I love this example❤‼️. It is very instructive to see a case where factorization over the integers fails, while the general Ferrari method leads to a reasonable simple solution because the associated cubic has an integer solution.
I first searched for an integer solution n, where n must be an integer factor of 15. No such n exist. Then I tried to factor into two quadratics with integer coefficients, (x^4 -24x-15)=(x^2 + px +q)(x^2-px-r)= x^4 +(q-r-p^2)x^2-(q+r)px-pr Here p, r must be factors of 15, both positive or both negative, with q-r a quadratic integer. Hence positive, i.e q>r. The set of possibilities are (p, q) = (-1,-15), (-3, -5), (5, 3), (15, 1). Note that (p, q) and (-q, -p) leads to the same value for p-q, so one only have to check half of the possibilities. Here I find p-q to be in the set {14, 2}, which does not contain any quadratic integers. Hence, this method also fails! Finally turn to the general method (as found by Ferrari): Rewrite the equation as (x^2 + P)^2 = 2Px^2 + 24x + P^2 + 15, and impose the condition that the right hand side should be a quadratic expression in x. This occurs when the discriminant vanishes. With 2P=Q this becomes the condition Q(Q^2 + 4*15) = 24^2. By searching for integer solutions to this cubic I find Q=6, i.e P=3. The original equation becomes (x^2 + 3)^2 = 6(x+2)^2. This leads to a factorization, with t=√6, (x^2 + tx + 3+2t)(x^2-tx+3-2t) The corresponding solutions becomes, x=(-t ± i √(4t+6))/2, x = (t ± √(4t-6))/2.
It's a pity you didn't finish the first method (Descartes' method for solving depressed quartics) because that is not difficult to do. At 4:32 you have arrived at a system of three equations in a, b, c which is (1) c − b = −24/a (2) c + b = a² (3) bc = −15 Subtracting (1) from (2) and adding (1) and (2) then gives (4) 2b = a² + 24/a (5) 2c = a² − 24/a Multiplying (4) and (5) now gives (6) (a² + 24/a)(a² − 24/a) = 4bc and substituting (3) in (6) we have (a² + 24/a)(a² − 24/a) = −60 a⁴ − 576/a² = −60 a⁶ − 576 = −60a² so we indeed end up with a bicubic equation (a cubic in a²) which is (7) a⁶ + 60a² − 576 = 0 As expected (because our quartic equation in x lacks a quadratic term) this is a depressed bicubic (that is, a cubic in a² which lacks a quartic term). Note that if we substitute (8) a² = 2k in (7) we get (9) k³ + 15k − 72 = 0 which is _exactly_ the cubic resolvent you obtained with the second method (Ferrari's method for solving any quartic equation, not just depressed quartics). Since k = 3 is a solution of (9) a² = 6 satisfies (7) in acordance with (8) and therefore a = √6 and a = −√6 are two of the solutions of (7). Of course (7) has four other solutions as well, corresponding to the two other solutions of (9), but we only need a single solution of (7) to get a factorization of our quartic polynomial x⁴ − 24x − 15 into two quadratics. Substituting a = √6 in (4) and (5) we find b = 3 + 2√6, c = 3 − 2√6 and therefore we get the factorization (10) x⁴ − 24x − 15 = (x² + √6·x + 3 + 2√6)(x² − √6·x + 3 − 2√6) Note that a = −√6 gives the same factorization except for the order of the two quadratic factors. Since the left hand side of (9) is strictly increasing on the real number line this equation can only have a single real solution. Therefore, the other two solutions of (9) are complex conjugates. This also means that a = √6 and a = −√6 are the only real solutions of (7), so its other four solutions are complex. Consequently, each of the four other solutions of (7) will give a factorization of x⁴ − 24x − 15 into two quadratics with _complex_ coefficients. This is quite understandable because the discriminant Δ₁ = −6 − 8√6 of the first quadratic x² + √6·x + 3 + 2√6 at the right hand side of (10) is negative whereas the discriminant Δ₂ = −6 + 8√6 of the second quadratic x² − √6·x + 3 − 2√6 at the right hand side of (10) is positive. Consequently, x⁴ − 24x − 15 has two real and two conjugate complex zeros. If a monic quartic polynomial in x has four distinct zeros x₁, x₂, x₃, x₄ then in accordance with the polynomial factoring theorem this quartic factors as (x − x₁)(x − x₂)(x − x₃)(x − x₄) This means that, disregarding the order of the two quadratics, there are exactly three ways in which a monic quartic with four distinct zeros can be factored into two quadratics because we can pair (x − x₁) with either (x − x₂) or (x − x₃) or (x − x₄) and then the remaining two factors will necessarily form the other quadratic factor. Of course, when we _do_ take the order of the two quadratic factors into account, this gives a total of 2·3 = 6 possible factorizations into two quadratics. Clearly, if a monic depressed quartic polynomial in x with four distinct zeros is to factor as (x² + ax + b)(x² − ax + c) then there will be 6 possible values for a which come as three pairs of opposites, so it is clear why Descartes' method results in a bicubic equation in a². If the monic depressed quartic with real coefficients and with four distinct zeros happens to have either two real and two complex zeros _or_ four complex zeros, then, disregarding the order of the quadratic factors, there can be only a single factorization into two quadratics with real coefficients because quadratics with real coefficients are only obtained by pairing two linear factors with conjugate complex zeros. Therefore, disregarding the order of the two quadratic factors (and other trival alterations such as multiplying each of the quadratics by a factor whose product is 1), x⁴ − 24x − 15 can only be factored into two quadratics with real coefficients as (x² + √6·x + 3 + 2√6)(x² − √6·x + 3 − 2√6).
The second mthod was cool!
I love this example❤‼️. It is very instructive to see a case where factorization over the integers fails, while the general Ferrari method leads to a reasonable simple solution because the associated cubic has an integer solution.
No ! You have 2 real solutions (what you find) and 2 complex conjugate solutions
Very nice
I first searched for an integer solution n, where n must be an integer factor of 15. No such n exist. Then I tried to factor into two quadratics with integer coefficients,
(x^4 -24x-15)=(x^2 + px +q)(x^2-px-r)=
x^4 +(q-r-p^2)x^2-(q+r)px-pr
Here p, r must be factors of 15, both positive or both negative, with q-r a quadratic integer. Hence positive, i.e q>r. The set of possibilities are (p, q) = (-1,-15), (-3, -5), (5, 3), (15, 1). Note that (p, q) and (-q, -p) leads to the same value for p-q, so one only have to check half of the possibilities. Here I find p-q to be in the set {14, 2}, which does not contain any quadratic integers. Hence, this method also fails!
Finally turn to the general method (as found by Ferrari): Rewrite the equation as
(x^2 + P)^2 = 2Px^2 + 24x + P^2 + 15,
and impose the condition that the right hand side should be a quadratic expression in x. This occurs when the discriminant vanishes. With 2P=Q this becomes the condition
Q(Q^2 + 4*15) = 24^2.
By searching for integer solutions to this cubic I find Q=6, i.e P=3. The original equation becomes
(x^2 + 3)^2 = 6(x+2)^2. This leads to a factorization, with t=√6,
(x^2 + tx + 3+2t)(x^2-tx+3-2t)
The corresponding solutions becomes,
x=(-t ± i √(4t+6))/2, x = (t ± √(4t-6))/2.
It's a pity you didn't finish the first method (Descartes' method for solving depressed quartics) because that is not difficult to do. At 4:32 you have arrived at a system of three equations in a, b, c which is
(1) c − b = −24/a
(2) c + b = a²
(3) bc = −15
Subtracting (1) from (2) and adding (1) and (2) then gives
(4) 2b = a² + 24/a
(5) 2c = a² − 24/a
Multiplying (4) and (5) now gives
(6) (a² + 24/a)(a² − 24/a) = 4bc
and substituting (3) in (6) we have
(a² + 24/a)(a² − 24/a) = −60
a⁴ − 576/a² = −60
a⁶ − 576 = −60a²
so we indeed end up with a bicubic equation (a cubic in a²) which is
(7) a⁶ + 60a² − 576 = 0
As expected (because our quartic equation in x lacks a quadratic term) this is a depressed bicubic (that is, a cubic in a² which lacks a quartic term). Note that if we substitute
(8) a² = 2k
in (7) we get
(9) k³ + 15k − 72 = 0
which is _exactly_ the cubic resolvent you obtained with the second method (Ferrari's method for solving any quartic equation, not just depressed quartics). Since k = 3 is a solution of (9) a² = 6 satisfies (7) in acordance with (8) and therefore a = √6 and a = −√6 are two of the solutions of (7). Of course (7) has four other solutions as well, corresponding to the two other solutions of (9), but we only need a single solution of (7) to get a factorization of our quartic polynomial x⁴ − 24x − 15 into two quadratics.
Substituting a = √6 in (4) and (5) we find b = 3 + 2√6, c = 3 − 2√6 and therefore we get the factorization
(10) x⁴ − 24x − 15 = (x² + √6·x + 3 + 2√6)(x² − √6·x + 3 − 2√6)
Note that a = −√6 gives the same factorization except for the order of the two quadratic factors. Since the left hand side of (9) is strictly increasing on the real number line this equation can only have a single real solution. Therefore, the other two solutions of (9) are complex conjugates. This also means that a = √6 and a = −√6 are the only real solutions of (7), so its other four solutions are complex.
Consequently, each of the four other solutions of (7) will give a factorization of x⁴ − 24x − 15 into two quadratics with _complex_ coefficients. This is quite understandable because the discriminant Δ₁ = −6 − 8√6 of the first quadratic x² + √6·x + 3 + 2√6 at the right hand side of (10) is negative whereas the discriminant Δ₂ = −6 + 8√6 of the second quadratic x² − √6·x + 3 − 2√6 at the right hand side of (10) is positive. Consequently, x⁴ − 24x − 15 has two real and two conjugate complex zeros.
If a monic quartic polynomial in x has four distinct zeros x₁, x₂, x₃, x₄ then in accordance with the polynomial factoring theorem this quartic factors as
(x − x₁)(x − x₂)(x − x₃)(x − x₄)
This means that, disregarding the order of the two quadratics, there are exactly three ways in which a monic quartic with four distinct zeros can be factored into two quadratics because we can pair (x − x₁) with either (x − x₂) or (x − x₃) or (x − x₄) and then the remaining two factors will necessarily form the other quadratic factor. Of course, when we _do_ take the order of the two quadratic factors into account, this gives a total of 2·3 = 6 possible factorizations into two quadratics.
Clearly, if a monic depressed quartic polynomial in x with four distinct zeros is to factor as (x² + ax + b)(x² − ax + c) then there will be 6 possible values for a which come as three pairs of opposites, so it is clear why Descartes' method results in a bicubic equation in a². If the monic depressed quartic with real coefficients and with four distinct zeros happens to have either two real and two complex zeros _or_ four complex zeros, then, disregarding the order of the quadratic factors, there can be only a single factorization into two quadratics with real coefficients because quadratics with real coefficients are only obtained by pairing two linear factors with conjugate complex zeros.
Therefore, disregarding the order of the two quadratic factors (and other trival alterations such as multiplying each of the quadratics by a factor whose product is 1), x⁴ − 24x − 15 can only be factored into two quadratics with real coefficients as (x² + √6·x + 3 + 2√6)(x² − √6·x + 3 − 2√6).
👍👏😊👏👍
I wasn't getting anywhere with either method, so I decided to just concede to you.
(x^2+3)^2-6x^2-9-24x=15...(x^2+3)^2=6x^2+24x+24=6(x+2)^2.... quindi le 4 soluzioni si trovano da x^2+3=√6(x+2)..e.x^2+3=-√6(x+2)
problem
x⁴-24 x = 15
x⁴ = 24x + 15
(x² + a/2)² -ax² -a² /4 = 24x + 15
Need ax² + 24x + (60+a²)/4 to be a perfect square.
ax² + 24x + (60+a²)/4 = (bx + c)²
b² x² +2bcx+ c² = ax² + 24x + (60+a²)/4
b² = a
2bc = 24
c² = (60+a²)/4
√a √(60+a²) = 24
576 = (60+a²)a
a³ + 60 a -576 = 0
Root is at a = 6.
This gives us
ax² + 24x + (60+a²)/4 = 6x² + 24x + 24
= 6 (x² + 4x + 4)
= 6 (x+2)²
= [√6(x+2) ]²
(x² + 3)² = [√6(x+2) ]²
Now we can use the difference of 2 squares form to solve.
[x² +3 - √6(x+2) ][x² +3 + √6(x+2) ] = 0
(x² - x√6 +3 -2√6 )(x² + x√6 +3 +2√6 ) = 0
By zero product principle,
x² - x√6 +3 -2√6 = 0
x = { √6 ± √[6-4(3 -2√6 )] } /2
Δ = 8√6 -6
x = √6/2 ± √(8√6 -6) /2
x² + x√6 +3 +2√6 = 0
x = {-√6 ± √[6-4(3 +2√6 )] } /2
Δ = -8√6 -6
x = -√6/2 ± i √(8√6 +6) / 2
answer
x ∈ { √6/2 - √(8√6 -6) /2,
√6/2 + √(8√6 -6) /2,
-√6/2 - i √(8√6 +6) / 2
-√6/2 + i √(8√6 +6) / 2 }
I did that in my head 🤥
😉😂
Prove it!
I kind of doubt that!
@@scottleung9587 (irony)