2. Polynomial Functions
  3. Equations of Polynomial Functions in Factored Form

Partial Solutions


  1. There is no solution provided for this question.
    1. The zeros \( -3, -\frac{1}{2},\) and \(\frac{5}{3} \) correspond to linear factors \( (x + 3), (2x + 1) \) and \( (3x - 5) \) respectively. Thus, the equation of the family of polynomials satisfying the given conditions is\[ y = k(x + 3)(2x + 1)(3x - 5), k \neq 0, k \in \mathbb{R} \]
    2. The zeros \( -2 \text{ (order 2) }, 1 \text{ (order 1) }, 5 \text{ (order 3) } \) correspond to factors \( (x + 2)^2, (x - 1) \) and \( (x - 5)^3 \) respectively. Thus, the equation of the family of polynomials satisfying the given conditions is\[ y = k(x + 2)^2(x - 1)(x - 5)^3, k \neq 0, k \in \mathbb{R} \]
    3. A point of inflection at \( (\frac{2}{3}, 0) \) implies that the quartic has a zero of order \( 3 \) at \( x = \frac{2}{3} \), corresponding to a factor of \( (3x - 2)^3 \). Since it is a quartic function, the zero at the origin must be order \( 1 \), corresponding to a factor of \( x \). Thus, the equation of the family of polynomials satisfying the given conditions is\[ y = kx(3x - 2)^3, k \neq 0, k \in \mathbb{R} \]
    4. An \( x \)-intercept at \( x = -4 \) corresponds to a factor of \( (x + 4) \). Since the function is cubic and has a turning point at \( (1, 0) \), it must have a zero of order \( 2 \) at \( x = 1 \), corresponding to a factor of \( (x - 1)^2 \). The end behaviour implies that the cubic has a negative leading coefficient. Thus, the equation of the family of polynomials satisfying the given condition is\[ y = k(x + 4)(x - 1)^2, k \lt 0, k \in \mathbb{R} \]
    5. Since the quartic has four distinct zeros, there will be four distinct linear factors. The zeros \( \pm \sqrt{5}, 1 \pm \sqrt{2} \) correspond to linear factors \( (x - \sqrt{5}), (x + \sqrt{5}), (x - (1 + \sqrt{2})), (x - (1 - \sqrt{2})) \) respectively. Thus, the equation of the family of polynomials satisfying the given conditions is\[\begin{align*} y &= k(x - \sqrt{5})(x + \sqrt{5})(x - (1 + \sqrt{2}))(x - (1 - \sqrt{2})) \\ &= k(x^2 - 5)((x - 1)^2 + 2) \\ &= k(x^2 - 5)(x^2 - 2x + 3), k \neq 0, k \in \mathbb{R} \end{align*}\]
  3. There is no solution provided for this question.
  4. Zeros at \( x = -\frac{1}{2}, 5 \) of order \( 1 \) correspond to factors \( (2x + 1), (x - 5) \) respectively; a zero at \( x = 2 \) of order \( 2 \) corresponds to a factor of \( (x - 2)^2 \). The equation of the quartic is then \( f(x) = a(2x + 1)(x - 5)(x - 2)^2 \). Since the \( y \)-intercept is \( 4 \),\[\begin{align*} f(0) & = 4 \\ a(1)^2(-5)(-2)^2 &= 4 \\ a &= -\tfrac{1}{5} \end{align*}\] Therefore, the equation of the quartic function is \( y = -\frac{1}{5}(2x + 1)(x - 5)(x - 2)^2 \).
  5. There is no solution provided for this question.
    1. \[\begin{align*} f(x) &= k(x - (-3 - \sqrt{5}))(x - (-3 + \sqrt{5})) \\ &= k((x + 3) + \sqrt{5})((x + 3) - \sqrt{5}) \\ &= k((x + 3)^2 - 5) \\ &= k(x^2 + 6x + 4) \end{align*}\]
    2. \[\begin{align*} f(x) &= kx(x - (1 - 2\sqrt{3}))(x - (1 + 2\sqrt{3})) \\ &= kx((x - 1) + 2\sqrt{3})((x - 1) - 2\sqrt{3}) \\ &= kx((x - 1)^2 - 12) \\ &= kx(x^2 - 2x - 11) \\ \end{align*}\]
    3. \[\begin{align*} f(x) &= k(x + 2)(x - 1)(x - 3i)(x + 3i) \\ &= k(x + 2)(x - 1)(x^2 + 9) \\ \end{align*}\]
    4. \[\begin{align*} f(x) &= k(x - (3 + \sqrt{2}))(x - (3 - \sqrt{2}))(x - (-4 + i\sqrt{3}))(x - (-4 - i\sqrt{3})) \\ &= k((x - 3) + \sqrt{2})((x - 3) - \sqrt{2})((x + 4) - i\sqrt{3})((x + 4) + i\sqrt{3}) \\ &= k((x - 3)^2 - 2)((x + 4)^2 + 3) \\ &= k(x^2 - 6x + 7)(x^2 + 8x + 19) \end{align*}\]Alternatively, using the formula \( y = x^2 - \left( \text{sum of zeros} \right)x + \left( \text{product of zeros} \right) \):
      • \( x = 3 \pm \sqrt{2} \) has sum \( 6 \), product \( 7 \), giving quadratic factor \( x^2 - 6x + 7 \).
      • \( x = -4 \pm i\sqrt{3} \) has sum \( -8 \), product \( 16 - (i^2)(\sqrt{3})^2 = 16 + 3 = 19 \), giving quadratic factor \( x^2 + 8x + 19 \).
      Therefore, \( f(x) = k(x^2 - 6x + 7)(x^2 + 8x + 19) \).
  7. There is no solution provided for this question.
  8. The graph of the function is shown.
    Graph of f(x)=0.25(x-2)^2(x+2)^2 w turning point (0,4)
    1. From the graph we see that \( f(x) \) is positive for \( \{ x \mid x \neq -2, 2, x \in \mathbb{R} \} \). There are no values of \( x \) for which \( f(x) \) is negative.
    2. From the graph, the function is increasing on the intervals \( (-2, 0) \cup (2, \infty) \) and decreasing on the intervals \( (-\infty, -2) \cup (0, 2) \).
  9. There is no solution provided for this question.
  10. Let \( g(x) \) be the equation of the transformed function. Using function notation to represent the described transformations on \( f(x) \),\[\begin{align*} g(x) &= -f\left(\frac{1}{2}(x - 3)\right) \\ &= -\left( \frac{1}{2}(x - 3) \right) \left( \frac{1}{2}(x - 3) + 1 \right) \left( \frac{1}{2}(x - 3) - 2\right) \\ &= -\left( \frac{1}{2}x - \frac{3}{2} \right) \left( \frac{1}{2}x - \frac{1}{2} \right) \left( \frac{1}{2}x - \frac{7}{2} \right) \\ &= -\frac{1}{8}(x - 3)(x - 1)(x - 7) \end{align*}\] By inspection, the zeros of \( g(x) \) occur at \( x = 1, 3, 7 \).
  11. There is no solution provided for this question.