It isn't so tricky when you think of real coefficient even power polynomials being a product of x- complex conjugate pairs and/or x - real root solutions. That means with all x - complex conjugate pairs will form quadratic x^2s of real coefficients a, b and c in each of the ax^2 + bx + cs that equal factors in that total even power polynomial. All polynomial solutions of nth order odd polynomials will at least have a real root factor and maybe more real root solutions but still factor to an even polynomial of lessor power n-1 that will have quadratic expression factorization(s). I believe in the root finding algorithm Muller's Method of a polynomial of order n root iterations programming that it searches for all the ax^2 + bx + c factors out of even power polynomials of greater than order 4.
x⁴ + 5x² + 9 = x⁴ + 6x² + 9 - x² = (x² + 3)² - x² = (x² + 3 + x)(x² + 3 - x) = (x² + x + 3)(x² - x + 3) As neither of those terms have positive discriminants, they cannot be factored further without extending into the complex plane.
Know: e^-1=e^-1, (a^b)^c=a^(bc), (a/b)^-c=(b^c)/(a^c) e=lim n →∞(1+1/n)^n, e^z=lim n →∞(1+z/n)^n Hence, sub e^-1 into the second and fifth equations lim n →∞(1+1/n)^-n=lim n →∞[1+(-1/n)]^n lim n →∞[(n+1)/n]^-n=lim n →∞[(n-1)/n]^n lim n →∞[n/(n+1)]^n=lim n →∞[(n-1)/n]^n Simplify the powers, so lim n →∞[n/(n+1)]=lim n →∞[(n-1)/n)]……(1) Now multiply the denominators. lim n →∞(n*n)=lim n →∞(n+1)(n-1) lim n →∞(n^2)=lim n →∞(n^2-1) 0=-1??? Now solve ……(1) via another way. lim n →∞[n/(n+1)]=lim n →∞(1-1/n) lim n →∞[n/(n+1)+1/n]=1 lim n →∞[(n*n)/n(n+1)+(n+1)/n(n+1)]=1 lim n →∞[(n^2+n+1)/(n^2+n)]=1 lim n →∞(n^2+n+1)=lim n →∞(n^2+n)……(2) 0=1??? Now solve ……(2) via another method. lim n →∞[1+1/(n^2+n)]=1 lim n →∞[1/(n^2+n)]=0 lim n →∞(n^2+n)=1/0……(3) lim n →∞(n^2+n-1/0)=0 Now use the quadratic formula, note {+\-}=+or-,{x\y}=multiplied by x or y n=[-1{+\-}sqrt(1-4*1*(-1/0))]/2 2n=-1{+\-}sqrt(1+4/0) (note x/0={+\-}∞if x=/=0) =-1{+\-}sqrt(1*0/0{+\-}4/0)=-1{+\-}2/0 or-1{+\-}2i/0 n=-1/2{+\-}{i\1}1/0 Sub……(3). n=-1/2{+\-}{i\1}(n^2+n) so if we take the+1, we have n^2=-1/2, ∞={+\-}i/sqrt(2)??? If we take the {+\-}i, we have n=-1/2+i*n^2+in or n=-1/2-i*n^2-in For the 1st equation, n+1/2=in(n+1), which converges to n=in(n+1),1=in+i, in=1-i, n=-i-1, ∞=-i-1??? For the 2nd, n+1/2=-in(n+1), which converges to n=-in(n+1), 1=-in-i, -in=1+i, n=-i+1 ∞=-1+i??? So in conclusion: 0={+\-}1, ∞=-i{+\-}1, ∞={+\-}i/sqrt(2) Please Help Me Lmao
@@ISoldßinLadensViagraOnEbayఔ ... Here is something that will blow your mind: f(x) = x/x equals 1 for all real x ... But for 0? For x=0 we have L'Hopital's Rule the limit f(x--> 0) of x/x being 1/1 = 1. But how!? We already proved any number multiplied by 0 equals 0: (1/x) 0 = 0 for all x defined in real numbers!? ... Except at x = 0!? ... So maybe the answer was with complex numbers defined by b of z=a + ib ... But even then under angular rotation of a= 0 + i0 of 2nπ where n can't be 0 or else 0 + i0 is not = 0 ?? we will have a 2nπ/(2nπ) = 1 situation from all 2kπ/(2nπ) even for all k/n integer rather than quotients results. 😬 OUCH!!
I've carefully studied how algebra was taught in the early 1900s, and this would be considered a routine problem during this error. What amazed me most was that they covered material many mathematics undergraduates would be flummoxed by.
Looking at the problem I think I might have said “let me make this easier on myself. If I just erase the x’s and treat it as 16+4y^2+y^4 it will be easier and then I can add the x back in the end.” That is because it seems fairly clear to me that it has to be x^2 and x^2 and not something crazy with odd powers.
My students would have been confused by your example. For the 16, since you know the answer, you use 4*4. But 1*16 and 2*8 are also possibilities. The student needs a way to keep track of all of the possibilities so they don’t miss one. As they get more experience, they can do more in their head, but at first they need a structure.
OK Thanks, but just one observation in the beginning of the video the factor solution presented has the plus sign missing in both terms of factoring expression. There is no plus "+" sign between 4xˆ2 and yˆ2. The correct in my point of view should be 4xˆ2 + yˆ2. The missing plus sign could let students think the correct factorization has an implicit multiply sign "*" between these two terms, which is wrong.Thanks again.
I think the original question just forgot to add the plus sign in the title question, as it is there in the text... there are quite a few typos in the questions, so that isn't too far of a stretch.
Sir , i am in 9th class please tell me how can i send you my question i made a equation(i have answer) but not getting good answer by solvig equations please tell , love from india 😊❤
@@BraedenHill536 Close? Those two literally only meet, if either x = 0 or y = 0. So they aren't even similar. 16x⁴ - 8x²y² + y⁴ = 16x⁴ + 4x²y² + y⁴ |-16x⁴ -y⁴ -8x²y² = 4x²y² |+8x²y² 12x²y² = 0 |:12 x²y² = 0 zero-product property x² = 0 → x = 0 y² = 0 → y = 0
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Factor this: x^4+5x^2+9
Answer here: ruclips.net/video/naLIrU5lcdc/видео.htmlsi=hNgbP8xlqJPG2Oa1
It isn't so tricky when you think of real coefficient even power polynomials being a product of x- complex conjugate pairs and/or x - real root solutions. That means with all x - complex conjugate pairs will form quadratic x^2s of real coefficients a, b and c in each of the ax^2 + bx + cs that equal factors in that total even power polynomial. All polynomial solutions of nth order odd polynomials will at least have a real root factor and maybe more real root solutions but still factor to an even polynomial of lessor power n-1 that will have quadratic expression factorization(s). I believe in the root finding algorithm Muller's Method of a polynomial of order n root iterations programming that it searches for all the ax^2 + bx + c factors out of even power polynomials of greater than order 4.
x^4+5x^2+9=(x^2-x+3)(x^2+x+3)
x⁴ + 5x² + 9 = x⁴ + 6x² + 9 - x²
= (x² + 3)² - x²
= (x² + 3 + x)(x² + 3 - x)
= (x² + x + 3)(x² - x + 3)
As neither of those terms have positive discriminants, they cannot be factored further without extending into the complex plane.
Know: e^-1=e^-1, (a^b)^c=a^(bc), (a/b)^-c=(b^c)/(a^c)
e=lim n →∞(1+1/n)^n, e^z=lim n →∞(1+z/n)^n
Hence, sub e^-1 into the second and fifth equations
lim n →∞(1+1/n)^-n=lim n →∞[1+(-1/n)]^n
lim n →∞[(n+1)/n]^-n=lim n →∞[(n-1)/n]^n
lim n →∞[n/(n+1)]^n=lim n →∞[(n-1)/n]^n
Simplify the powers, so lim n →∞[n/(n+1)]=lim n →∞[(n-1)/n)]……(1)
Now multiply the denominators. lim n →∞(n*n)=lim n →∞(n+1)(n-1)
lim n →∞(n^2)=lim n →∞(n^2-1)
0=-1???
Now solve ……(1) via another way.
lim n →∞[n/(n+1)]=lim n →∞(1-1/n)
lim n →∞[n/(n+1)+1/n]=1
lim n →∞[(n*n)/n(n+1)+(n+1)/n(n+1)]=1
lim n →∞[(n^2+n+1)/(n^2+n)]=1
lim n →∞(n^2+n+1)=lim n →∞(n^2+n)……(2)
0=1???
Now solve ……(2) via another method.
lim n →∞[1+1/(n^2+n)]=1
lim n →∞[1/(n^2+n)]=0
lim n →∞(n^2+n)=1/0……(3)
lim n →∞(n^2+n-1/0)=0
Now use the quadratic formula, note {+\-}=+or-,{x\y}=multiplied by x or y
n=[-1{+\-}sqrt(1-4*1*(-1/0))]/2
2n=-1{+\-}sqrt(1+4/0) (note x/0={+\-}∞if x=/=0) =-1{+\-}sqrt(1*0/0{+\-}4/0)=-1{+\-}2/0 or-1{+\-}2i/0
n=-1/2{+\-}{i\1}1/0
Sub……(3).
n=-1/2{+\-}{i\1}(n^2+n)
so if we take the+1, we have n^2=-1/2, ∞={+\-}i/sqrt(2)???
If we take the {+\-}i, we have n=-1/2+i*n^2+in or n=-1/2-i*n^2-in
For the 1st equation, n+1/2=in(n+1), which converges to n=in(n+1),1=in+i, in=1-i, n=-i-1, ∞=-i-1???
For the 2nd, n+1/2=-in(n+1), which converges to n=-in(n+1), 1=-in-i, -in=1+i, n=-i+1 ∞=-1+i???
So in conclusion: 0={+\-}1, ∞=-i{+\-}1, ∞={+\-}i/sqrt(2)
Please Help Me Lmao
@@ISoldßinLadensViagraOnEbayఔ ... Here is something that will blow your mind: f(x) = x/x equals 1 for all real x ... But for 0? For x=0 we have L'Hopital's Rule the limit f(x--> 0) of x/x being 1/1 = 1. But how!? We already proved any number multiplied by 0 equals 0: (1/x) 0 = 0 for all x defined in real numbers!? ... Except at x = 0!? ... So maybe the answer was with complex numbers defined by b of z=a + ib ... But even then under angular rotation of a= 0 + i0 of 2nπ where n can't be 0 or else 0 + i0 is not = 0 ?? we will have a 2nπ/(2nπ) = 1 situation from all 2kπ/(2nπ) even for all k/n integer rather than quotients results. 😬 OUCH!!
I've carefully studied how algebra was taught in the early 1900s, and this would be considered a routine problem during this error. What amazed me most was that they covered material many mathematics undergraduates would be flummoxed by.
@IonRuby Yep, that was an error, and I should have typed 'era".
Just curious In what grade/grade level would students be expected to be able to factor this? 7th, 8th or 9th
This would be probably Algebra 2 level, which for most students is between 9th and 11th grade.
Not a bad example to show why factoring in two variables is hard.
Looking at the problem I think I might have said “let me make this easier on myself. If I just erase the x’s and treat it as 16+4y^2+y^4 it will be easier and then I can add the x back in the end.” That is because it seems fairly clear to me that it has to be x^2 and x^2 and not something crazy with odd powers.
My students would have been confused by your example. For the 16, since you know the answer, you use 4*4. But 1*16 and 2*8 are also possibilities. The student needs a way to keep track of all of the possibilities so they don’t miss one. As they get more experience, they can do more in their head, but at first they need a structure.
Hey, I saw the poster of the number E on your wall in one of your videos, I wonder where I can find it?
OK Thanks, but just one observation in the beginning of the video the factor solution presented has the plus sign missing in both terms of factoring expression. There is no plus "+" sign between 4xˆ2 and yˆ2. The correct in my point of view should be 4xˆ2 + yˆ2. The missing plus sign could let students think the correct factorization has an implicit multiply sign "*" between these two terms, which is wrong.Thanks again.
I think the original question just forgot to add the plus sign in the title question, as it is there in the text... there are quite a few typos in the questions, so that isn't too far of a stretch.
Er, the conclusion you get at the end doesn't match the problem the original poster had
Sir , i am in 9th class please tell me how can i send you my question i made a equation(i have answer) but not getting good answer by solvig equations please tell , love from india 😊❤
I did it!
My answer was (2x+y)²(2x-y)²
If you multiply out the terms that gives: 16x^4 - 8x^2y^2 +y^4, not 16x^4 + 4x^2y^2 + y^4. Close but not correct
@@BraedenHill536 Close? Those two literally only meet, if either x = 0 or y = 0. So they aren't even similar.
16x⁴ - 8x²y² + y⁴ = 16x⁴ + 4x²y² + y⁴ |-16x⁴ -y⁴
-8x²y² = 4x²y² |+8x²y²
12x²y² = 0 |:12
x²y² = 0
zero-product property
x² = 0 → x = 0
y² = 0 → y = 0
@@m.h.6470 You know what I meant
how tf you you you so smart ??!?1!?!!! i am dumb please make video about permutation and combination i beg you sir