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

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

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

$P(1) = 0$, 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: $60/6 = 10$.

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

Factor as $(2\sin x + 1)(\sin x - 1) = 0$. $\sin x = 1$ gives $x = \pi/2$; $\sin x = -\tfrac{1}{2}$ gives $x = 7\pi/6,\, 11\pi/6$. 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 $1/6 \ne 0$, so the four points are NOT coplanar — they form a non-degenerate tetrahedron.

$f'(x) = 3x^2 - 3$ 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.