written wrong. "3 / x+5" is NOT the correct answer to the exercise. And you should not feel like you're falling behind just cause you can't seem to get to the same answer written on the screen. People make mistakes and that's okay, it happens. As Prof. Dave wrote "-14" instead of "+14" in the first exercise, he also wrote something wrong in the second exercise, or simply solved it wrong. I stayed at this exercise trying to get to the same answer from yesterday. Took me so long to get to the incorrect answer until I realized it's just something there that's written wrong. Just move on, there's nothing wrong with you. Keep studying, cause if you've got to the point of reading explenations to one problem of one random math video on RUclips it means that you're dedicated. And you should stay that way. You're definitely on the right track.
OMG, you can't imagine how much your comment helped me and encouraged me , I'm sorry if my language was bad but I'm not a native speaker to English , maybe you will not even read it but I wanted to express how thankful I am for your comment
Tsk-tsk-tsk! Well, I figured out that there should have been a +14 instead a ‒14 in the first expression because there is no way that a+b=9 and a×b=‒14. However, I thought I might be wrong and tried to work it the other way round, from the solution to the original. It didn’t work, so I was even surer about the mistake. The second one was trickier. I tried to work it from solution to the original expression and there was no way for it to work. I am sorry that I didn’t read the comments before starting doing the maths here.
At 5:27, I feel like I'm missing the point of getting the lowest common denominator? Why would you not cancel out terms from each expression first and then work from there? I.e (x-5)(x+1) / (x+9)(x+1), cancel out (x+1), leaving (x-5)/(x-9), and do the same to (x+2) in the other term, so you have [(x-5)/(x-9)] - [(x-3)/(x+1)] and work with that?
I just spent 30 minutes tearing my diminishing hair out - grrrrr...........and I had been telling my Math teacher wife just this morning that this guy must be a genius or polymath of some kind. She kind of sneered at me!
Good question. Generally speaking, it is preferable to expand numerators, and factor denominators. This helps put your expressions in the most calculus-friendly form. Consider: 6*(x + 2)*(x + 3)/(x^3 + 10*x^2 + 29*x + 20) You could take a derivative of the original expression without simplifying it, it will be tedious to do so. To integrate it without preprocessing, is impossible. I deliberately set it up, to factor with no removable factors. The denominator factors as: (x + 1)*(x + 4)*(x + 5). Expanding the top, and factoring the bottom, you get: (6*x^2 + 30*x + 36)/[(x + 1)*(x + 4)*(x + 5)] There is an algebraic trick to simplify it further, so it is much more calculus-friendly, called partial fractions. Set up the solution in the form of; A/(x + 1) + B/(x + 4) + C/(x + 5) Equate to the original expression, match corresponding terms, and solve an equation system for A, B, and C. For simple linear factors, Heaviside coverup works as a shortcut. To find A, cover up (x + 1), and let x = -1 to make the covered-up term equal zero. Evaluate what remains of the original expression, to find A. at x = -1: A = 6*((-1)^2 + 5*-1 + 6)/[(-1 + 4)*(-1 + 5)] = 1 I'll leave it as practice for you to find the others, which are B = -4 & C = 9. This means the expression is: 1/(x + 1) - 4/(x + 4) + 9/(x + 5) Now it looks MUCH simpler, and easier to use for calculus. It's now a simple case of the power rule and trivial case of the chain rule for differentiation. And for integration, it's just natural log saving the day, at the special case where the reverse power rule fails. The derivative: -1/(x + 1)^2 + 4/(x + 4)^2 - 9/(x + 5)^2 And the integral: ln(|x + 1|) - 4*ln(|x + 4|) + 9*ln(|x + 5|) + C (this C has nothing to do with the C from earlier)
Historical reasons. Because division is a lot more tedious to do by hand than multiplication, it is preferable to simplify the denominator as much as possible. Because all you need to do is multiply by 1 in a fancy way, to do so, it is the preferred method of simplifying a number or expression with roots in its denominator. With calculators, it is no longer as much of an issue, because 1/sqrt(2) is just as easy to enter into a calculator as sqrt(2)/2. However, if you are writing a computer program that will run millions of calculations, simplifying your expression to the most computationally efficient version is important for the efficiency of your program, which is why it is still a valuable skill to anticipate this issue. sqrt(2)/2 will compute a lot faster than 1/sqrt(2). There are some applications where it is of interest to deliberately move a radical expression to the denominator, such as integration techniques.
Given: (x - 2)/(x + 5) Add zero in a fancy way (i.e. 0 = 5 - 5) to the numerator, so we can form a term we can cancel. (x + 5 - 5 - 2)/(x + 5) Regroup the fraction: (x + 5)/(x + 5) - 7/(x + 5) Since the first part of the fraction equals 1, everywhere except at x = -5, we can replace it with 1. The second term will take care of the forbidden value of x=-5, because -5 is also forbidden for it as well. Thus it simplifies to: 1 - 7/(x + 5)
@@Carrymejane I think I misread the OP. I thought the OP was asking about simplifying (x-2)/(x+5). Anyway, Dave's example simplifies with similar principles. Given: 3*x/(x + 5) - 6/(x - 2) Change to common denominators and add: 3*x*(x - 2)/[(x - 2)(x + 5)] - 6*(x + 5)/[(x - 2)(x + 5)] [3*x*(x - 2) - 6*(x + 5)]/[(x - 2)(x + 5)] Expand numerator, and factor out the common factor of 3: 3*(x^2 - 4*x - 10)/((x-2)(x + 5)) This is the simplest form as a single fraction. Numerators "prefer" to be expanded, while denominators "prefer" to be factored. You may also pull out common constants out in front, where practical, like the 3.
@@Carrymejane You can take this further, if you want. For calculus applications, it's preferred to have single factors in each denominator, instead of a product of factors. Since the degree on the top is 2, and the degree on the bottom is also 2, we can expand it so it becomes a stand-alone constant, and a sum of partial fractions if we want. Dave has a video on partial fractions, but he doesn't cover my preferred shortcut, Heaviside Coverup. In this case, it helps us, to look at the original given form. Had you been given the single solid fraction we just derived, you can expand the denominator, and manipulate the numerator so you can form a term we can cancel. This brings it to the form: 3*(x^2 + 3*x - 10)/(x^2 + 3*x - 10) - 21*x/[(x-2)(x + 5)] Which allows us to cancel common factors in the first term and get: 3 - 21*x/[(x-2)(x + 5)] Partial fractions and Heaviside cover-up, can simplify it from here. For us, let's look back at the original form: 3*x/(x + 5) - 6/(x - 2) The second term is already as simplified as we can get it. The first term, has a linear term on top of a linear term. We can manipulate it, to form a term we can cancel, by adding zero in a fancy way: 3*x/(x + 5) = 3*(x + 5 - 5)/(x + 5) = 3*(x+5)/(x+5) - 15/(x + 5) Construct complete form, equal to the original expression: 3 - 15/(x + 5) - 6/(x - 2) Now, we have the factors untangled from each other. It is now just a sum of elementary functions in as close to their pure forms as possible, and it is simple to do Calculus operations on it.
It was infact way coolerr. Haha, I think he just went ahead put the 3x - 6 together on the numerator. Then factoring out 3(x-2) over (x+5)(x-2) resulting in 3/(x+5). :(
None of the videos thusfar have actually explained "factoring" outside the context of quadratics; it's getting less intuitive. I feel like I'm falling behind again.
There are formulas for factoring ANY cubic and ANY quartic, but they are complicated, and seldom taught outside a math degree. Mathologer has an excellent video on making sense of the cubic formula. Usually, if you're working with a cubic or quartic in an academic example, it will be deliberately set up to have rational roots, and you can find them with the rational roots theorem. Given any polynomial in the form of: a*x^n + b*x^(n - 1) + ... (other terms) ... + k Its roots will add to -b/a, and multiply to k/a for even polynomials, and multiply to -k/a for odd polynomials. This means, roots of the polynomial will be (integer factors of k)/(integer factors of a), and knowing the two properties above will help narrow them down from those options. Once you positively identify one rational root, you can use polynomial division to reduce the polynomial to a lower degree. This way, a cubic can factor as a linear term times a quadratic term, and the quadratic term has a simple formula for it.
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Guys, I'm so sorry, the first checking comprehension should have a + 14 rather than - 14 in the second fraction, that makes it work out nicely. And the second problem is also looking strange and you may have to disregard it. I can't believe I let these errors slide, rest assured the other math comprehensions are not like this!
Whew! Thanks! was stuck on those questions for quite a while
Thanks Dave, honestly thought I was loosing my mind :) Cheers and thanks for the amazing content!
Damn I was stuck like 2 hours solving this. My bad for not reading the comment section 😂
I solved the 2nd question and come up with an answer of
3(x^2 - 4x - 10)
all over
(x+5) (x-2)
Thats why I'm so confused
Thank Christ, I was stuck on the 9x+14 part for ages and couldn't figure out what I was doing wrong. Do you have other checking comprehension examples for this section?
SOLUTION TO PROBLEM 2:
I hope you find this comment, because I'm here to say that if you weren't able to solve the second problem is cause it's written wrong. "3 / x+5" is NOT the correct answer to the exercise. And you should not feel like you're falling behind just cause you can't seem to get to the same answer written on the screen. People make mistakes and that's okay, it happens. As Prof. Dave wrote "-14" instead of "+14" in the first exercise, he also wrote something wrong in the second exercise, or simply solved it wrong. I stayed at this exercise trying to get to the same answer from yesterday. Took me so long to get to the incorrect answer until I realized it's just something there that's written wrong. Just move on, there's nothing wrong with you. Keep studying, cause if you've got to the point of reading explenations to one problem of one random math video on RUclips it means that you're dedicated. And you should stay that way. You're definitely on the right track.
Thanks for clarifying
Thanks for the encouragement too
I spent half an hour on exercises, I thought I was dumb ;DD
OMG, you can't imagine how much your comment helped me and encouraged me , I'm sorry if my language was bad but I'm not a native speaker to English , maybe you will not even read it but I wanted to express how thankful I am for your comment
life is now worth living
it takes me days and lots of research... then I see on the comment line and the question was wrong. ain't regret it tho. it makes me learn more.
Hahahaaha same but not days just 3 in a half hours
The second one , i guess its wrong
COMPREHENSION #1 SOLUTION
Bearing in mind that '- 14' should have been '+ 14'...
((x^2 + 3x - 4)/(x^2 + 4x - 21)) ÷ ((x^2 + x - 2)/(x^2 + 9x + 14))
= ((x + 4)(x - 1))/(x + 7)(x - 3)) ÷ ((x + 2)(x - 1)/(x + 2)(x + 7)) - factorization.
= ((x + 4)(x - 1))/(x + 7)(x - 3)) * ((x + 2)(x + 7)/(x + 2)(x - 1)) - changing from ÷ to * and flipping the fraction on the right.
= (x + 4) (x - 1) (x + 2) (x + 7) / (x + 7) (x - 3) (x + 2) (x - 1) - shown more simply as 1 fraction with terms combined (multiply numerators together and denominators together).
= (x + 4) -(x - 1)- -(x + 2)- -(x + 7)- / -(x + 7)- (x - 3) -(x + 2)- -(x - 1)- - cancelling matching factors on the top and bottom.
= (x + 4) / (x - 3)
Thanks man! I having trouble to factorize it due to the sign swap.
Damn💀
its -14 not +
@@arwinkumar3736 yeah, professor dave himself said that he wrote it wrong and it indeed should have been +14
can you please explain the 3rd step " shown more simply as 1 fraction with terms combined"
i didnt get it
Tsk-tsk-tsk!
Well, I figured out that there should have been a +14 instead a ‒14 in the first expression because there is no way that a+b=9 and a×b=‒14. However, I thought I might be wrong and tried to work it the other way round, from the solution to the original. It didn’t work, so I was even surer about the mistake.
The second one was trickier. I tried to work it from solution to the original expression and there was no way for it to work.
I am sorry that I didn’t read the comments before starting doing the maths here.
yeah, i came across with the same problem, it should´ve been +14, well, it´s just a little mistake he made.
humans make mistakes
I passed my test because of him. glad i subscribed
5:47
we can just multiply the Numerator an the Denominator by x
so it will give us a direct answer to the simplifying
Best math teacher on internet🎉
At 5:27, I feel like I'm missing the point of getting the lowest common denominator? Why would you not cancel out terms from each expression first and then work from there? I.e (x-5)(x+1) / (x+9)(x+1), cancel out (x+1), leaving (x-5)/(x-9), and do the same to (x+2) in the other term, so you have
[(x-5)/(x-9)] - [(x-3)/(x+1)] and work with that?
I just spent 30 minutes tearing my diminishing hair out - grrrrr...........and I had been telling my Math teacher wife just this morning that this guy must be a genius or polymath of some kind. She kind of sneered at me!
wait why cant we just cancel out 1+ square root 2 at 8:06?
You'd end up the same of how you started.
Solution 2:
- (3x) / (x + 5) - (6) / (x - 2)
- (x - 2) * ( 3x) / (x - 2) * ( x + 5) - (6) * (x + 5) / (x - 2) * (x + 5)
- (3x^2 - 6x) - (6x + 30) / (x - 2) * (x + 5)
- (3x^2 - 6x - 6x - 30) / (x - 2) * (x + 5)
Result: (3x^2 - 12x - 30) / (x - 2) * (x + 5)
Same as: 3(x^2 - 4x - 10) / (x - 2) * (x + 5)
same! this question seems to have a problem
now what?
4:44 Why don't we FOIL the denominator?
Good question. Generally speaking, it is preferable to expand numerators, and factor denominators. This helps put your expressions in the most calculus-friendly form.
Consider:
6*(x + 2)*(x + 3)/(x^3 + 10*x^2 + 29*x + 20)
You could take a derivative of the original expression without simplifying it, it will be tedious to do so. To integrate it without preprocessing, is impossible.
I deliberately set it up, to factor with no removable factors. The denominator factors as:
(x + 1)*(x + 4)*(x + 5).
Expanding the top, and factoring the bottom, you get:
(6*x^2 + 30*x + 36)/[(x + 1)*(x + 4)*(x + 5)]
There is an algebraic trick to simplify it further, so it is much more calculus-friendly, called partial fractions. Set up the solution in the form of;
A/(x + 1) + B/(x + 4) + C/(x + 5)
Equate to the original expression, match corresponding terms, and solve an equation system for A, B, and C. For simple linear factors, Heaviside coverup works as a shortcut. To find A, cover up (x + 1), and let x = -1 to make the covered-up term equal zero. Evaluate what remains of the original expression, to find A.
at x = -1: A = 6*((-1)^2 + 5*-1 + 6)/[(-1 + 4)*(-1 + 5)] = 1
I'll leave it as practice for you to find the others, which are B = -4 & C = 9. This means the expression is:
1/(x + 1) - 4/(x + 4) + 9/(x + 5)
Now it looks MUCH simpler, and easier to use for calculus. It's now a simple case of the power rule and trivial case of the chain rule for differentiation. And for integration, it's just natural log saving the day, at the special case where the reverse power rule fails.
The derivative:
-1/(x + 1)^2 + 4/(x + 4)^2 - 9/(x + 5)^2
And the integral:
ln(|x + 1|) - 4*ln(|x + 4|) + 9*ln(|x + 5|) + C
(this C has nothing to do with the C from earlier)
@@carultch thanks!
@@carultch u saved someone lives again
Thank you so much sir for the lovely video
Thanks for your videos, sir. I learned a lot from them. 😊😊
The checking comprehension questions are wrong for this video though so be careful.
Could someone solve problem #2? I left with 3(x^2 - 4x - 10) over (x + 5) (x - 2) :(
I have the same problem, i understood maybe there is an error in this one also...
@@gianfrancobruschetta2423, well that was the only result actually (I guess)
Professor, is the second problem wrong too? I really can't solve it. Can you give the explanation on this one.
www.symbolab.com/solver/step-by-step/%5Cfrac%7B3x%7D%7Bx%2B5%7D%20-%20%5Cfrac%7B6%7D%7Bx-2%7D
the second question is wrong
@@alexci3563 thank you very much sir
@Alex Chi Thank you kind sir
yeah for me also
I didn’t check the comments and was trying to solve for when it was -14 all three ways. Kinda funny
1:42
Why don't we like radicals in the denominators of fractions?
Historical reasons. Because division is a lot more tedious to do by hand than multiplication, it is preferable to simplify the denominator as much as possible. Because all you need to do is multiply by 1 in a fancy way, to do so, it is the preferred method of simplifying a number or expression with roots in its denominator. With calculators, it is no longer as much of an issue, because 1/sqrt(2) is just as easy to enter into a calculator as sqrt(2)/2. However, if you are writing a computer program that will run millions of calculations, simplifying your expression to the most computationally efficient version is important for the efficiency of your program, which is why it is still a valuable skill to anticipate this issue. sqrt(2)/2 will compute a lot faster than 1/sqrt(2).
There are some applications where it is of interest to deliberately move a radical expression to the denominator, such as integration techniques.
@@carultchwoww!!
perfect
doing internet homework . remember its incorrect unsolveble questions
ty
I think the #2 is 3x-6/(x+5) (x-2) not
3x/x+5 - 6/x-2
Did anybody solved problem 2 and figure out the mistake ? I am stuck at making factors,my denominatot is (x-2)(x+5)
Given:
(x - 2)/(x + 5)
Add zero in a fancy way (i.e. 0 = 5 - 5) to the numerator, so we can form a term we can cancel.
(x + 5 - 5 - 2)/(x + 5)
Regroup the fraction:
(x + 5)/(x + 5) - 7/(x + 5)
Since the first part of the fraction equals 1, everywhere except at x = -5, we can replace it with 1. The second term will take care of the forbidden value of x=-5, because -5 is also forbidden for it as well. Thus it simplifies to:
1 - 7/(x + 5)
@@carultchi want to learn more, seems like it's different subject title from we just learn in here, what is this?
@@Carrymejane I think I misread the OP. I thought the OP was asking about simplifying (x-2)/(x+5).
Anyway, Dave's example simplifies with similar principles.
Given:
3*x/(x + 5) - 6/(x - 2)
Change to common denominators and add:
3*x*(x - 2)/[(x - 2)(x + 5)] - 6*(x + 5)/[(x - 2)(x + 5)]
[3*x*(x - 2) - 6*(x + 5)]/[(x - 2)(x + 5)]
Expand numerator, and factor out the common factor of 3:
3*(x^2 - 4*x - 10)/((x-2)(x + 5))
This is the simplest form as a single fraction. Numerators "prefer" to be expanded, while denominators "prefer" to be factored. You may also pull out common constants out in front, where practical, like the 3.
@@Carrymejane You can take this further, if you want. For calculus applications, it's preferred to have single factors in each denominator, instead of a product of factors. Since the degree on the top is 2, and the degree on the bottom is also 2, we can expand it so it becomes a stand-alone constant, and a sum of partial fractions if we want. Dave has a video on partial fractions, but he doesn't cover my preferred shortcut, Heaviside Coverup.
In this case, it helps us, to look at the original given form. Had you been given the single solid fraction we just derived, you can expand the denominator, and manipulate the numerator so you can form a term we can cancel. This brings it to the form:
3*(x^2 + 3*x - 10)/(x^2 + 3*x - 10) - 21*x/[(x-2)(x + 5)]
Which allows us to cancel common factors in the first term and get:
3 - 21*x/[(x-2)(x + 5)]
Partial fractions and Heaviside cover-up, can simplify it from here.
For us, let's look back at the original form:
3*x/(x + 5) - 6/(x - 2)
The second term is already as simplified as we can get it. The first term, has a linear term on top of a linear term. We can manipulate it, to form a term we can cancel, by adding zero in a fancy way:
3*x/(x + 5) = 3*(x + 5 - 5)/(x + 5) = 3*(x+5)/(x+5) - 15/(x + 5)
Construct complete form, equal to the original expression:
3 - 15/(x + 5) - 6/(x - 2)
Now, we have the factors untangled from each other. It is now just a sum of elementary functions in as close to their pure forms as possible, and it is simple to do Calculus operations on it.
#1 is cool, can anyone solve #2?
It was infact way coolerr. Haha, I think he just went ahead put the 3x - 6 together on the numerator. Then factoring out 3(x-2) over (x+5)(x-2) resulting in 3/(x+5). :(
I appreciate the work, but those mistakes really messed me up. 1 or 2 hours down the drain trying to figure out what did I get wrong.
Somebody help me with the second one plss🥺🥺🥺
Read the comment, the questions happen to be wrong on the comprehension.
👍
❤❤❤
Maybe you'll get more views if you put subtitle? Maybe some Asian will like it. Try Indian And maybe add Indonesia
None of the videos thusfar have actually explained "factoring" outside the context of quadratics; it's getting less intuitive. I feel like I'm falling behind again.
There are formulas for factoring ANY cubic and ANY quartic, but they are complicated, and seldom taught outside a math degree. Mathologer has an excellent video on making sense of the cubic formula.
Usually, if you're working with a cubic or quartic in an academic example, it will be deliberately set up to have rational roots, and you can find them with the rational roots theorem.
Given any polynomial in the form of:
a*x^n + b*x^(n - 1) + ... (other terms) ... + k
Its roots will add to -b/a, and multiply to k/a for even polynomials, and multiply to -k/a for odd polynomials. This means, roots of the polynomial will be (integer factors of k)/(integer factors of a), and knowing the two properties above will help narrow them down from those options. Once you positively identify one rational root, you can use polynomial division to reduce the polynomial to a lower degree. This way, a cubic can factor as a linear term times a quadratic term, and the quadratic term has a simple formula for it.
He knows a lot about kinds of stuff
Professor Dave confuses )
💌