1. Problem: For the function $f(x) = -2x^3$, analyze the behavior as $x \to -\infty$. The leading term dominates the behavior of the polynomial for large $|x|$. Since the leading term is $-2x^3$, and $x^3$ is negative when $x$ is negative, multiplying by $-2$ makes it positive large. So as $x \to -\infty$, $x^3 \to -\infty$, thus $-2x^3 = -2 \times (-\infty) = +\infty$. Therefore, $f(x) \to +\infty$ as $x \to -\infty$, not $-\infty$. 2. Problem: If $x = 2 + i$ is a root of a polynomial with real coefficients, then $x = 2 - i$ is also a root. This is true because complex roots of polynomials with real coefficients come in conjugate pairs. 3. Problem: If $p(x)$ divided by $(x+1)$ has remainder 2, then $p(-1) = 2$. By the Remainder Theorem, the remainder when dividing by $(x - a)$ is $p(a)$. Here, dividing by $(x+1)$ means $a = -1$, so remainder = $p(-1) = 2$. 4. Problem: Is it possible for a 5th degree polynomial to have 3 nonreal roots and 2 real roots? Yes. Complex roots come in conjugate pairs, so 3 nonreal roots means one root is real or multiplicity involved. But since degree is 5, having 3 nonreal roots (which must be in pairs) is impossible unless one is repeated or real. Actually, 3 nonreal roots means at least 2 pairs, so 3 nonreal roots is impossible. But if counting multiplicity, 3 nonreal roots can be 2 complex conjugates plus one real root counted as nonreal? No. So 3 nonreal roots is impossible for real-coefficient polynomials. So the statement is false. 5. Problem: For $g(x) = 3x^2 + 2x - x + 1$, is $x = \frac{1}{3}$ a zero? Simplify $g(x)$: $$g(x) = 3x^2 + (2x - x) + 1 = 3x^2 + x + 1$$ Evaluate at $x = \frac{1}{3}$: $$g\left(\frac{1}{3}\right) = 3\left(\frac{1}{3}\right)^2 + \frac{1}{3} + 1 = 3 \times \frac{1}{9} + \frac{1}{3} + 1 = \frac{1}{3} + \frac{1}{3} + 1 = \frac{2}{3} + 1 = \frac{5}{3} \neq 0$$ So $x=\frac{1}{3}$ is not a zero. 6. Problem: Given roots $(2 + \sqrt{5}), (2 - \sqrt{5}), 3i$, find the polynomial with these roots. Since $3i$ is a root, its conjugate $-3i$ must also be a root for real coefficients. The polynomial is: $$p(x) = (x - (2 + \sqrt{5}))(x - (2 - \sqrt{5}))(x - 3i)(x + 3i)$$ Multiply conjugate pairs: $$(x - (2 + \sqrt{5}))(x - (2 - \sqrt{5})) = (x - 2)^2 - (\sqrt{5})^2 = (x - 2)^2 - 5 = x^2 - 4x + 4 - 5 = x^2 - 4x - 1$$ $$(x - 3i)(x + 3i) = x^2 - (3i)^2 = x^2 - (-9) = x^2 + 9$$ So: $$p(x) = (x^2 - 4x - 1)(x^2 + 9)$$ Expand: $$p(x) = x^4 + 9x^2 - 4x^3 - 36x - x^2 - 9 = x^4 - 4x^3 + 8x^2 - 36x - 9$$ 7. Problem: Find roots or factor $f(x) = 3x^4 + 5x^3 - 25x^2 - 45x - 18$ Try Rational Root Theorem candidates: factors of 18 over factors of 3: $\pm1, \pm2, \pm3, \pm6, \pm9, \pm18, \pm\frac{1}{3}, \pm\frac{2}{3}, \pm\frac{3}{3}=\pm1$ etc. Test $x = -3$: $$f(-3) = 3(-3)^4 + 5(-3)^3 - 25(-3)^2 - 45(-3) - 18 = 3(81) + 5(-27) - 25(9) + 135 - 18 = 243 - 135 - 225 + 135 - 18 = 0$$ So $x = -3$ is a root. Divide $f(x)$ by $(x + 3)$: Using synthetic division: -3 | 3 5 -25 -45 -18 | -9 12 39 18 ---------------- 3 -4 -13 -6 0 Quotient: $3x^3 - 4x^2 - 13x - 6$ Try to factor $3x^3 - 4x^2 - 13x - 6$ further. Try $x=2$: $$3(8) - 4(4) - 13(2) - 6 = 24 - 16 - 26 - 6 = -24 \neq 0$$ Try $x = -1$: $$3(-1) - 4(1) - 13(-1) - 6 = -3 - 4 + 13 - 6 = 0$$ So $x = -1$ is a root. Divide $3x^3 - 4x^2 - 13x - 6$ by $(x + 1)$: -1 | 3 -4 -13 -6 | -3 7 6 -------------- 3 -7 -6 0 Quotient: $3x^2 - 7x - 6$ Factor $3x^2 - 7x - 6$: Find two numbers that multiply to $3 \times (-6) = -18$ and add to $-7$: $-9$ and $2$. Rewrite: $$3x^2 - 9x + 2x - 6 = 3x(x - 3) + 2(x - 3) = (3x + 2)(x - 3)$$ So full factorization: $$f(x) = (x + 3)(x + 1)(3x + 2)(x - 3)$$ 8. Problem: Factor $g(x) = 12x^5 + 2x^4 - 4x^3$ Factor out common term: $$g(x) = 2x^3(6x^2 + x - 2)$$ Factor quadratic: Find two numbers that multiply to $6 \times (-2) = -12$ and add to $1$: $4$ and $-3$. Rewrite: $$6x^2 + 4x - 3x - 2 = 2x(3x + 2) - 1(3x + 2) = (2x - 1)(3x + 2)$$ So: $$g(x) = 2x^3(2x - 1)(3x + 2)$$