Tagged Questions

74 views

Show that $\sqrt{2}$ is irrational using integer root theorem

Show that $\sqrt{2}$ is irrational using integer root theorem. Let $P(x)=x^2-2$. Since $\sqrt{2}$ is a root of this polynomial, had it been a rational (suppose $\sqrt{2}=\frac{p}{q}$) no, by ...
152 views

Existence of rational sequence such that a polynomial is split over $\Bbb{Q}$

Does there exist a sequence $(a_n)_{n\in \Bbb{N}}$ of rationals such that for all $n\in \Bbb{N}$, $a_n\neq 0$ and the polynomial $a_0+a_1X+\cdots+a_nX^n$ is split over $\Bbb{Q}$? I was asked this ...
40 views

Help solve rational expression

I need help solving this rational expression. Divide $$\frac{4x^4 + 6x^3 + 3x - 1}{2x^2 + 1}$$ How do you solve this problem? Where do I start?
50 views

Is there a polynomial $f\in \mathbb Q[x]$ such that $f(x)^2=g(x)^2(x^2+1)$

I was asked the following question: $g\in \mathbb Q[x]$ is a polynomial (not the zero polynomial). Find $f \in \mathbb Q[x]$ such that $f(x)^2=g(x)^2(x^2+1)$ or show that such an $f$ does not exist. ...
1k views

Prove that $x^3 + x^2 = 1$ has no rational solutions?

Is this enough for a proof?: $$x^3+x^2 = 1$$ I would factor and get: $x^2(x+1) = 1$ I would show that $x = \sqrt1$, which is irrational but then do I have to show more? $x+1=1$ which gives me $x=0$ ...
77 views

Two quartic polynomials to be made a square?

Given two generally non-square quartic polynomials that are to be simultaneously made squares for particular values of $x$, $$c_1x^4+c_2x^3+c_3x^2+c_4x+c_5 = y_1^2$$ ...
271 views

Why is $x^3-5x$ injective on the rationals?

I've found the statement on the internet that the polynomial $x^3-5x$ is injective on the rational numbers, but without any comments on how to prove it. I think it means it must be easy, but I don't ...
for what value of $a$ has equation rational roots?
Suppose that we have following quadratic equation containing some constant $a$ $$ax^2-(1-2a)x+a-2=0.$$ We have to find all integers $a$,for which this equation has rational roots. First I have ...
The title pretty much asks my question: Does $f\in\mathbb{Q}[x]$ such that $$f(x)=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3),$$ where $\alpha_1, \alpha_2, \alpha_3\in\mathbb{R}\setminus\mathbb{Q}\$ ...