Fun with polynomials!

[JimC]

Solve me! (exact values required)


x3 + 11x2 + 33x + 27 = 0

[JimC]

Bump with a hint!

It factorises to a simple linear factor times a quadratic with irrational solutions...

[Seabass]

I'd have been able to solve your equation easily back in the early 90s. Now it just evokes feelings of inadequacy and self-loathing.

[cronus]

x=-3

x=-4- SQR(7)

x=SQR(7) -4

Sorry no mathtype installed on this machine.

[JimC]

x=-3

x=-4- SQR(7)

x=SQR(7) -4

Sorry no mathtype installed on this machine.

Correct!

I'd have been able to solve your equation easily back in the early 90s. Now it just evokes feelings of inadequacy and self-loathing.

My work here is done! :twisted:

Another one coming later - might make it one with 4 complex solutions...

[JimC]

OK, the following polynomial is to be solved over C. It has 4 complex solutions (in 2 conjugate pairs, naturally...)


x4 + 4x3 + 11x2 + 14x + 15 = 0

[cronus]

x=0.5( -3-i(SQR(11)))

x=0.5(-1-i(SQR(11)))

x=0.5(-3+i(SQR(11)))

x=0.5(-1+i(SQR(11)))

[Calilasseia]

Here is my solution,, with full working.

The polynomial equation to solve is a4+4x3+11x2+14x+15=0. Since we are told that this has complex solutions c1, c2, c3 and c4, such that c1=c2*, and c3=c4*, where z* denotes the complex conjugate of z, this means that the quartic equation must be factorisable into two quadratic factors Q1(x) and Q2(x), such that the solutions of Q1(x)=0 are c1 and c2, and the solutions of Q2(x)=0 are c3 and c4.

Therefore, we write the quartic in the form (x2+ax+b)(x2+cx+d)=0, and multiply out, giving:

x4+(a+c)x3+(b+d+ac)x2+(ad+bc)x+bd=0

Equating coefficients, we have:

[1] a+c=4
[2] b+d+ac=11
[3] ad+bc=14
[4] bd=15

Taking [4] first, 15 can be factorised as either 3×5 or 15×1. Let us choose b=3, d=5, and see if this choice leads to a consistent solution.

Substituting these values into [3] yields 5a+3c=14. Multiplying [1] by 3 throughout, and solving the resulting simultaneous equations, we have:

[5] 5a+3c=14
[6] 3a+3c=12

Subtracting [6] from [5] gives 2a=2, therefore a=1. Substituting this result into [1] gives c=3.

These values form a consistent solution. We therefore have a=1, b=3, c=3, d=5, yielding:

[7] Q1(x) = x2+x+3
[8] Q2(x) = x2+3x+5

and the quartic factorises as:

(x2+x+3)(x2+3x+5)=0

From the quadratic formula, the roots of the equation Q1(x)=0 are:

[-1±(1-12)]/2

which gives as our first roots:

[9] c1 = (-½+11½i)
[10] c2 = (-½-11½i)

Likewise, the roots of the equation Q2(x)=0 are:

[-3±(9-20)]/2

which gives as our second roots:

[11] c3 = (-3/2+11½i)
[12] c4 = (-3/2-11½i)

EDIT: Bugger, someone got there first. But without providing full working. I'm sure JimC will see fit to award bonus points for full working being provided.

[cronus]

I could follow the route of precise working but then what happens should I meet a absolutely novel problem? My working is bits and pieces. No precision at all. Check the answer, though.

[Svartalf]

2 (x-1) = y
y-1=x

Does that qualify?
Funny enough, I instinctively knew the value of x+y but had trouble demonstrating it.

[MiM]

http://www.quickmath.com/webMathematica ... /basic.jsp

[JimC]

Very good, Cali and Scrumple!

And yes, Cali's methodology was faultless; that is definitely the way to proceed with these beasties...

[JimC]

Just adding that, it is much easier to create these than to solve them, which of course, is closely related to the fundamental principles of cryptography...

[Calilasseia]

Meanwhile, necromancing the thread, here's some more fun with polynomials.

Theorem. Any four consecutive numbers, multiplied together, produces a multiple of 24.

We start by recognising the following elementary facts:

[1] Any integer a that is a multiple of some other integer N, can be written as a = Nk, where k is some other integer:

[2] If two integers a and b are multiples of some other integer N, and can thus be written a = Nk and b = Nm, then their sum and difference are also multiples of N, viz:

(a+b) = (Nk+Nm) = N(k+m)

(a-b) = (Nk-Nm) = N(k-m)

These will become useful as we proceed.

We now recognise that any four consecutive numbers take the form n, n+1, n+2, n+3. The theorem states that:

n(n+1)(n+2)(n+3) mod 24 = 0

or equivalently, that:

n(n+1)(n+2)(n+3) = 24K, where K is an integer.

Proof by induction takes place as follows. Begin by establishing the result for n=1:

1×2×3×4 = 24, which is clearly 24×1.

Assume the result is true for n=j:

A = j(j+1)(j+2)(j+3) = 24K

We expand the multiplication to generate a polynomial:

A = j(j+1)(j+2)(j+3) = j4+6j3+11j2+6j

Now, for n=j+1, we have:

B = (j+1)(j+2)(j+3)(j+4) = j4+10j3+35j2+50j+24

Now, we subtract the two polynomials. If the two quantities are indeed multiples of 24, then their difference will also be a multiple of 24. Thus we have:

C = (B-A) = (j4+10j3+35j2+50j+24) - (j4+6j3+11j2+6j)

= 4j3+24j2+44j+24

Is this divisible by 24? Well, it's divisibly by 4, which can be factored out as follows:

C = 4(j3+6j2+11j+6)

This in turn can be seen, by comparison with A above, to be equal to:

C = 4(j+1)(j+2)(j+3)

For any integer choice of j, at least one of the factors above will be even, and thus divisible by 2. Since we have three consecutive integers comprising the factors, at least one of the factors will also be divisible by 3, because any three consecutive integers will always contain a multiple of three within that collection. Therefore, the factors:

(j+1)(j+2)(j+3)

will be divisible by 6.

Since C is both divisible by 4 and by 6, C is divisible by 24, and both A and B are therefore divisible by 24.

Since we have proven the case for n=1, the above establishes that for any n=j that the theorem is true, the theorem is true also for any n=j+1, by induction, the theorem is true for all positive n. A similar procedure working backwards to n=j-1 establishes the same result for all negative n, and of course, the theorem is also true for n=0, since 0 = 24×0.

QED.

[JimC]