Daily math problems for August 15, 2026 — Calculus, Linear Function, Polynomials & more

$P(-\mathrm{\infty })=-\mathrm{\infty }$, $P(-1)=-1+3+1=3>0$, $P(1)=1-3+1=-1<0$, $P(\mathrm{\infty })=+\mathrm{\infty }$. The sign sequence $-,+,-,+$ shows three sign changes, hence three distinct real roots.

$m={\displaystyle \frac{11-2}{4-1}}={\displaystyle \frac{9}{3}}=3$.

$P(1)=0$, so $X-1$ divides $P$. By Horner: $P(X)=(X-1)(X^2-4)=(X-1)(X-2)(X+2)$.

Set the terms as $5-r,5,5+r$. Then $(5-r{)}^2+25+(5+r{)}^2=50+2r^2+25=83$, so $2r^2=8\Rightarrow r=2$. The largest term is $5+2=7$.

The three edges from origin are $(3,0,0),(0,4,0),(0,0,5)$. Determinant: $3\cdot 4\cdot 5=60$. Volume: $60\mathrm{/}6=10$.

$(x^2-4x)+(y^2+6y)=-4\Rightarrow (x-2{)}^2-4+(y+3{)}^2-9=-4\Rightarrow (x-2{)}^2+(y+3{)}^2=9$. Radius $=3$.

Factor as $(2\sin x+1)(\sin x-1)=0$. $\sin x=1$ gives $x=\pi \mathrm{/}2$; $\sin x=-\frac{1}{2}$ gives $x=7\pi \mathrm{/}6,11\pi \mathrm{/}6$. Total: 3 solutions.

R2 − R1: $y+2z=8$. R3 − R1: $3y+8z=30$. From the first: $y=8-2z$. Substituting: $3(8-2z)+8z=30\Rightarrow 24+2z=30\Rightarrow z=3$. Then $y=2$, $x=1$. Solution: $(1,2,3)$.

The tetrahedron with these vertices has volume $1\mathrm{/}6\mathrm{\ne }0$, so the four points are NOT coplanar — they form a non-degenerate tetrahedron.

$f^{\mathrm{\prime }}(x)=3x^2-3$ vanishes at $x=\pm 1$. $f(-1)=-1+3+1=3>0$ and $f(1)=1-3+1=-1<0$. The local max is positive and the local min is negative, so the cubic crosses the $x$-axis three times.