Daily math problems for August 21, 2026 — Algebra, Analytic Geometry, Complex Numbers & more

$X^2+1$ has no real roots, so it can't factor into linear pieces over $\mathbb{R}$. It's irreducible over $\mathbb{R}$. (Over $\mathbb{C}$, it factors as $(X-i)(X+i)$.)

The constant term in slope-intercept form is the $y$-intercept: $b=5$.

$\overline{a+bi}=a-bi$, so $\overline{2-3i}=2+3i$.

The cancellation law states: $a\ast b=a\ast c\Rightarrow b=c$. (Even without inverses, this property may or may not hold; in a group it always does.)

$\mathrm{\Delta }=16-4=12$, $a=-1$, so $f_{\max }=-{\displaystyle \frac{\mathrm{\Delta }}{4a}}=-{\displaystyle \frac{12}{-4}}=3$. The image is $(-\mathrm{\infty },3]$.

$\mathrm{\mid }z-1\mathrm{\mid }=\mathrm{\mid }z-(-1)\mathrm{\mid }$ describes the set of points equidistant from $1$ and $-1$. That's the perpendicular bisector of the segment between them, i.e. the imaginary axis $\mathrm{Re}(z)=0$.

Since $x\in \text{\hspace{0.17em}⁣}({\displaystyle \frac{\pi }{2}},\pi )$, $\cos x<0$. From ${\sin }^{2}x+{\cos }^{2}x=1$: ${\cos }^{2}x=1-{\displaystyle \frac{16}{25}}={\displaystyle \frac{9}{25}}$, so $\cos x=-{\displaystyle \frac{3}{5}}$.

Then $\sin 2x=2\cdot {\displaystyle \frac{4}{5}}\cdot \text{\hspace{0.17em}⁣}(-{\displaystyle \frac{3}{5}})=-{\displaystyle \frac{24}{25}}$.

$2x-1=-x+5\Rightarrow 3x=6\Rightarrow x=2$, $y=3$. Intersection: $(2,3)$.

Outer: ${\sec }^2(x^2)$. Inner derivative: $2x$. So $f^{\mathrm{\prime }}(x)=2x{\sec }^2(x^2)$.

On $[0,1]$: $1\le 1+x\le 2$, hence ${\displaystyle \frac{1}{2}}\le {\displaystyle \frac{1}{1+x}}\le 1$. So ${\displaystyle \frac{x^n}{1+x}}\le x^n$, giving $I_n\le {\int }_0^1{x}^{n}dx={\displaystyle \frac{1}{n+1}}$.