Building Polynomials from Zeros - Math 3

1. If the polynomial $g(x) = 2x^3 - 12x^2 + 22x - 12$ can be written as $g(x) = 2(x - a)(x - b)(x - c)$ where $a \leq b \leq c$, what is the value of $b^2$?
   Answer: $4$ We have $g(x) = 2x^3 - 12x^2 + 22x - 12 = 2(x^3 - 6x^2 + 11x - 6)$. Let $h(x) = x^3 - 6x^2 + 11x - 6$, so $g(x) = 2h(x)$. We need to find the zeros of $h(x)$. By the rational root theorem, possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 6$. Testing $x = 1$: $h(1) = 1 - 6 + 11 - 6 = 0$ ✓. So $(x - 1)$ is a factor. Using polynomial division: $h(x) = (x - 1)(x^2 - 5x + 6) = (x - 1)(x - 2)(x - 3)$. Therefore, $g(x) = 2(x - 1)(x - 2)(x - 3)$. The zeros are $a = 1$, $b = 2$, and $c = 3$ (since $1 \leq 2 \leq 3$). Thus, $b^2 = 2^2 = 4$.
2. The polynomial $f(x) = 2x^4 - 8x^3 - 2x^2 + 32x - 24$ can be written in factored form as $f(x) = 2(x-a)(x-b)(x-c)(x-d)$ where $a, b, c, d$ are the zeros. If one of the zeros is $x = 3$, what is the sum of all four zeros?
   Answer: $4$ For a monic polynomial $x^4 + px^3 + qx^2 + rx + s$, the sum of the zeros equals $-p$ by Vieta's formulas. First, we factor out the leading coefficient: $f(x) = 2x^4 - 8x^3 - 2x^2 + 32x - 24 = 2(x^4 - 4x^3 - x^2 + 16x - 12)$. For the monic polynomial $g(x) = x^4 - 4x^3 - x^2 + 16x - 12$, the sum of zeros is $-(-4) = 4$. The fact that one zero is $x = 3$ confirms our polynomial but doesn't change the sum calculation. Choice A would be the sum if the coefficient of $x^3$ were $-2$. Choice C would be the sum if we incorrectly used $-(-6)$. Choice D would be the sum if we incorrectly used the coefficient $-8$ directly without factoring out the $2$.
3. A polynomial function has zeros at $x = -1$ (multiplicity 3) and $x = 2$ (multiplicity 2). If the leading coefficient is negative and the degree is 5, which statement about the end behavior is correct?
   Answer: As $x \to -\infty$, $f(x) \to +\infty$ and as $x \to +\infty$, $f(x) \to -\infty$ The polynomial has the form $f(x) = a(x+1)^3(x-2)^2$ where $a < 0$ and the degree is $3 + 2 = 5$ (odd). For polynomials with odd degree and negative leading coefficient, the end behavior is: as $x \to -\infty$, $f(x) \to +\infty$ and as $x \to +\infty$, $f(x) \to -\infty$. This is because the leading term is $ax^5$ with $a < 0$. Choice A describes even degree with positive leading coefficient. Choice C describes odd degree with positive leading coefficient. Choice D describes even degree with negative leading coefficient.
4. A quartic polynomial $P(x)$ has zeros at $x = -2, 1, 3, 5$ and satisfies $P(0) = 30$. What is $P(2)$?
   Answer: $-12$ Since $P(x)$ has zeros at $x = -2, 1, 3, 5$, we can write $P(x) = a(x+2)(x-1)(x-3)(x-5)$ for some constant $a$. Using $P(0) = 30$: $P(0) = a(0+2)(0-1)(0-3)(0-5) = a(2)(-1)(-3)(-5) = a(2)(-15) = -30a$. Since $P(0) = 30$, we have $-30a = 30$, so $a = -1$. Therefore, $P(x) = -(x+2)(x-1)(x-3)(x-5)$. Now we calculate $P(2) = -(2+2)(2-1)(2-3)(2-5) = -(4)(1)(-1)(-3) = -(4)(3) = -12$. Choice A is incorrect because $x = 2$ is not a zero. Choice B gives the positive value. Choice D is twice the correct magnitude.
5. Consider the polynomial $h(x) = (x - a)(x - b)(x - c)$ where $a < b < c$. If $h(x)$ has a local maximum at $x = 2$ and a local minimum at $x = 6$, what is $a + b + c$?
   Answer: $12$ For the cubic polynomial $h(x) = (x - a)(x - b)(x - c)$, we have $h'(x) = (x - b)(x - c) + (x - a)(x - c) + (x - a)(x - b)$. The critical points occur where $h'(x) = 0$. Given that there's a local maximum at $x = 2$ and a local minimum at $x = 6$, these are the two critical points of the cubic. For a cubic with positive leading coefficient, the critical points occur in the order: local maximum, then local minimum as $x$ increases. So $h'(2) = 0$ and $h'(6) = 0$. We can write $h'(x) = 3x^2 - 2(a + b + c)x + (ab + bc + ca)$. Since the critical points are $x = 2$ and $x = 6$, we have $h'(x) = 3(x - 2)(x - 6) = 3(x^2 - 8x + 12) = 3x^2 - 24x + 36$. Comparing coefficients: $-2(a + b + c) = -24$, so $a + b + c = 12$.
6. The polynomial $g(x) = 3x^3 - 12x^2 - 15x + 60$ has three real zeros. If one zero is $x = 4$, what is the product of the other two zeros?
   Answer: $-5$ First, factor out the leading coefficient: $g(x) = 3(x^3 - 4x^2 - 5x + 20)$. Let $h(x) = x^3 - 4x^2 - 5x + 20$. Since $x = 4$ is a zero, we can factor out $(x-4)$: $h(x) = (x-4)(x^2 + bx + c)$ for some $b$ and $c$. Using polynomial division or substitution: $h(x) = (x-4)(x^2 - 5) = (x-4)(x^2 + 0x - 5)$. Wait, let me verify: $(x-4)(x^2 - 5) = x^3 - 5x - 4x^2 + 20 = x^3 - 4x^2 - 5x + 20$ ✓. So the zeros of $h(x)$ are $x = 4$ and the zeros of $x^2 - 5 = 0$, which are $x = \pm\sqrt{5}$. The product of the other two zeros is $\sqrt{5} \cdot (-\sqrt{5}) = -5$. Choice B would be the result if both zeros were positive. Choice C would be the result if we incorrectly factored as $x^2 - 3$. Choice D would be the product $\sqrt{3} \cdot \sqrt{3} = 3$ from $x^2 - 3$.
7. A polynomial $f(x)$ of degree $4$ has zeros at $x = -3$, $x = 1$, and $x = 5$, where the zero at $x = 1$ has multiplicity $2$. If $f(x)$ has a $y$-intercept of $30$, what is the leading coefficient?
   Answer: $-2$ Since $f(x)$ has degree 4 with zeros at $x = -3$, $x = 1$ (multiplicity 2), and $x = 5$, we can write $f(x) = a(x + 3)(x - 1)^2(x - 5)$ where $a$ is the leading coefficient. The $y$-intercept is $f(0) = 30$. Substituting: $f(0) = a(0 + 3)(0 - 1)^2(0 - 5) = a(3)(1)(-5) = -15a = 30$. Solving for $a$: $a = 30/(-15) = -2$. Therefore, the leading coefficient is $-2$.
8. If $g(x) = (x-2)^2(x+3)(x-1)$ and $h(x) = k \cdot g(x)$ where $k$ is a nonzero constant, which statement is always true regardless of the value of $k$?
   Answer: $h(x)$ has the same zeros as $g(x)$ When you see a function multiplied by a constant, think about how that transformation affects the key features of the original function. Here, $h(x) = k \cdot g(x)$ represents a vertical scaling of $g(x)$. The zeros of a function occur where the function equals zero. For $h(x) = k \cdot g(x)$, we need $k \cdot g(x) = 0$. Since $k$ is nonzero, this happens exactly when $g(x) = 0$. The zeros of $g(x)$ occur when $(x-2)^2(x+3)(x-1) = 0$, giving us $x = 2$ (with multiplicity 2), $x = -3$, and $x = 1$. These same values make $h(x) = 0$, so choice D is correct. Choice A is wrong because the $y$-intercept changes. When $x = 0$, $g(0) = (0-2)^2(0+3)(0-1) = 4 \cdot 3 \cdot (-1) = -12$, but $h(0) = k \cdot (-12) = -12k$. Unless $k = 1$, the $y$-intercepts differ. Choice B is incorrect because multiplying by $k$ scales all $y$-values, including maximum values. If $k = 2$, every maximum doubles; if $k = -1$, maxima become minima. Choice C is wrong because the end behavior can reverse. Since $g(x)$ has degree 4 with positive leading coefficient, $g(x) \to +\infty$ as $x \to \pm\infty$. But if $k < 0$, then $h(x) \to -\infty$ as $x \to \pm\infty$. Remember: Multiplying a function by a constant preserves zeros but changes almost everything else. The zeros are the "anchor points" that remain fixed under vertical scaling.

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