No, I think you may be looking at the line where I evaluated when x = -1. The original equation (in red) has coefficients 1 -1 -10 -8 which is what I used in the synthetic division.
Thanks again for these videos, i have a question, if the divisor was a quadradic, can i still use the remainder theorem to find my remainder or will i have to use long division to find the remainders.
So sorry … novice TH-camr mistake! I’m sure you can still understand from what I was saying where the function would go. If not go to Desmos and enter the function and it will give you the exact sketch 😊
At 18:08 why did you use the remainder theorem again when you already used it with the initial expression x^4+2x^3-23x^2-24x+144. Was it because you needed to find the zeros to sketch the graph or am I missing something?
In this question you are factoring the quartic function in order to graph it. We use the FACTOR theorem testing possible factors, so if f(3) = 0 the (x-3) is a factor etc. Finding one factor at a time we work to find all possible roots. In this case there are 2 double roots, one double root at x = 3 and one double root at x = -4. Does that help?
@@mshavrotscanadianuniversit6234 If the question did not ask to graph and instead, asked to simply divide using synthetic division, would you have written the division statement like this: x^4+2x^3-23x^2-24x+144=(x-3)(x^3+5x^2-8x-48)?
Doesn't the remainder theorem in a way contradict the restriction we use in long division? If the divisor is for example x+4, we put the restriction as x≠ -4. But here in this video, we are substituting the whole equation with the same ≠-4. Why is that?
You are not creating a rational function but rather simply using division to factor. For example if I had x^2 -2x and I asked you to factor it you would say x(x - 2) but would not say that x could not be equal to zero.
I cannot solve this question: factor the following polynomial using the factor theorem 4x^4-7x^3-80x^2-21x+270 I figured out (x+2) is a factor and after synthetic division I got the following polynomial: 4x^3-x^2-78x+135 Now I'm confused on how to factor it correctly as I tried to many times but it won't match the answer at the back of the textbook
Aha! I see it should have been a plus before the cubic term. I’ll get back to you shortly. Just making breakfast 😊. Did you try f(-5) to see if it is a zero?
@@mshavrotscanadianuniversit6234 No I have not. I was taking too long on this question so I left it to ask you! I think a solution would help me comprehend solving it better as the textbook only provides the answer
Okay, so you have no problem with synthetic division or finding one of the factors using the factor theorem and the first part of your solution is perfect. All you needed to do was to repeat what you started and keep going until you have a quadratic function, which you can either factor or use the quadratic formula to solve. Here is the worked problem showing all steps: mshavrot.pbworks.com/f/p%20177%206d.pdf
My average in math skyrocketed after I discovered this Channel
That’s amazing! Good for you!!!
This channel is a blessing and the teacher is an angel.
And you are made in Heaven! ❤️ Thank you 😊
at 15:25 why are you finding factors for a quartic function? Do you need to find factors even if the function isn't quartic?
You find factors so that you can find the zeroes of the function and be able to graph it using them and the other clues that I list.
The graph at 9:48 is cut off in the video :(
Would you like me to graph it for you? Looks like I forgot to shift the paper up 😳
@@mshavrotscanadianuniversit6234 If that's not too much trouble, yes please!
mshavrot.pbworks.com/w/file/fetch/146425371/question%20from%203.6.pdf
Don't forget that you can always check your answers on DESMOS : )
@@mshavrotscanadianuniversit6234 Thank you so much! Have a nice day
Amazing Video! Very clear and at an amazing pace. God bless.
Thank you! Are you preparing for the fall?
@@mshavrotscanadianuniversit6234Yup! And your videos have been making studying so much easier 😁
Good for you! You will be able to teach the teacher! 😂
Thank you for your video :)
How come at 8:05
The leading coefficient is now positive when it was previously negative?
Wasn’t it -1 -1 + 10 - 8
So why is it now 1 - 1 + 10 - 8?
No, I think you may be looking at the line where I evaluated when x = -1. The original equation (in red) has coefficients 1 -1 -10 -8 which is what I used in the synthetic division.
@@mshavrot_math oh I see, thank you miss!
Thanks again for these videos, i have a question, if the divisor was a quadradic, can i still use the remainder theorem to find my remainder or will i have to use long division to find the remainders.
There is a way to do it, but it is lengthy. Much faster to simply use long division.
thank you very much for your clear and helpful explanation, now i feel confident for my test
Confidence is half the battle! Good luck!!
please i cannot see the first graph you did it was below the camera
Send me an email and I’ll draw you a picture OR you can use Desmos online
Sorry!
@@mshavrotscanadianuniversit6234 thank you sooooo much i did it don't worry
Another banger video queen
Thanks ❤️
What is if the first term doesnt have a coefficient of 1?
Fortunately it's not in the curriculum : )
9:48 where graph ; - ;
So sorry … novice TH-camr mistake! I’m sure you can still understand from what I was saying where the function would go. If not go to Desmos and enter the function and it will give you the exact sketch 😊
@@mshavrotscanadianuniversit6234 I still understood Ofc! You are an amazing teacher and I wish I were my teacher at school😭
I’ll be your virtual teacher ❤️
Miss how would you factor 4x^3+4x^2-x-1
Use the factor theorem . I am just returning to Canada so if you are still stuck let me know.
At 18:08 why did you use the remainder theorem again when you already used it with the initial expression x^4+2x^3-23x^2-24x+144. Was it because you needed to find the zeros to sketch the graph or am I missing something?
In this question you are factoring the quartic function in order to graph it. We use the FACTOR theorem testing possible factors, so if f(3) = 0 the (x-3) is a factor etc. Finding one factor at a time we work to find all possible roots. In this case there are 2 double roots, one double root at x = 3 and one double root at x = -4. Does that help?
@@mshavrotscanadianuniversit6234 If the question did not ask to graph and instead, asked to simply divide using synthetic division, would you have written the division statement like this: x^4+2x^3-23x^2-24x+144=(x-3)(x^3+5x^2-8x-48)?
Yes that is true, but normally you would be asked to factor fully.
Doesn't the remainder theorem in a way contradict the restriction we use in long division? If the divisor is for example x+4, we put the restriction as x≠ -4. But here in this video, we are substituting the whole equation with the same ≠-4. Why is that?
you from Mr Lee summer school right??
You are not creating a rational function but rather simply using division to factor. For example if I had x^2 -2x and I asked you to factor it you would say x(x - 2) but would not say that x could not be equal to zero.
@@mshavrotscanadianuniversit6234 Alright thank you!!
@@classicnnamdi Nope! Mr. Mason's summer school.
What happens in the L.C is greater than 1 ?
I’m not sure what you are asking … could you point to a time in the video?
I can't believe I just discovered this
Spread the word and save your friends! 😂
I cannot solve this question: factor the following polynomial using the factor theorem
4x^4-7x^3-80x^2-21x+270
I figured out (x+2) is a factor and after synthetic division I got the following polynomial:
4x^3-x^2-78x+135
Now I'm confused on how to factor it correctly as I tried to many times but it won't match the answer at the back of the textbook
You must have copied the question down incorrectly ... can you refer to the question page and number?
@@mshavrotscanadianuniversit6234 Page 177 question 6d (and 6f is also similar) within the Nelson Advanced Functions textbook
Aha! I see it should have been a plus before the cubic term. I’ll get back to you shortly. Just making breakfast 😊. Did you try f(-5) to see if it is a zero?
@@mshavrotscanadianuniversit6234 No I have not. I was taking too long on this question so I left it to ask you! I think a solution would help me comprehend solving it better as the textbook only provides the answer
Okay, so you have no problem with synthetic division or finding one of the factors using the factor theorem and the first part of your solution is perfect. All you needed to do was to repeat what you started and keep going until you have a quadratic function, which you can either factor or use the quadratic formula to solve. Here is the worked problem showing all steps:
mshavrot.pbworks.com/f/p%20177%206d.pdf