Find multiple roots (repeated roots) of a polynomial depending on the real parameter "a"

[vilgeforc5]

Hello everyone! I have got this task:
Find multiple roots (repeated roots) of a polynomial $$?3−(1−2?)?^2−(?^2−2?)?+?^2$$ depending on the real parameter a.
My idea was to factorize the polynomial using the Euclidean algorithm, but I was unable to do it , because it's becoming larger and larger every single step and I don't know what to do..Could anyone help me or give me a hint?

 

[Steven G]

We need to see your work so we know how to offer you help. If you had read the forum's posting guidelines you would have received help by now.

 

[BigBeachBanana]

For a cubic to have 3 distinct roots, it must satisfy these conditions:
1) Two real critical points points, $x_1,x_2$
2) $f(x_1)\times f(x_2)<0$.
Visual aid:

Find the critical points.
What values of a will sastify this condition $f(x_1)\times f(x_2)<0?$

 

[blamocur]

Hello everyone! I have got this task:
Find multiple roots (repeated roots) of a polynomial $$?3−(1−2?)?^2−(?^2−2?)?+?^2$$ depending on the real parameter a.
My idea was to factorize the polynomial using the Euclidean algorithm, but I was unable to do it , because it's becoming larger and larger every single step and I don't know what to do..Could anyone help me or give me a hint?

I believe your polynomial is divisible by $t-1$.

 

[topsquark]

I believe your polynomial is divisible by $t-1$.

Interesting. That does force a value of a that will work. How did you find it?

-Dan

 

[Steven G]

t=1 is not a root.
The corner is getting crowded tonight.

 

[BigBeachBanana]

t=1 is not a root.
The corner is getting crowded tonight.

It is for $a=0$.

 

[blamocur]

t=1 is not a root.
The corner is getting crowded tonight.

Ooops -- I blame this on the colored words puzzle (https://www.freemathhelp.com/forum/threads/colours.134689/)

 

[Cubist]

Hello everyone! I have got this task:
Find multiple roots (repeated roots) of a polynomial $$?3−(1−2?)?^2−(?^2−2?)?+?^2$$ depending on the real parameter a.
My idea was to factorize the polynomial using the Euclidean algorithm, but I was unable to do it , because it's becoming larger and larger every single step and I don't know what to do..Could anyone help me or give me a hint?

If you want to have a repeated root then working out the derivative might be helpful, can you see why?

t=1 is not a root.
The corner is getting crowded tonight.

Well spotted! I was in complete agreement with (t-1) being a factor

 

[topsquark]

The interesting thing about t - 1, though, is if you divide by it you get a remainder which suggests that you set a = 0, which then leads to a double root. I think it's a very interesting approach to the problem so I think blamocur should at least be allowed to sit while he's in the corner.

-Dan

 

[BigBeachBanana]

Somehow I misread "repeated roots" for "distinct roots" in the OP.
For a cubic to have repeated roots, it must satisfy these conditions:
1) Two real critical points: $x_1,x_2$
2) $f(x_1)\times f(x_2)=0 \implies f(x_1)=0$ or $f(x_2)=0$
Visual aid:



I got a=0 as one of the three solutions, so I think @blamocur statement is justified.

 

[blamocur]

Somehow I misread "repeated roots" for "distinct roots" in the OP.
For a cubic to have repeated roots, it must satisfy these conditions:
1) Two real critical points: $x_1,x_2$
2) $f(x_1)\times f(x_2)=0 \implies f(x_1)=0$ or $f(x_2)=0$
Visual aid:

View attachment 33116

I got a=0 as one of the three solutions, so I think @blamocur statement is justified.

I might get a permission to nap in the corner after all.

 

[JeffM]

Interesting. That does force a value of a that will work. How did you find it?

-Dan

We have a specific cubic defined by a parameter a

$$f(t) = t^3 + (1 - 2a)t^2 + (a^2 - 2a)t + a^2, \text { where } a \in \mathbb R.$$
Generally, a cubic with a leading coefficient of one can be factored into three linear terms

$$f(t) = (t - p)(t - q)(t - r) = t^3 - (p + q + r)t^2 + (pq + pr + qr)t - pqr$$.
Now if p is real and p = q = r, this general form reduces to

$$f(t) = t^3 - 3pt^2 + 3p^2t - p^3 \implies p = \dfrac{1 - 2a}{3}, \ p^2 = \dfrac{a^2 - 2a}{3}, \text { and } p^3 = a^2.\\ \therefore \left ( \dfrac{1 - 2a}{3} \right )^2 = \dfrac{a^2 - 2a}{3} \implies 1 - 4a + 4a^2 = \dfrac{9}{3} * (a^2 - 2a) = 3a^2 - 6a \implies \\ a^2 + 2a + 1 = 0 \implies a = \dfrac{-1 \pm \sqrt{4 - 4}}{2} = - \dfrac{1}{2} \implies \\ p = \dfrac{1 - 2(-0.5)}{3} = \dfrac{2}{3}.\\ \text {However, there is an additional constraint, namely } p^3 = a^2. \text {But } p^3 = \left (\dfrac{2}{3} \right)^3 = \dfrac{8}{27} \ne \dfrac{1}{4} = \left ( - \dfrac{1}{2} \right )^2 = a^2.$$There is no real a that gives a real root of multiplicity 3.

We can use the same type of analysis to find what (if any) are the real values of a that result in a real root of multiplicity a.