Polynomials

A polynomial has only whole-number powers of $x$. Only $x^3 - 2x + 1$ qualifies; the others contain $x^{\frac{1}{2}}$, $x^{-1}$, $x^{-2}$ or $\sqrt{x}$, which are not whole-number powers.

The degree is the highest power of $x$, which is $5$. (4 is the leading coefficient, not the degree.)

Volume $= x(x + 1)(x + 3)$. First $(x + 1)(x + 3) = x^2 + 4x + 3$, then multiply by $x$: $x^3 + 4x^2 + 3x$.

Substitute $x = 2$: $2^3 - 3(2) + 4 = 8 - 6 + 4 = 6$. (Forgetting the $-3x$ term gives 8.)

Let the integers be $n$, $n + 1$ and $n + 2$, so $n(n + 1)(n + 2) = 210$. Testing near the cube root of $210$ (about $6$): $5 \times 6 \times 7 = 210$.

Add like terms: $(2x^2 + x^2) - x + 3 = 3x^2 - x + 3$. ($2x^4$ wrongly multiplied the $x^2$ terms.)

$x(x^2 - 3x + 1) + 2(x^2 - 3x + 1) = x^3 - 3x^2 + x + 2x^2 - 6x + 2 = x^3 - x^2 - 5x + 2$.

By the remainder theorem the remainder is $P(2) = 8 - 8 + 6 - 5 = 1$. (No long division needed.)

The factor theorem gives $P(1) = 0$ and $P(2) = 0$. So $1 + a + b + 6 = 0$ (i.e. $a + b = -7$) and $8 + 4a + 2b + 6 = 0$ (i.e. $2a + b = -7$). Subtracting the first from the second gives $a = 0$, then $b = -7$.

By the factor theorem, $(x - a)$ is a factor when $P(a) = 0$. $P(1) = 1 - 7 + 6 = 0$, so $(x - 1)$ is a factor. ($P(-1) = 12 \ne 0$, so $(x + 1)$ is not.)

The factor theorem states $(x - a)$ is a factor exactly when $P(a) = 0$. For $(x - 3)$ that means $P(3) = 0$. ($P(-3) = 0$ would correspond to the factor $(x + 3)$.)

Since $P(1) = 0$, $(x - 1)$ is a factor; dividing leaves $x^2 - 5x + 6 = (x - 2)(x - 3)$. So $P(x) = (x - 1)(x - 2)(x - 3)$ (roots 1, 2, 3 multiply to $6$, matching the constant).

Set the two expressions equal: $x^3 = 9x$, so $x^3 - 9x = 0$ and $x(x - 3)(x + 3) = 0$, giving $x = 0,\ 3$ or $-3$. (Dividing by $x$ would lose the $x = 0$ intersection.)

The factor theorem gives $P(2) = 0$: $8 + 2k - 10 = 0$, so $2k = 2$ and $k = 1$.

Since $P(1) = 0$, $(x - 1)$ is a factor. Dividing gives $x^2 - x - 6 = (x - 3)(x + 2)$, so the roots are $1,\ 3$ and $-2$. Their sum is $1 + 3 - 2 = 2$ (which also matches $-\dfrac{b}{a}$).

$P(x) = (x - 1)(x + 2)(x - 3)$, and the constant term is $P(0) = (-1)(2)(-3) = 6$. (Two negatives make the product positive.)

Using the roots, write $P(x) = a(x + 2)(x - 1)(x - 3)$. The graph passes through $(0, 12)$, so $P(0) = a(2)(-1)(-3) = 6a = 12$, giving $a = 2$.

For large $x$ the leading term $-x^3$ dominates everything else. Since $-x^3 \to -\infty$ as $x \to \infty$, the graph falls without bound. (The $2x^2$ term grows more slowly and cannot reverse this.)

A degree-3 polynomial has at most 3 roots. Because its ends go in opposite directions (one to plus infinity, one to minus infinity), the graph must cross the $x$-axis at least once, so there is always at least 1 real root.

The curve cuts the $x$-axis at $x = -1, 2$ and $3$, giving factors $(x + 1)(x - 2)(x - 3)$. For large positive $x$ the graph rises, so the leading coefficient is positive (not the reflected $-(\dots)$ in option with a minus sign). (Roots with the wrong signs, a repeated root, or a fourth factor are ruled out by the three simple crossings.)

Multiplying powers adds the degrees: the leading terms multiply to give $x^3 \times x^2 = x^5$, so the product has degree $5$. (6 multiplied the degrees instead of adding.)

$P(x) = (x - 1)(x^2 + x + 2) + 3 = x^3 + x - 2 + 3 = x^3 + x + 1$ (using $P(x) = D(x)Q(x) + R$).

$(2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$, and $-4(2 + \sqrt{3}) = -8 - 4\sqrt{3}$. Adding, with the $+1$: $(7 + 4\sqrt{3}) + (-8 - 4\sqrt{3}) + 1 = 0$. (So $2 + \sqrt{3}$ is a root.)

Group: $x^2(x - 3) - 4(x - 3) = (x - 3)(x^2 - 4) = (x - 3)(x - 2)(x + 2)$. (Check: roots $3, 2, -2$ multiply to $-12$, matching the constant $12$ with the sign rule.)

$P(x) = (x + 1)(x^2 - 4) = (x + 1)(x - 2)(x + 2) = 0$, so $x = -1,\ 2$ or $-2$. ($x^2 - 4$ is a difference of two squares, giving two roots, not one.)