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

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

$|z - 1| = |z - (-1)|$ 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 $\operatorname{Re}(z) = 0$.

Since $x \in \!\left(\dfrac{\pi}{2}, \pi\right)$, $\cos x < 0$. From $\sin^2 x + \cos^2 x = 1$: $\cos^2 x = 1 - \dfrac{16}{25} = \dfrac{9}{25}$, so $\cos x = -\dfrac{3}{5}$.

Then $\sin 2x = 2 \cdot \dfrac{4}{5} \cdot \!\left(-\dfrac{3}{5}\right) = -\dfrac{24}{25}$.

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

Outer: $\sec^2(x^2)$. Inner derivative: $2x$. So $f'(x) = 2x\sec^2(x^2)$.

On $[0, 1]$: $1 \le 1 + x \le 2$, hence $\dfrac{1}{2} \le \dfrac{1}{1 + x} \le 1$. So $\dfrac{x^n}{1 + x} \le x^n$, giving $I_n \le \int_0^1 x^n \, dx = \dfrac{1}{n + 1}$.