Wanna learn a different method? Factor x^4-4x^3+2x^2-11x+12 By rational zero theorem: ruclips.net/video/yx2RetjV1Bo/видео.html By grouping: ruclips.net/video/55ufNfFofzY/видео.html The double-cross method: ruclips.net/video/B8dCd6PkHMY/видео.html
Okay, then tell me how'd you find the roots using this expression or would cancel the terms when presented in the numerator or denominator of a rational expression using your factorization of this expression?
A quartic can be factored into the product of two quadratics with rational coefficients if and only if a linear combination of the roots of the form a+b-c-d is rational which corresponds to the cubic polynomial having the squares of a+b-c-d, a+c-b-d and a+d-b-c as roots has a rational root that is the square of a rational. For this one the cubic has 1156/25 as a root which is the square of 34/5 so the quartic can be factorized into two quadratics with integer coefficient! If the 3 roots satisfy this property then there are actually three factorizations into two quadratics with integer coefficient which corresponds to the fact that the quartic only has rational roots.
They are formulas for third and fourth power linear polynomials .Rational root theorem will tell tell you if any degree polynomial has rational roots. In other words if polynomial of degree n Ax^n+Bx^(n-1)+Cx^(n-2) ......Z=0 with integer coefficients has at least one rational root it must be a factor (ax+b) with integer coefficients where a is a factor of A and b is a factor of Z. If you can find one real factor you can divide and reduce the degree by one. The binomial theorem and pascal's triangle can help you choose coefficients and substitutions for sums and differences of squares and cubes and make alternating terms disappear. You can also try some complex analysis where you substitute a complex variable a+bi for x and split it into at least two simultaneous equations of real and imaginary parts. Then there is linear algabra wher you can solve sytems of equations using matrices, determinants and inverses.
Here is an alternative method to do this, actually Ferrari's method for solving quartic equations adapted to factoring a quartic polynomial. The quartic polynomial to factor is 5x⁴ − 26x³ − 11x² − 10x + 6 and we are given that this polynomial has no rational zeros, but it does factor into two quadratics with integer coefficients. The task is to find this factorization. To avoid having to deal with either irrational or fractional coefficients later on we first multiply this polynomial by _four times the coefficient of the leading term_ which is 4·5 = 20. Of course, when we are done we will have to divide the result again by 20 to obtain the factorization of the original quartic. Multiplying by 20 we have 100x⁴ − 520x³ − 220x² − 200x + 120 Now we have 100x⁴ = (10x²)² and 520x³ = 2·(10x²)·26x which is twice the product of 10x² and 26x. Therefore, we can complete the square with respect to the quartic and cubic terms by adding and subtracting (26x)² = 676x² since, applying the identity a² − 2·a·b + b² = (a − b)², we have 100x⁴ − 520x³ + 676x² = (10x²)² − 2·(10x²)·26x + (26x)² = (10x² − 26x)² so we get (10x² − 26x)² − 676x² − 220x² − 200x + 120 (10x² − 26x)² − 896x² − 200x + 120 (10x² − 26x)² − (896x² + 200x − 120) We have now written our quartic polynomial as a _difference of two terms_ where the first term is a perfect square but the second term is not. The difference of two terms will remain the same if we add the same quantity to each of the two terms. The idea is to add something to both terms in such a way that the first term will _remain_ a perfect square and the second term will _become_ a perfect square as well. We want to create a difference of two squares because we can then use the difference of two squares identity a² − b² = (a + b)(a − b) to factor our polynomial into two quadratics. Now, if we take _any_ number λ and add 2λ(10x² − 26x) + λ² = 20λx² − 52λx + λ² to both terms, then the first term will _remain_ a perfect square because, applying the identity a² + 2·a·b + b² = (a + b)², we then have (10x² − 26x)² + 2·(10x² − 26x)·λ + λ² = (10x² − 26x + λ)². Adding 2λ(10x² − 26x) + λ² = 20λx² − 52λx + λ² to each of the two terms of (10x² − 26x)² − (896x² + 200x − 120) we therefore get (10x² − 26x + λ)² − ((20λ + 896)x² + (200 − 52λ)x + (λ² − 120)) Since the first term (10x² − 26x + λ)² is a perfect square regardless of the value of λ we are free to choose λ in such a way that the second term (20λ + 896)x² + (200 − 52λ)x + (λ² − 120), which is a quadratic polynomial in x, will also become a perfect square. A quadratic polynomial ax² + bx + c (a ≠ 0) is a perfect square, that is, the square of a linear polynomial in x, if and only if this quadratic has two coinciding and therefore identical zeros. This is the case if and only if the _discriminant_ b² − 4ac of the quadratic is equal to zero. For our quadratic polynomial (20λ + 896)x² + (200 − 52λ)x + (λ² − 120) we have a = 20λ + 896, b = 200 − 52λ, c = λ² − 120 so this will be a perfect square if and only if λ satisfies (200 − 52λ)² − 4(20λ + 896)(λ² − 120) = 0 This is a cubic equation in λ but we _do not need to solve this equation_ considering that our quartic polynomial has a factorization into two quadratics with integer coefficients. If there exists a value of λ that makes (20λ + 896)x² + (200 − 52λ)x + (λ² − 120) the square (mx + n)² = m²x² + 2mnx + n² of a linear polynomial mx + n with _integer_ coefficients m and n, then both the coefficient 20λ + 896 of x² and the constant term λ² − 120 must be squares of integers. Clearly, if λ² − 120 is to be the square of an integer then λ² must be an integer. But λ itself cannot be irrational because 20λ + 896 also needs to be the square of an integer, and therefore an integer. So, if there exists a value of λ which makes both 20λ + 896 and λ² − 120 the square of an integer, then this value of λ must be an integer. But how do we go about finding integer values of λ which will make both 20λ + 896 and λ² − 120 the square of an integer? Well, if λ² − 120 is the square of some integer μ for some integer λ, that is, λ² − 120 = μ², then 120 = λ² − μ² must be a difference of two squares of integers. Now, any product of two quantities p and q can be rewritten as a difference of squares since pq = ((p + q)/2 + (p − q)/2)((p + q)/2 − (p − q)/2) = ((p + q)/2)² − ((p − q)/2)² Clearly, if (p + q)/2 and (p − q)/2 are integers then their sum p and their difference q must be integers and, moreover, p and q must be integers of like parity if (p + q)/2 and (p − q)/2 are to be integers. Since 120 = 2³·3·5 is even this number evidently cannot be written as the product of two odd integers but we can write 120 as the product of two even integers in several ways, and using the identity above we have 120 = 60·2 = (31 + 29)(31 − 29) = 31² − 29² 120 = 30·4 = (17 + 13)(17 − 13) = 17² − 13² 120 = 20·6 = (13 + 7)(13 − 7) = 13² − 7² 120 = 12·10 = (11 + 1)(11 − 1) = 11² − 1² which exhausts all possibilities of writing 120 as a difference of two squares of integers. Evidently if λ² − 120 is the square of some integer μ for some integer λ and therefore 120 = λ² − μ² then λ must be one of the numbers ±11, ±13, ±17, ±31 so we need to test at most eight integer values of λ to see if this will make 20λ + 896 a square as well. Starting with λ = 11 and λ = −11 we have 20λ + 896 = 20·11 + 896 = 1116 and 20·(−11) + 896 = 776 which are not squares of integers since 33² = 1089 < 1116 < 34² = 1156 and 27² = 729 < 776 < 28² = 784. But with λ = 13 we have 20λ + 896 = 20·13 + 896 = 1156 = 34², as required. With λ = 13 our polynomial (10x² − 26x + λ)² − ((20λ + 896)x² + (200 − 52λ)x + (λ² − 120)) becomes (10x² − 26x + 13)² − (1156x² − 476x + 49) which is (10x² − 26x + 13)² − (34x − 7)² and we have converted our quartic polynomial into a difference of two squares. Using the difference of two squares identity a² − b² = (a + b)(a − b) this can be written as (10x² − 26x + 13 + 34x − 7)(10x² − 26x + 13 − 34x + 7) which is (10x² + 8x + 6)(10x² − 60x + 20) and we have factored our quartic into two quadratics. Now note that the first quadratic has a factor 2 and that the second quadratic has a factor 10 we can take out to get 20(5x² + 4x + 3)(x² − 6x + 2) and since we started out by multiplying the polynomial to factor by 20 this means that the original quartic polynomial 5x⁴ − 26x³ − 11x² − 10x + 6 factors as (5x² + 4x + 3)(x² − 6x + 2)
@@bimokayoba The equation 4x⁴ − 12x³ + 5x² + 6x − 64 = 0 is easy to solve. First observe that 4x⁴ = (2x²)² is the square of 2x² and that 12x³ = 2·2x²·3x is twice the product of 2x² and 3x so we can complete the square with respect to 4x⁴ − 12x³ by adding and subtracting the square (3x)² = 9x² of 3x. This gives 4x⁴ − 12x³ + 9x² − 9x² + 5x² + 6x − 64 = 0 which can be written as (2x² − 3x)² − 4x² + 6x − 64 = 0 Next observe that −4x² + 6x = −2(2x² − 3x) so we can rewrite the equation as (2x² − 3x)² −2(2x² − 3x) − 64 = 0 Now note that (2x² − 3x)² is the square of 2x² − 3x and that 2(2x² − 3x) = 2·(2x² − 3x)·1 is twice the product of 2x² − 3x and 1 so we can again complete the square, this time with respect to (2x² − 3x)² − 2(2x² − 3x), by adding and subtracting the square 1² of 1. This gives (2x² − 3x)² −2(2x² − 3x) + 1² − 1 − 64 = 0 which can be written as (2x² − 3x − 1)² − 65 = 0 or (2x² − 3x − 1)² = 65 Of course, the solution of the equation is now easy since this gives 2x² − 3x − 1 = √65 ⋁ 2x² − 3x − 1 = −√65 2x² − 3x − (1 + √65) = 0 ⋁ 2x² − 3x − (1 − √65) = 0 and then we only need to solve these two quadratic equations using the quadratic formula, which gives x = (3 + √(17 + 8√65))/4 ⋁ x = (3 + √(17 + 8√65))/4 ⋁ x = (3 + √(17 − 8√65))/4 ⋁ x = (3 − √(17 − 8√65))/4 The first two solutions are real but since 17 − 8√65 < 0 the last two solutions are complex so we have x = (3 + √(17 + 8√65))/4 ⋁ x = (3 + √(17 + 8√65))/4 ⋁ x = (3 + i√(−17 + 8√65))/4 ⋁ x = (3 − i√(−17 + 8√65))/4 and your equation is solved.
If I set this expression to zero and note that there are two sign changes, there should be at least two real roots. I can use fixed-point iteration or Newton-Raphson iteration to obtain numerical values for these two roots. Then synthetically divide to obtain a quadratic, from which the remaining two roots may be obtained. Using these four roots I can then obtain the desired factors of the original expression if I really needed them.
There is absolutely a quartic discriminant. Every polynomial has a discriminant. But It is complicated and alone won’t tell you if the quartic in factorable. You CAN tell using the resolvent if you turn the quartic into a depressed quartic (without a cubic term). This is not super simple to do, but can be done.
I had a class about factor polinomials like this (others too) and taught to my a class we have another way which is like factoring another polynomial. You reach a polynomial that its quadratic term is close to that polynomial then factor. (????)
@bimokayoba The equation 4x⁴ − 12x³ + 5x² + 6x − 64 = 0 is easy to solve. First observe that 4x⁴ = (2x²)² is the square of 2x² and that 12x³ = 2·2x²·3x is twice the product of 2x² and 3x so we can complete the square with respect to 4x⁴ − 12x³ by adding and subtracting the square (3x)² = 9x² of 3x. This gives 4x⁴ − 12x³ + 9x² − 9x² + 5x² + 6x − 64 = 0 which can be written as (2x² − 3x)² − 4x² + 6x − 64 = 0 Next observe that −4x² + 6x = −2(2x² − 3x) so we can rewrite the equation as (2x² − 3x)² −2(2x² − 3x) − 64 = 0 Now note that (2x² − 3x)² is the square of 2x² − 3x and that 2(2x² − 3x) = 2·(2x² − 3x)·1 is twice the product of 2x² − 3x and 1 so we can again complete the square, this time with respect to (2x² − 3x)² − 2(2x² − 3x), by adding and subtracting the square 1² of 1. This gives (2x² − 3x)² −2(2x² − 3x) + 1² − 1 − 64 = 0 which can be written as (2x² − 3x − 1)² − 65 = 0 or (2x² − 3x − 1)² = 65 Of course, the solution of the equation is now easy since this gives 2x² − 3x − 1 = √65 ⋁ 2x² − 3x − 1 = −√65 2x² − 3x − (1 + √65) = 0 ⋁ 2x² − 3x − (1 − √65) = 0 and then we only need to solve these two quadratic equations using the quadratic formula, which gives x = (3 + √(17 + 8√65))/4 ⋁ x = (3 + √(17 + 8√65))/4 ⋁ x = (3 + √(17 − 8√65))/4 ⋁ x = (3 − √(17 − 8√65))/4 The first two solutions are real but since 17 − 8√65 < 0 the last two solutions are complex so we have x = (3 + √(17 + 8√65))/4 ⋁ x = (3 + √(17 + 8√65))/4 ⋁ x = (3 + i√(−17 + 8√65))/4 ⋁ x = (3 − i√(−17 + 8√65))/4 and your equation is solved.
I’m genuinely not trying to insult anyone, it’s only a bit of roasting, but I love how these mathematicians can do integrals and trigonometry without looking, then ask them to do a (fairly) simple multiplication, they freeze.
if you mean getting the roots, then probably the rational roots theorem but I'm too lazy to check. if you mean factoring it to two quadratics, then i have no idea
@@codexcursorsno. It says in the title that there are no rational zeroes here. However it is possible to factor into two quadratics in other ways just much much more complicated.
Wanna learn a different method? Factor x^4-4x^3+2x^2-11x+12
By rational zero theorem: ruclips.net/video/yx2RetjV1Bo/видео.html
By grouping: ruclips.net/video/55ufNfFofzY/видео.html
The double-cross method: ruclips.net/video/B8dCd6PkHMY/видео.html
I'm a big fan of you! I'm 15 years old and you inpires me! Thank you for your channel! Hello from Brazil!
Tomorrow I will have 15
Xd
Eu sou do Brasil, também!!! Eu tenho 11 anos. Bom ver mais gente de nossa nação!!!
@@pimpaovieira No, I am from Ecuador
Way too complicated. Heres how to factor 5x^4 + -26x^3 -11x^2 -10x +6: ANSWER: 1(5x^4 + -26x^3 -11x^2 -10x +6)
didnt quite finish, could have gone for 1(1(5x^4 + -26x^3 -11x^2 -10x +6)) smh
Kkkkkkkk
Okay, then tell me how'd you find the roots using this expression or would cancel the terms when presented in the numerator or denominator of a rational expression using your factorization of this expression?
r/whoosh@@abdulkhadarainur4348
@@abdulkhadarainur4348 simple. if you multiply the polynomial by 1 enough times, it will vanish off to the right, into infinity. so the root is 1
A quartic can be factored into the product of two quadratics with rational coefficients if and only if a linear combination of the roots of the form a+b-c-d is rational which corresponds to the cubic polynomial having the squares of a+b-c-d, a+c-b-d and a+d-b-c as roots has a rational root that is the square of a rational. For this one the cubic has 1156/25 as a root which is the square of 34/5 so the quartic can be factorized into two quadratics with integer coefficient! If the 3 roots satisfy this property then there are actually three factorizations into two quadratics with integer coefficient which corresponds to the fact that the quartic only has rational roots.
I just learned this in my class sometimes ago
Well duh
Thank you for making maths so enjoyable!!!
They are formulas for third and fourth power linear polynomials .Rational root theorem will tell tell you if any degree polynomial has rational roots. In other words if polynomial of degree n Ax^n+Bx^(n-1)+Cx^(n-2) ......Z=0 with integer coefficients has at least one rational root it must be a factor (ax+b) with integer coefficients where a is a factor of A and b is a factor of Z. If you can find one real factor you can divide and reduce the degree by one. The binomial theorem and pascal's triangle can help you choose coefficients and substitutions for sums and differences of squares and cubes and make alternating terms disappear. You can also try some complex analysis where you substitute a complex variable a+bi for x and split it into at least two simultaneous equations of real and imaginary parts. Then there is linear algabra wher you can solve sytems of equations using matrices, determinants and inverses.
Here is an alternative method to do this, actually Ferrari's method for solving quartic equations adapted to factoring a quartic polynomial. The quartic polynomial to factor is
5x⁴ − 26x³ − 11x² − 10x + 6
and we are given that this polynomial has no rational zeros, but it does factor into two quadratics with integer coefficients. The task is to find this factorization.
To avoid having to deal with either irrational or fractional coefficients later on we first multiply this polynomial by _four times the coefficient of the leading term_ which is 4·5 = 20. Of course, when we are done we will have to divide the result again by 20 to obtain the factorization of the original quartic. Multiplying by 20 we have
100x⁴ − 520x³ − 220x² − 200x + 120
Now we have 100x⁴ = (10x²)² and 520x³ = 2·(10x²)·26x which is twice the product of 10x² and 26x. Therefore, we can complete the square with respect to the quartic and cubic terms by adding and subtracting (26x)² = 676x² since, applying the identity a² − 2·a·b + b² = (a − b)², we have 100x⁴ − 520x³ + 676x² = (10x²)² − 2·(10x²)·26x + (26x)² = (10x² − 26x)² so we get
(10x² − 26x)² − 676x² − 220x² − 200x + 120
(10x² − 26x)² − 896x² − 200x + 120
(10x² − 26x)² − (896x² + 200x − 120)
We have now written our quartic polynomial as a _difference of two terms_ where the first term is a perfect square but the second term is not. The difference of two terms will remain the same if we add the same quantity to each of the two terms. The idea is to add something to both terms in such a way that the first term will _remain_ a perfect square and the second term will _become_ a perfect square as well. We want to create a difference of two squares because we can then use the difference of two squares identity a² − b² = (a + b)(a − b) to factor our polynomial into two quadratics.
Now, if we take _any_ number λ and add 2λ(10x² − 26x) + λ² = 20λx² − 52λx + λ² to both terms, then the first term will _remain_ a perfect square because, applying the identity a² + 2·a·b + b² = (a + b)², we then have (10x² − 26x)² + 2·(10x² − 26x)·λ + λ² = (10x² − 26x + λ)².
Adding 2λ(10x² − 26x) + λ² = 20λx² − 52λx + λ² to each of the two terms of (10x² − 26x)² − (896x² + 200x − 120) we therefore get
(10x² − 26x + λ)² − ((20λ + 896)x² + (200 − 52λ)x + (λ² − 120))
Since the first term (10x² − 26x + λ)² is a perfect square regardless of the value of λ we are free to choose λ in such a way that the second term (20λ + 896)x² + (200 − 52λ)x + (λ² − 120), which is a quadratic polynomial in x, will also become a perfect square.
A quadratic polynomial ax² + bx + c (a ≠ 0) is a perfect square, that is, the square of a linear polynomial in x, if and only if this quadratic has two coinciding and therefore identical zeros. This is the case if and only if the _discriminant_ b² − 4ac of the quadratic is equal to zero. For our quadratic polynomial (20λ + 896)x² + (200 − 52λ)x + (λ² − 120) we have a = 20λ + 896, b = 200 − 52λ, c = λ² − 120 so this will be a perfect square if and only if λ satisfies
(200 − 52λ)² − 4(20λ + 896)(λ² − 120) = 0
This is a cubic equation in λ but we _do not need to solve this equation_ considering that our quartic polynomial has a factorization into two quadratics with integer coefficients.
If there exists a value of λ that makes (20λ + 896)x² + (200 − 52λ)x + (λ² − 120) the square (mx + n)² = m²x² + 2mnx + n² of a linear polynomial mx + n with _integer_ coefficients m and n, then both the coefficient 20λ + 896 of x² and the constant term λ² − 120 must be squares of integers.
Clearly, if λ² − 120 is to be the square of an integer then λ² must be an integer. But λ itself cannot be irrational because 20λ + 896 also needs to be the square of an integer, and therefore an integer. So, if there exists a value of λ which makes both 20λ + 896 and λ² − 120 the square of an integer, then this value of λ must be an integer.
But how do we go about finding integer values of λ which will make both 20λ + 896 and λ² − 120 the square of an integer? Well, if λ² − 120 is the square of some integer μ for some integer λ, that is, λ² − 120 = μ², then 120 = λ² − μ² must be a difference of two squares of integers. Now, any product of two quantities p and q can be rewritten as a difference of squares since
pq = ((p + q)/2 + (p − q)/2)((p + q)/2 − (p − q)/2) = ((p + q)/2)² − ((p − q)/2)²
Clearly, if (p + q)/2 and (p − q)/2 are integers then their sum p and their difference q must be integers and, moreover, p and q must be integers of like parity if (p + q)/2 and (p − q)/2 are to be integers. Since 120 = 2³·3·5 is even this number evidently cannot be written as the product of two odd integers but we can write 120 as the product of two even integers in several ways, and using the identity above we have
120 = 60·2 = (31 + 29)(31 − 29) = 31² − 29²
120 = 30·4 = (17 + 13)(17 − 13) = 17² − 13²
120 = 20·6 = (13 + 7)(13 − 7) = 13² − 7²
120 = 12·10 = (11 + 1)(11 − 1) = 11² − 1²
which exhausts all possibilities of writing 120 as a difference of two squares of integers.
Evidently if λ² − 120 is the square of some integer μ for some integer λ and therefore 120 = λ² − μ² then λ must be one of the numbers ±11, ±13, ±17, ±31 so we need to test at most eight integer values of λ to see if this will make 20λ + 896 a square as well.
Starting with λ = 11 and λ = −11 we have 20λ + 896 = 20·11 + 896 = 1116 and 20·(−11) + 896 = 776 which are not squares of integers since 33² = 1089 < 1116 < 34² = 1156 and 27² = 729 < 776 < 28² = 784. But with λ = 13 we have 20λ + 896 = 20·13 + 896 = 1156 = 34², as required.
With λ = 13 our polynomial
(10x² − 26x + λ)² − ((20λ + 896)x² + (200 − 52λ)x + (λ² − 120))
becomes
(10x² − 26x + 13)² − (1156x² − 476x + 49)
which is
(10x² − 26x + 13)² − (34x − 7)²
and we have converted our quartic polynomial into a difference of two squares. Using the difference of two squares identity a² − b² = (a + b)(a − b) this can be written as
(10x² − 26x + 13 + 34x − 7)(10x² − 26x + 13 − 34x + 7)
which is
(10x² + 8x + 6)(10x² − 60x + 20)
and we have factored our quartic into two quadratics. Now note that the first quadratic has a factor 2 and that the second quadratic has a factor 10 we can take out to get
20(5x² + 4x + 3)(x² − 6x + 2)
and since we started out by multiplying the polynomial to factor by 20 this means that the original quartic polynomial
5x⁴ − 26x³ − 11x² − 10x + 6
factors as
(5x² + 4x + 3)(x² − 6x + 2)
Damn that so much simpler!
I got no words to express this
Help, how to solve that:
4x⁴ - 12x³ + 5x² + 6x - 64 = 0
I tried eight ways but no one works 😭😢
@@bimokayoba The equation
4x⁴ − 12x³ + 5x² + 6x − 64 = 0
is easy to solve. First observe that 4x⁴ = (2x²)² is the square of 2x² and that 12x³ = 2·2x²·3x is twice the product of 2x² and 3x so we can complete the square with respect to 4x⁴ − 12x³ by adding and subtracting the square (3x)² = 9x² of 3x. This gives
4x⁴ − 12x³ + 9x² − 9x² + 5x² + 6x − 64 = 0
which can be written as
(2x² − 3x)² − 4x² + 6x − 64 = 0
Next observe that −4x² + 6x = −2(2x² − 3x) so we can rewrite the equation as
(2x² − 3x)² −2(2x² − 3x) − 64 = 0
Now note that (2x² − 3x)² is the square of 2x² − 3x and that 2(2x² − 3x) = 2·(2x² − 3x)·1 is twice the product of 2x² − 3x and 1 so we can again complete the square, this time with respect to (2x² − 3x)² − 2(2x² − 3x), by adding and subtracting the square 1² of 1. This gives
(2x² − 3x)² −2(2x² − 3x) + 1² − 1 − 64 = 0
which can be written as
(2x² − 3x − 1)² − 65 = 0
or
(2x² − 3x − 1)² = 65
Of course, the solution of the equation is now easy since this gives
2x² − 3x − 1 = √65 ⋁ 2x² − 3x − 1 = −√65
2x² − 3x − (1 + √65) = 0 ⋁ 2x² − 3x − (1 − √65) = 0
and then we only need to solve these two quadratic equations using the quadratic formula, which gives
x = (3 + √(17 + 8√65))/4 ⋁ x = (3 + √(17 + 8√65))/4 ⋁ x = (3 + √(17 − 8√65))/4 ⋁ x = (3 − √(17 − 8√65))/4
The first two solutions are real but since 17 − 8√65 < 0 the last two solutions are complex so we have
x = (3 + √(17 + 8√65))/4 ⋁ x = (3 + √(17 + 8√65))/4 ⋁ x = (3 + i√(−17 + 8√65))/4 ⋁ x = (3 − i√(−17 + 8√65))/4
and your equation is solved.
Best method out of all I have seen before❤❤
how do you gess 6 must be 3*2 when is 7 other posibilities?
If I set this expression to zero and note that there are two sign changes, there should be at least two real roots. I can use fixed-point iteration or Newton-Raphson iteration to obtain numerical values for these two roots. Then synthetically divide to obtain a quadratic, from which the remaining two roots may be obtained. Using these four roots I can then obtain the desired factors of the original expression if I really needed them.
There is absolutely a quartic discriminant. Every polynomial has a discriminant. But It is complicated and alone won’t tell you if the quartic in factorable. You CAN tell using the resolvent if you turn the quartic into a depressed quartic (without a cubic term). This is not super simple to do, but can be done.
Well, we can use the same method he used to eliminate the square term in the cubic equation. I have tried it using a general equation.
If the resolvent cubic has a rational root, especially a perfect square, you can factor the quartic.
How did he do last part can anyone EXPLAIN
THANK YOU so many people are cowards like Brian Mclogan and just use a leading coefficient of 1
😂😂
bro how did you write last step of the ans as product of factors
I had a class about factor polinomials like this (others too) and taught to my a class we have another way which is like factoring another polynomial.
You reach a polynomial that its quadratic term is close to that polynomial then factor. (????)
If you didn't get a solution with 6=3 x 2, would you have tried 6=6 x 1?
@bimokayoba The equation
4x⁴ − 12x³ + 5x² + 6x − 64 = 0
is easy to solve. First observe that 4x⁴ = (2x²)² is the square of 2x² and that 12x³ = 2·2x²·3x is twice the product of 2x² and 3x so we can complete the square with respect to 4x⁴ − 12x³ by adding and subtracting the square (3x)² = 9x² of 3x. This gives
4x⁴ − 12x³ + 9x² − 9x² + 5x² + 6x − 64 = 0
which can be written as
(2x² − 3x)² − 4x² + 6x − 64 = 0
Next observe that −4x² + 6x = −2(2x² − 3x) so we can rewrite the equation as
(2x² − 3x)² −2(2x² − 3x) − 64 = 0
Now note that (2x² − 3x)² is the square of 2x² − 3x and that 2(2x² − 3x) = 2·(2x² − 3x)·1 is twice the product of 2x² − 3x and 1 so we can again complete the square, this time with respect to (2x² − 3x)² − 2(2x² − 3x), by adding and subtracting the square 1² of 1. This gives
(2x² − 3x)² −2(2x² − 3x) + 1² − 1 − 64 = 0
which can be written as
(2x² − 3x − 1)² − 65 = 0
or
(2x² − 3x − 1)² = 65
Of course, the solution of the equation is now easy since this gives
2x² − 3x − 1 = √65 ⋁ 2x² − 3x − 1 = −√65
2x² − 3x − (1 + √65) = 0 ⋁ 2x² − 3x − (1 − √65) = 0
and then we only need to solve these two quadratic equations using the quadratic formula, which gives
x = (3 + √(17 + 8√65))/4 ⋁ x = (3 + √(17 + 8√65))/4 ⋁ x = (3 + √(17 − 8√65))/4 ⋁ x = (3 − √(17 − 8√65))/4
The first two solutions are real but since 17 − 8√65 < 0 the last two solutions are complex so we have
x = (3 + √(17 + 8√65))/4 ⋁ x = (3 + √(17 + 8√65))/4 ⋁ x = (3 + i√(−17 + 8√65))/4 ⋁ x = (3 − i√(−17 + 8√65))/4
and your equation is solved.
I’m genuinely not trying to insult anyone, it’s only a bit of roasting, but I love how these mathematicians can do integrals and trigonometry without looking, then ask them to do a (fairly) simple multiplication, they freeze.
Is there another method to reach the answer
if you mean getting the roots, then probably the rational roots theorem but I'm too lazy to check.
if you mean factoring it to two quadratics, then i have no idea
@@codexcursorsno. It says in the title that there are no rational zeroes here. However it is possible to factor into two quadratics in other ways just much much more complicated.
Why didn't you use synthetic division?
naive
Because there are no rational roots to begin with; if you start by finding all possible rational roots (p/q) and test them, none of them will work.
Noob
what if the system of equations ends up something like: a + b = -6, 3a + 3b = -18
What is the -11x^2 for if the answer would be in a form that does not use it in any way?
Doesnt he have to factor the x^2 terms too by factoring or quadratic formula at least for a complete factor?
Thxxx
I love when he says "Ladies and Gentlemen"
It didn't work for me
Yes same
Can u factorize it plssss
x³+x+1
It will help me so much( I don't even sure if this is possible)
(a+b)^3=3ab(a+b)+ (a^3+b^3)
x^3=x-1
3ab=-1
a^3+b^3=-1
27a^3*b^3= - 1
a^3+b^3= - 1
27uv= - 1
u+v=-1
.
v= - 1- u
27u(-1-u)= - 1
this is not going to be nice
x = (1/2 (sqrt(93) - 9))^(1/3)/3^(2/3) - (2/(3 (sqrt(93) - 9)))^(1/3)
x = -(1/2 (sqrt(93) - 9))^(1/3)/(2 3^(2/3)) + 1/(2^(2/3) (3 (sqrt(93) - 9))^(1/3)) + i (-3^(1/6)/(2^(2/3) (sqrt(93) - 9)^(1/3)) - (sqrt(93) - 9)^(1/3)/(2 2^(1/3) 3^(1/6)))
x = -(1/2 (sqrt(93) - 9))^(1/3)/(2 3^(2/3)) + 1/(2^(2/3) (3 (sqrt(93) - 9))^(1/3)) + i (3^(1/6)/(2^(2/3) (sqrt(93) - 9)^(1/3)) + (sqrt(93) - 9)^(1/3)/(2 2^(1/3) 3^(1/6)))
This video needs some explaining 😭 😭
Using your method can’t factor this quartic: X^4+2X^3+4X^2+8X+16 = 0
Your polynomial gives the roots of x^5-2^5=0 expect the real one.
Hi there
Thanks for your reply
X = 2 doesn’t satisfy the equation
So you telling me math is just try and error
Where's a? where's b?😭😭
What I have to do for catch you attention
I don’t think I am free for subscribe
The discriminant formula for a quartic polynomial en.wikipedia.org/wiki/Discriminant#Degree_4
I guess we can't expect quartic equations also to have the "can do in head" discriminant
Sir I am your big fan from India.sir plz give me your mail Id so that I can send you beautiful problems on mathematics