Recent content by Fernando Revilla

Right. :)

If we prove that $\det A_{\Phi}\ne 0$ then, $\text{rank }A_{\Phi}=5$. So, $$\text{rank }A_{\Phi}=5=\dim \text{Im }f =\dim V\Rightarrow \text{Im }f=V\Rightarrow f\text{ is surjective.}$$ On the other hand, $$\dim \text{Im }f=5\Rightarrow \dim \ker f=\dim V-\dim \text{Im }f=0\Rightarrow f\text{...

Sorry, I had a distraction error. :)

Better, $t=\dfrac{1}{u}$. Then, $$\displaystyle\int\frac{1}{u\sqrt{u^2+1}} du =\ldots =-\int\frac{1}{\sqrt{t^2+1}} dt=\ldots $$ EDIT: See answer #7.

So far so good. :) EDIT: See answer #7.

I,ve decided to complete the solution: Applying the Ruffini algorithm to $Q_1(x,y)$: $$\begin{array}{r|rrrrr} & 30 & -86y & 0 & 100y^3 \\ \dfrac{5}{3}y & & 50y & -60y^2 & -100y^3 \\ \hline & 30 & -36y & -60y^2 & 0 \end{array}$$ $$\Rightarrow...

Using the Ruffini Algorithm for the root $x=-\dfrac{3}{2}y$ $$\begin{array}{r|rrrrr} & 30 & -41y & -129y^2 & 100y^3 & 150y^4 \\ -\dfrac{3}{2}y & & -45y & 129y^2 & 0 & -150y^4 \\ \hline & 30 & -86y & 0 & 100y^3 & 0\end{array}$$ This implies...

Consider $$P_y(x)=30x^4+(-41y)x^3+(-129y^2)x^2+(100y^3)x+150y^4$$ as a polynomial in $x$ and $y$ as a parameter Suppose there are roots of the form $x=ky$. Then, $$P_y(ky)=\left(30k^4-41k^3-129k^2+100k+150\right)y^4=0.$$ Consider the equation $$30k^4-41k^3-129k^2+100k+150=0.$$ Using the theorem...

(1) You write $$\dfrac{2}{3}b^5 - \dfrac{1}{6}b^3+\dfrac{4}{9}b^2 - 1=\dfrac{1}{18}\left((2b+1)(2b-1)\left(3b^3 + \dfrac{2(2b + 3)(2b-3)}{(2b+1)(2b-1)}\right)\right)$$ What happens when you substitute $b=\pm\dfrac{1}{2}$? (2) The title of your thread Factoring polynomial, usually means to...

My solution:

In these problems, we are supposed to provide a closed expression. For example if $p(x)=x^2-2$, then $1.5^2-2=0.25\approx 0.$ A closed expression is $p(x)=(x-\sqrt{2})(x+\sqrt{2})$, not $p(x)=(x-1.5)(x+1.5).$

According to your solution, $b=\pm\dfrac{1}{2}$ are roots. Have you checked them?

As Jameson says, a great factorization is impossible for this polynomial. We can write $$p(x)=(2/3)x^5-(1/6)x^3+(4/9)x^2-1=\frac{1}{18}(\underbrace{12x^5-3x^3+8x^2-18}_{q(x)}).$$ We have $q(1)=\ldots <0$ and $q(2)=\ldots>0,$ so by Bolzano's theorem there is a root $\xi\in (1,2).$ It is not...

Yes, you can use https://en.wikipedia.org/wiki/Lagrange_polynomial Here, you have an online calculator: Lagrange Interpolating Polynomial Calculator - Online Software Tool

Another way: