Worksheet 04 - Finding Zeros of Polynomial Functions

Section 04: Polynomial Functions

Problem 1: Finding Zeros Using the Quadratic Formula (x)

For each function below:

  1. Find all zeros (roots) of the function
  2. Use your calculator’s table function to create a value table in the “interesting” range
  3. Sketch the graph of the function
  1. \(f(x) = 2x^{2} + 2x - 12\)
  2. \(f(x) = 3x^{2} - 13.5x + 6\)
  3. \(f(x) = -2x^{2} - 2x + 4\)
  4. \(f(x) = 0.5x^{2} - 2x - 10.5\)

a) \(f(x) = 2x^{2} + 2x - 12\)

Using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

  • \(a = 2, b = 2, c = -12\)
  • \(x = \frac{-2 \pm \sqrt{4 - 4(2)(-12)}}{2(2)} = \frac{-2 \pm \sqrt{4 + 96}}{4} = \frac{-2 \pm \sqrt{100}}{4} = \frac{-2 \pm 10}{4}\)

Zeros: \(x_1 = 2; x_2 = -3\)

b) \(f(x) = 3x^{2} - 13.5x + 6\)

  • \(a = 3, b = -13.5, c = 6\)
  • \(x = \frac{13.5 \pm \sqrt{182.25 - 72}}{6} = \frac{13.5 \pm \sqrt{110.25}}{6} = \frac{13.5 \pm 10.5}{6}\)

Zeros: \(x_1 = 4; x_2 = 0.5\)

c) \(f(x) = -2x^{2} - 2x + 4\)

  • \(a = -2, b = -2, c = 4\)
  • \(x = \frac{2 \pm \sqrt{4 - 4(-2)(4)}}{2(-2)} = \frac{2 \pm \sqrt{4 + 32}}{-4} = \frac{2 \pm 6}{-4}\)

Zeros: \(x_1 = 1; x_2 = -2\)

d) \(f(x) = 0.5x^{2} - 2x - 10.5\)

  • \(a = 0.5, b = -2, c = -10.5\)
  • \(x = \frac{2 \pm \sqrt{4 - 4(0.5)(-10.5)}}{2(0.5)} = \frac{2 \pm \sqrt{4 + 21}}{1} = \frac{2 \pm 5}{1}\)

Zeros: \(x_1 = 7; x_2 = -3\)

Problem 2: Finding Zeros by Factoring (xx)

For each function below:

  1. Find all zeros by factoring out common terms
  2. Use your calculator’s table function to create a value table in the “interesting” range
  3. Sketch the graph of the function
  1. \(f(x) = 3x^{3} - 12x\)
  2. \(f(x) = -x^{3} - 2x^{2}\) c)b\(f(x) = 3x^{3} + 2x^{2} - 8x\)
  3. \(f(x) = -3x^{4} + 6x^{3}\)
  4. \(f(x) = 3x^{5} - 12x^{3}\)

a) \(f(x) = 3x^{3} - 12x\)

Factor out \(3x\): \[3x(x^{2} - 4) = 3x(x - 2)(x + 2) = 0\]

Zeros: \(x_1 = 0; x_2 = 2; x_3 = -2\)

b) \(f(x) = -x^{3} - 2x^{2}\)

Factor out \(-x^{2}\): \[-x^{2}(x + 2) = 0\]

Zeros: \(x_1 = 0\) (multiplicity 2); \(x_2 = -2\)

c) \(f(x) = 3x^{3} + 2x^{2} - 8x\)

Factor out \(x\): \[x(3x^{2} + 2x - 8) = 0\]

For \(3x^{2} + 2x - 8 = 0\): \[x = \frac{-2 \pm \sqrt{4 + 96}}{6} = \frac{-2 \pm 10}{6}\]

Zeros: \(x_1 = 0; x_2 = \frac{4}{3}; x_3 = -2\)

d) \(f(x) = -3x^{4} + 6x^{3}\)

Factor out \(-3x^{3}\): \[-3x^{3}(x - 2) = 0\]

Zeros: \(x_1 = 0\) (multiplicity 3); \(x_2 = 2\)

e) \(f(x) = 3x^{5} - 12x^{3}\)

Factor out \(3x^{3}\): \[3x^{3}(x^{2} - 4) = 3x^{3}(x - 2)(x + 2) = 0\]

Zeros: \(x_1 = 0\) (multiplicity 3); \(x_2 = 2; x_3 = -2\)

Problem 3: Finding Zeros by Substitution (xxx)

For each function below:

  1. Use substitution (\(u = x^{2}\)) to find all zeros
  2. Use your calculator’s table function to create a value table in the “interesting” range
  3. Sketch the graph of the function

Remember that after solving for \(u\), you need to solve \(x^{2} = u\) for each positive value of \(u\) (negative values of \(u\) have no real solutions since \(x^{2}\) cannot be negative).

  1. \(f(x) = 0.5x^{4} - 3x^{2} + 4\)
  2. \(f(x) = 0.2x^{4} + 0.8x^{2} - 9\)

a) \(f(x) = 0.5x^{4} - 3x^{2} + 4\)

Let \(u = x^{2}\): \[0.5u^{2} - 3u + 4 = 0\]

Multiply by 2: \[u^{2} - 6u + 8 = 0\]

Factor: \[(u - 4)(u - 2) = 0\]

So \(u_1 = 4\) and \(u_2 = 2\)

Since \(x^{2} = u\):

  • From \(x^{2} = 4\): \(x = \pm 2\)
  • From \(x^{2} = 2\): \(x = \pm \sqrt{2}\)

Zeros: \(x_1 = 2; x_2 = -2; x_3 = \sqrt{2} \approx 1.414; x_4 = -\sqrt{2} \approx -1.414\)

b) \(f(x) = 0.2x^{4} + 0.8x^{2} - 9\)

Let \(u = x^{2}\): \[0.2u^{2} + 0.8u - 9 = 0\]

Multiply by 5: \[u^{2} + 4u - 45 = 0\]

Using the quadratic formula: \[u = \frac{-4 \pm \sqrt{16 + 180}}{2} = \frac{-4 \pm \sqrt{196}}{2} = \frac{-4 \pm 14}{2}\]

So \(u_1 = 5\) and \(u_2 = -9\)

Since \(x^{2} = z\) must be non-negative, only \(z_1 = 5\) is valid:

  • From \(x^{2} = 5\): \(x = \pm \sqrt{5}\)

Zeros: \(x_1 = \sqrt{5} \approx 2.236; x_2 = -\sqrt{5} \approx -2.236\)

Note: The value \(z_2 = -9\) does not yield real solutions since \(x^{2}\) cannot be negative.

Problem 4: Mixed Practice (xx)

Choose the appropriate method (quadratic formula, factoring, or substitution) to find all zeros of each function:

  1. \(f(x) = x^{2} + 7x + 12\)
  2. \(f(x) = 2x^{3} - 8x\)
  3. \(f(x) = x^{4} - 5x^{2} + 4\)
  4. \(f(x) = -4x^{2} + 12x - 9\)

a) \(f(x) = x^{2} + 7x + 12\)

Method: Factoring or quadratic formula

Factor: \((x + 3)(x + 4) = 0\)

Zeros: \(x_1 = -3; x_2 = -4\)

b) \(f(x) = 2x^{3} - 8x\)

Method: Factoring

Factor out \(2x\): \[2x(x^{2} - 4) = 2x(x - 2)(x + 2) = 0\]

Zeros: \(x_1 = 0; x_2 = 2; x_3 = -2\)

c) \(f(x) = x^{4} - 5x^{2} + 4\)

Method: Substitution

Let \(z = x^{2}\): \[z^{2} - 5z + 4 = 0\] \[(z - 4)(z - 1) = 0\]

So \(z_1 = 4\) and \(z_2 = 1\)

  • From \(x^{2} = 4\): \(x = \pm 2\)
  • From \(x^{2} = 1\): \(x = \pm 1\)

Zeros: \(x_1 = 2; x_2 = -2; x_3 = 1; x_4 = -1\)

d) \(f(x) = -4x^{2} + 12x - 9\)

Method: Quadratic formula or factoring

Using the quadratic formula:

\[x = \frac{-12 \pm \sqrt{144 - 144}}{-8} = \frac{-12 \pm 0}{-8} = \frac{3}{2}\]

Or factor: \(-4(x^{2} - 3x + \frac{9}{4}) = -4(x - \frac{3}{2})^{2} = 0\)

Zero: \(x = \frac{3}{2} = 1.5\) (multiplicity 2)

Problem 5: Application - Projectile Motion (xxx)

A ball is thrown upward from a platform. Its height \(h\) (in meters) above the ground after \(t\) seconds is given by:

\[h(t) = -5t^{2} + 20t + 15\]

  1. At what time(s) is the ball at ground level (height = 0)? Show your work using the quadratic formula.
  2. What is the maximum height reached by the ball? (Hint: This occurs at the vertex of the parabola, at \(t = -\frac{b}{2a}\))
  3. At what time does the ball reach its maximum height?
  4. For how long is the ball above 30 meters? (Solve \(h(t) = 30\))

a) Ground level: \(h(t) = 0\)

\[-5t^{2} + 20t + 15 = 0\]

Divide by -5: \[t^{2} - 4t - 3 = 0\]

Using the quadratic formula: \[t = \frac{4 \pm \sqrt{16 + 12}}{2} = \frac{4 \pm \sqrt{28}}{2} = \frac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}\]

  • \(t_1 = 2 - \sqrt{7} \approx -0.65\) seconds (not physically meaningful - before launch)
  • \(t_2 = 2 + \sqrt{7} \approx 4.65\) seconds

Answer: The ball hits the ground at approximately 4.65 seconds.

b) Maximum height

Time at maximum: \(t = -\frac{b}{2a} = -\frac{20}{2(-5)} = 2\) seconds

Maximum height: \[h(2) = -5(4) + 20(2) + 15 = -20 + 40 + 15 = 35 \text{ meters}\]

c) Time at maximum height

Answer: 2 seconds (from part b)

d) Time above 30 meters

Solve \(h(t) = 30\): \[-5t^{2} + 20t + 15 = 30\] \[-5t^{2} + 20t - 15 = 0\]

Divide by -5: \[t^{2} - 4t + 3 = 0\] \[(t - 1)(t - 3) = 0\]

The ball is at 30 meters at \(t = 1\) second and \(t = 3\) seconds.

Answer: The ball is above 30 meters for 2 seconds (from \(t = 1\) to \(t = 3\)).