It's tough to overestimate how much essential algebra is in this great video lesson. You start out using the remainder theorem, then go to the rational roots theorem, the 3 solutions found demonstrate the fundamental theorem of algebra, and the graph shows the intermediate value theorem in action.
You could justify your choice for pursuing a negative x solution by appealing to Descartes Rule of Signs. This rule says ffor this cubic equation that there is a negative real root, and there could be two positive roots or no positive roots. So, since one negative solution is guaranteed if the equation has a rational root, then try x = -3 or x = -9 first. Also, given the coefficients of the x³, x² and the x and the constant terms are pretty similar in magnitude, then the x³ term is going to grow quickly compared two the square and linear term, which favours trying x = -3 as our first choice.
You are very welcome Izyl! I'm sure you are an awesome student 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and all the best😃 Keep smiling😊 Have a very happy and blessed New Year!
PreMath I've been watched most of the vids to make my vacation productive and I can say this channel is my partner in most of my lectures,that I can use as a future Engineer
Thank you AARON DEMETERIO once again. I'm also on vacation and have decided to make quality vid tutorials. I'm blessed and honored to have an audience like you. Please share my channel with your friends as well. Kind regards
PreMath I will share it with my co-students to find out that there's TH-cam TH-cam channel like this.. Again feel my sincere thank you and may you continue to create more educational vids.
Some Hindu/Indians recorded video on youtube for it You have equation a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}=0 Firstly you want to get rid od x^2 term so you can decompose it to the sum of binomial powers a_{3}(x+a_{2}/(3a_{3}))^3+a_{3}p(x+a_{2}/(3a_{3}))+a_{3}q=a_{3}x^3+a_{2}x^2+a_{1}x+a_{0} In this step Horner's rule is helpful if you do it repeatedly You have equation in the form y^{3}+py+q = 0 Let y = u+v (u+v)^3+p(u+v)+q = 0 u^3+3u^2v+3uv^2+v^3+p(u+v)+q = 0 u^3+v^3 + 3(u+v)uv+p(u+v)+q = 0 u^3 + v^3 + q + 3(u+v)(uv+p/3) = 0 ---------------------------------------------------------- u^3 + v^3 + q = 0 3(u+v)(uv+p/3)=0 Here we are not in the ring with divisors of zero so product will be zero if at least one factor is zero but we assumed that y = u + v so we can not equate it to zero otherwise we lose generality u^3 + v^3 + q =0 uv+p/3 = 0 u^3 + v^3 = -q uv = -p/3 u^3 + v^3 = -q u^3v^3 = -p^3/27 Now this system is in fact Vieta formulas for quadratic with the roots u^3 and v^3 so we can immediately write this quadratic t^2+qt-p^3/27=0 u^3 = (-q - sqrt(q^2+4p^3/27)/2 v^3 = (-q + sqrt(q^2+4p^3/27))/2 u^3 = -q/2 - sqrt(q^2/4 + p^3/27) v^3 = -q/2 + sqrt(q^2/4 + p^3/27) In complex numbers there are three cube roots This gives you nine possible ways to create pair (u,v) but only three of them will work Your pair (u,v) must solve following system of equations u^3 + v^3 = -q uv = -p/3 so to choose correct pair (u , v) you must check equation uv = -p/3 If u and v satisfies this equation then you will choose correct pair (u,v)
Thank you so much Hemanshu Parekh for taking the time to leave this comment. I'm glad you liked it! Your feedback is always appreciated. Please keep sharing my channel with your family and friends. Take care dear and all the best 😃
These laws at the beggining doesn't make any sense. In another video, the equation x^3 + 11x^2 - 25x + 13 is presented with -1 as a solution. If you put that "alternate coefficient sums should be the same in order for x= -1", the solution should not be -1. Hopung for some kind of answer. :/
I think it's something like: this alternative sums law tells you that one of the roots is -1 as long as the alternative sums of the coefficients equals to -1. But it doesn't mean that every equation that has -1 as its root works in the light of that law.
*ALTERNATE METHOD* Fast forwarding to the quadratic formula. We can use the Discriminant Rule for the quadratic equation instead of the former. x²-2x+3 D= (-2)²-4(1)(3) D=4-12 D= -8 It turns out that there are no real solutions on the quadratic equation since D
It's tough to overestimate how much essential algebra is in this great video lesson. You start out using the remainder theorem, then go to the rational roots theorem, the 3 solutions found demonstrate the fundamental theorem of algebra, and the graph shows the intermediate value theorem in action.
Thanks MathGives YouPower for the feedback. Thank you so much for taking the time to leave this comment. Kind regards :)
Very clear and easy to understand. Thanks for the video
You could justify your choice for pursuing a negative x solution by appealing to Descartes Rule of Signs.
This rule says ffor this cubic equation that there is a negative real root, and there could be two positive roots or no positive roots. So, since one negative solution is guaranteed if the equation has a rational root, then try x = -3 or x = -9 first.
Also, given the coefficients of the x³, x² and the x and the constant terms are pretty similar in magnitude, then the x³ term is going to grow quickly compared two the square and linear term, which favours trying x = -3 as our first choice.
Thankyou so much for teaching a step by step process!
You are very welcome Izyl! I'm sure you are an awesome student 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and all the best😃 Keep smiling😊
Have a very happy and blessed New Year!
It's really help me in my advance study. So easy to understand.
Great! Thanks AARON DEMETERIO for the feedback. Your input is always appreciated. I'm sure you are an awesome student. Keep it up :)
PreMath I've been watched most of the vids to make my vacation productive and I can say this channel is my partner in most of my lectures,that I can use as a future Engineer
Thank you AARON DEMETERIO once again. I'm also on vacation and have decided to make quality vid tutorials. I'm blessed and honored to have an audience like you. Please share my channel with your friends as well. Kind regards
PreMath I will share it with my co-students to find out that there's TH-cam TH-cam channel like this.. Again feel my sincere thank you and may you continue to create more educational vids.
There must be a better solution instead of trial and error?
Exactly what I said
Can we use Lagrange’s Resolvent in this case?
Professor, could you solve it with cardano’s formula in another video? Thank you sir
Some Hindu/Indians recorded video on youtube for it
You have equation
a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}=0
Firstly you want to get rid od x^2 term so you can decompose it to the sum of binomial powers
a_{3}(x+a_{2}/(3a_{3}))^3+a_{3}p(x+a_{2}/(3a_{3}))+a_{3}q=a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}
In this step Horner's rule is helpful if you do it repeatedly
You have equation in the form
y^{3}+py+q = 0
Let y = u+v
(u+v)^3+p(u+v)+q = 0
u^3+3u^2v+3uv^2+v^3+p(u+v)+q = 0
u^3+v^3 + 3(u+v)uv+p(u+v)+q = 0
u^3 + v^3 + q + 3(u+v)(uv+p/3) = 0
----------------------------------------------------------
u^3 + v^3 + q = 0
3(u+v)(uv+p/3)=0
Here we are not in the ring with divisors of zero so product will be zero if at least one factor is zero
but we assumed that y = u + v so we can not equate it to zero otherwise we lose generality
u^3 + v^3 + q =0
uv+p/3 = 0
u^3 + v^3 = -q
uv = -p/3
u^3 + v^3 = -q
u^3v^3 = -p^3/27
Now this system is in fact Vieta formulas for quadratic with the roots u^3 and v^3
so we can immediately write this quadratic
t^2+qt-p^3/27=0
u^3 = (-q - sqrt(q^2+4p^3/27)/2
v^3 = (-q + sqrt(q^2+4p^3/27))/2
u^3 = -q/2 - sqrt(q^2/4 + p^3/27)
v^3 = -q/2 + sqrt(q^2/4 + p^3/27)
In complex numbers there are three cube roots
This gives you nine possible ways to create pair (u,v)
but only three of them will work
Your pair (u,v) must solve following system of equations
u^3 + v^3 = -q
uv = -p/3
so to choose correct pair (u , v)
you must check equation uv = -p/3
If u and v satisfies this equation then you will choose correct pair (u,v)
x= -3 satisfy the equation hence (x+3) is a factor of the expression.
Second factor is (x^2-2x+3)
Solve it quadratic formula
On the graph, where is i, the unit circle?
sir kindly solve the equation 18x^3-21x^2+8x-2=0
Thank you so much
Thanks! Cheers! 😀
You are awesome. Keep it up 👍
What is the remainder is not 0?
I'm still looking how to solve that one
it means that the number you putted is not a root of the equation. if you want to learn more, check out the briot-ruffini method.
thank you sir
#CubicEquation #factoring #factor
a = 1
b = -2
c = 3
You should use the term "Real number" solution and "Complex number" solution in your videos as the solutions are all "Real" solutions.
Reals cross the x-axis. Complex solutions have dips, a max followed by min or a min followed by a max, that do not cross the x-axis.
#syntheticdivision #longdivision
sir you are gate one
A + B + C + D = 8
Nice
Thank you so much Hemanshu Parekh for taking the time to leave this comment. I'm glad you liked it! Your feedback is always appreciated. Please keep sharing my channel with your family and friends. Take care dear and all the best 😃
These laws at the beggining doesn't make any sense. In another video, the equation x^3 + 11x^2 - 25x + 13 is presented with -1 as a solution. If you put that "alternate coefficient sums should be the same in order for x= -1", the solution should not be -1. Hopung for some kind of answer. :/
I think it's something like: this alternative sums law tells you that one of the roots is -1 as long as the alternative sums of the coefficients equals to -1. But it doesn't mean that every equation that has -1 as its root works in the light of that law.
-1 is not a solution to x^3 + 11x^2 - 25x + 13 = 0 Plug in -1 instead of x...
#QuadraticEquation
plz reply sir 🙏
*ALTERNATE METHOD*
Fast forwarding to the quadratic formula.
We can use the Discriminant Rule for the quadratic equation instead of the former.
x²-2x+3
D= (-2)²-4(1)(3)
D=4-12
D= -8
It turns out that there are no real solutions on the quadratic equation since D
Square root of -8 = error
imaginary numbers involved
x^(2) - 2x + 3 = 0
#imaginarynumber #iota #i
Nc bro
x = -3
yes