191 votes
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Why can a quadratic equation have only 2 roots?

Suppose there are three distinct roots $x,y,z$. One has $$\begin{cases}ax^2+bx+c=0\\ay^2+by+c=0\\az^2+bz+c=0\end{cases}\Rightarrow\begin{cases}a(x^2-y^2)+b(x-y)=0\\a(x^2-z^2)+b(x-z)=0\end{cases}\...
Piquito's user avatar
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139 votes
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Is it impossible to perfectly fit a polynomial to a trigonometric function on a closed interval?

Yes, it is impossible. Pick any point in the interior of the interval, and any polynomial. If you differentiate the polynomial repeatedly at that point, you will eventually get only zeroes. This ...
Arthur's user avatar
  • 200k
134 votes

Why does the discriminant in the Quadratic Formula reveal the number of real solutions?

Think about it geometrically $-$ then compute. Everyone knows $x^2$ describes a parabola with its apex at $(0,0)$. By adding a parameter $\alpha$, we can move the parabola up and down: $x^2+\alpha$ ...
M. Winter's user avatar
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117 votes

What actually is a polynomial?

A polynomial (in one variable) is an expression of the form $$ p(x) = a_0+a_1x+a_2x^2+\ldots+a_nx^n$$ where the coefficients $a_i$ are some kind of number (or more generally they're elements of a Ring)...
spaceisdarkgreen's user avatar
103 votes
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Solution to the equation of a polynomial raised to the power of a polynomial.

Denote $a=x^2-7x+11.$ The equation becomes $a^{a-5}=1,$ or equivalently* $$a^a=a^5,$$ which has in $\mathbb{R}$ the solutions $a\in \{ {5,1,-1}\}.$ Solving the corresponding quadratic equations we get ...
user376343's user avatar
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93 votes
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Why does a polynomial with real, simple roots change its sign between its roots?

Firstly, to simplify the problem, start by re-numbering all the $a_{i}$’s from least to greatest. Think of the behavior at $x=a_n$ notice how the polynomial will look like $$(\text{pos numb})(\text{...
Snacc's user avatar
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89 votes
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Are polynomials with the same roots identical?

No, they are not. For instance, $2x^2-2$ and $x^2-1$ have the same roots, yet they are not identical. And, depending on what you mean by "the same roots", we have that $x^2-2x+1$ and $x-1$ have the ...
Arthur's user avatar
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81 votes
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Every polynomial with real coefficients is the sum of cubes of three polynomials

We have that the following identity holds $$(x+1)^3+2(-x)^3+(x-1)^3=6x.$$ Hence $$\left(\frac{f(x)+1}{6^{1/3}}\right)^{3}+\left(\frac{-f(x)}{3^{1/3}}\right)^{3}+ \left(\frac{f(x)-1}{6^{1/3}}\right)^{3}...
Robert Z's user avatar
  • 146k
81 votes

What actually is a polynomial?

There are lots of good answers here and they are all essentially correct, even though they are different! I will try to contribute another, which is somewhat more abstract than the others. I normally ...
Ethan Bolker's user avatar
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76 votes
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Is there any geometric intuition for the factorials in Taylor expansions?

Yes. There is a geometric explanation. For simplicity, let me take $x=0$ and $h=1$. By the Fundamental Theorem of Calculus (FTC), $$ f(1)=f(0)+\int_{0}^{1}dt_1\ f'(t_1)\ . $$ Now use the FTC for the $...
Abdelmalek Abdesselam's user avatar
74 votes

Can two integer polynomials touch in an irrational point?

What about $(x^2-2)^2$ and $0$? If you want both coordinates to be irrational, you can add something like $x^3$ to both. I hope this helps $\ddot\smile$ Images courtesy of WolframAlpha.
dtldarek's user avatar
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74 votes

Find $xy+yz+zx$ given systems of three homogenous quadratic equations for $x, y, z$

We can obtain $yz+zx+xy=2$ simply by finding the values of $x$, $y$, and $z$. We are given $$y^2+yz+z^2=1,\qquad(1)$$$$z^2+zx+x^2=4,\qquad(2)$$$$x^2+xy+y^2=3,\,\qquad(3)$$with $x,y,z>0$. ...
John Bentin's user avatar
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73 votes

Irreducible polynomial which is reducible modulo every prime

If $-1$ is a square in $\Bbb F_p$ (which includes the case $p=2$), say $a^2=-1$, then we have $$X^4+1=X^4-a^2=(X^2+a)(X^2-a).$$ If $p$ is odd and $2$ is a square in $\Bbb F_p$, say $2=b^2$, then we ...
Hagen von Eitzen's user avatar
71 votes
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Monic polynomials whose roots are their remaining coefficients

For the quadratic case $x^2+ax+b=0$, by Viète's formulas we have $$-a=a+b\qquad b=ab$$ The first formula implies $b=-2a$. Substituting this into the second equation gives $-2a=-2a^2$ or $a=a^2$, so $a=...
Parcly Taxel's user avatar
68 votes

Is it impossible to perfectly fit a polynomial to a trigonometric function on a closed interval?

We don't even need to differentiate many times. Just note that $f'' = -f$ is satisfied by $f = \cos$ but not if $f$ is a non-zero polynomial function because $f''$ has lower degree than $f$. (This ...
user21820's user avatar
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62 votes
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Why does the discriminant in the Quadratic Formula reveal the number of real solutions?

Because for $a\neq0$ we obtain: $$ax^2+bx+c=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right).$$ Now, we see that if $b^2-4ac<0$ then $\...
Michael Rozenberg's user avatar
57 votes
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Does convergence of polynomials imply that of its coefficients?

Consider the sequence of polynomials $$ p_n(x)=\left\{ \begin{array}{@{}ll@{}} (1-x)^n, & \text{if}\ n\equiv 0 \mod 2 \\ x^n, & \text{if}\ n\equiv 1 \mod 2 \end{array}\right. $$ ...
BHT's user avatar
  • 2,215
56 votes

Why can a quadratic equation have only 2 roots?

I think derivation of quadratic formula is not enough.... Yes it is. The derivation is of the form if $ax^2+bx+c=0$, then $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. The derivation is a proof if you pay ...
djechlin's user avatar
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55 votes
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Compute polynomial $p(x)$ if $x^5=1,\, x\neq 1$ [reducing mod $\textit{simpler}$ multiples]

The key idea employed here is the method of simpler multiples - a very widely used technique. Note that $\,Q\,$ has a "simpler" multiple $\,QR = x^5\!-\!1,\,$ so we can first reduce $P$ ...
Bill Dubuque's user avatar
55 votes
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Is there an easier way to prove the Fundamental Theorem of Algebra for polynomials with real coefficients?

The fundamental theorem of algebra is exactly as difficult for real vs. complex coefficients. The reason is that if $f(x) = f_0 + \dots + f_n x^n$ is any polynomial with complex coefficients, then the ...
Qiaochu Yuan's user avatar
53 votes
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Reversing an integer's digits is multiplicative for small digits

Suppose we have a 2 digit number $x$. We can write is in terms of its digits $a$ and $b$. When we attempt to square this number, we get an interesting result. $$x^2=(10a+b)^2=100a^2+10(2ab)+b^2$$ We ...
Kajelad's user avatar
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53 votes

Find $xy+yz+zx$ given systems of three homogenous quadratic equations for $x, y, z$

To my surprise, this problem can be solved using geometry. Identify the Euclidean plane with complex plane $\mathbb{C}$. and let $\omega = e^{\frac{2\pi}{3}i}$ be the cubic root of unity. Consider ...
achille hui's user avatar
52 votes

Does Newton's method converge for all polynomials?

Here is a plot of the function $$f(x) = 49 x^7+31 x^6-10 x^5-41 x^4+37 x^3-21 x^2-9 x+12$$ in the region of interest. The blue curve is $f$. The orange lines represent the forward orbit of $x_0 = \...
heropup's user avatar
  • 137k
51 votes

What actually is a polynomial?

Note: in this answer I will try to motivate the definition which is used in more advanced contexts such as "abstract algebra". This may go beyond what is in a typical pre-algebra book, but I hope it ...
Carl Mummert's user avatar
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51 votes
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Showing that roots of a quadratic polynomial are of opposite signs.

A general quadratic with roots $\alpha$ and $\beta$ can be written $$ x^2-(\alpha+\beta)x+\alpha\beta=0 $$ The last, constant term is the product of the roots. In your case that equals $-3$. The roots ...
PM.'s user avatar
  • 5,259
51 votes
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Why are we allowed to cancel fractions in limits?

Simply because we are dealing with values $x\neq 1$ in this case, thus for algebraic rule we are allowed to cancel out $$\lim_{x\to 1} \frac{x^4-1}{x-1}=\lim_{x\to 1} \frac{\color{red}{(x-1)}(x^3 + x^...
user's user avatar
  • 155k
51 votes
Accepted

Is the notion "If a polynomial has small coefficients (relative to the exponent), then it has small roots" true?

There exist estimates for the size of the largest root. The most general go back to the idea that $z$ is not a root of $$ p(z)=a_nz^n+a_{n-1}+...+a_1z+a_0 $$ if $|z|>R>0$ with an outer root ...
Lutz Lehmann's user avatar
48 votes
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Why can you find the roots a of polynomial by factoring it?

It's not an assumption. It is a general fact that if $ab = 0$, then either $a = 0$ or $b = 0$. The roots of a polynomial are the numbers for which that polynomial evaluates to $0$, so to find the ...
A. Thomas Yerger's user avatar
47 votes
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Value of $(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)$ if $z^4-2z^3+z^2+z-7=0$ for $z=\alpha$, $\beta$, $\gamma$, $\delta$

There is no need to use Vieta's formulas. Let $$f(z)=z^4-2z^3+z^2+z-7=(z-\alpha)(z-\beta)(z-\gamma)(z-\delta).$$ Then, since $(i-a)(-i-a)=-i^2+a^2=1+a^2$, it follows that $$(\alpha^2+1)(\beta ^2+1)(\...
Robert Z's user avatar
  • 146k

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