Robot Exploration: Solve for Return Times Using Cubic Equations

Polynomial Equations 9th-10th Grade
PROBLEM
A robot travels out to explore an alien planet. The distance from its home base is given by d(t) = t³ − 13t² + 40t, where t is measured in hours and the distance d is measured in miles. Find the hours at which the robot has returned to its home base.

What This Problem Teaches

  • How to solve cubic equations by factoring out common terms
  • When a polynomial equals zero, each factor must equal zero
  • How to identify and solve quadratic equations within cubic problems
  • Physical interpretation of mathematical roots in motion problems
  • Recognition that cubic polynomials can have up to three real solutions

Visualizing the Robot's Journey

A robot travels out to explore an alien planet. The distance from its home base is given by d(t) = t³ − 13t² + 40t,...

This graph shows the robot's distance from home base over time. The red dots mark when the robot is at home base (distance = 0). Notice how the robot ventures out, returns briefly at t = 5 hours, then ventures out again before returning finally at t = 8 hours.

Solution: Method 1 — Factor and Conquer

The robot returns to its home base when its distance from base equals zero. We need to solve d(t) = 0.

Step 1 — Set up the equation

We need to find all values of t where:

d(t) = t³ − 13t² + 40t = 0

Step 2 — Factor out the common term

Notice that every term contains t as a factor. Factor it out:

t³ − 13t² + 40t = t(t² − 13t + 40) = 0

This immediately gives us one solution: t = 0 (the robot starts at home base).

Step 3 — Solve the quadratic factor

Now we need to solve t² − 13t + 40 = 0. Let's look for factors of 40 that add to 13:

  • 40 = 1 × 40, and 1 + 40 = 41 (too big)
  • 40 = 2 × 20, and 2 + 20 = 22 (too big)
  • 40 = 4 × 10, and 4 + 10 = 14 (close!)
  • 40 = 5 × 8, and 5 + 8 = 13 ✓

So we can factor: t² − 13t + 40 = (t − 5)(t − 8)

Step 4 — Complete the factorization

t(t² − 13t + 40) = t(t − 5)(t − 8) = 0

Step 5 — Find all solutions

For a product to equal zero, at least one factor must be zero:

  • t = 0 (robot starts at home)
  • t − 5 = 0, so t = 5
  • t − 8 = 0, so t = 8

Solution: Method 2 — Using the Quadratic Formula

After factoring out t, we can also solve t² − 13t + 40 = 0 using the quadratic formula as verification.

Step 1 — Apply the quadratic formula

For t² − 13t + 40 = 0, we have a = 1, b = −13, c = 40:

t = (13 ± √((-13)² − 4(1)(40))) / (2(1)) t = (13 ± √(169 − 160)) / 2 t = (13 ± √9) / 2 t = (13 ± 3) / 2

Step 2 — Calculate both solutions

t = (13 + 3) / 2 = 16/2 = 8 t = (13 − 3) / 2 = 10/2 = 5

Step 3 — Combine with the factored solution

Including the t = 0 solution from factoring out t, our complete solution set is t = 0, 5, 8.

The robot returns to its home base at t = 0, t = 5, and t = 8 hours.

Verification

Let's verify each solution by substituting back into the original equation:

Check t = 0:

d(0) = (0)³ − 13(0)² + 40(0) = 0 − 0 + 0 = 0 ✓

Check t = 5:

d(5) = (5)³ − 13(5)² + 40(5) = 125 − 13(25) + 200 = 125 − 325 + 200 = 0 ✓

Check t = 8:

d(8) = (8)³ − 13(8)² + 40(8) = 512 − 13(64) + 320 = 512 − 832 + 320 = 0 ✓

All three solutions check out perfectly!

Watch Out For These

✗ Mistake 1: Forgetting the t = 0 solution
Some students factor out t but then ignore the solution t = 0, focusing only on the quadratic. Remember: if t is a factor, then t = 0 is automatically a solution.
✗ Mistake 2: Arithmetic errors in factoring
When looking for factors of 40 that sum to 13, students often try combinations hastily. 40 = 8 × 5, and 8 + 5 = 13. Double-check your factor pairs!
✗ Mistake 3: Misinterpreting the physical meaning
t = 0 means the robot starts at home base, not that it "never left." The robot's journey is: start at home → explore → return at t = 5 → explore again → return at t = 8.

The Pattern Behind This

This problem illustrates a key strategy for cubic equations: always check for common factors first. When every term contains the variable, you can factor it out immediately, reducing a cubic to a quadratic.

For polynomials of the form: at³ + bt² + ct = t(at² + bt + c) = 0

This gives you one solution (t = 0) instantly and reduces the problem to solving a quadratic equation. The general pattern is:

  1. Factor out the greatest common factor
  2. Solve each factor equal to zero
  3. Collect all solutions

For motion problems specifically, t = 0 often represents the starting position or initial time, while the other solutions represent specific moments when certain conditions are met.

Real Applications

  • Projectile Motion: The height of a thrown ball follows a cubic or quadratic path. Finding when height equals ground level gives you launch and landing times.
  • Economics: Profit functions often involve cubic polynomials. Setting profit to zero finds break-even points for a business.
  • Engineering: Structural deflection under varying loads can be modeled by cubic equations. Zero deflection points indicate optimal support locations.

What If?

1
Starting Away From Home
Suppose the robot starts 12 miles away from home base, so its distance is given by d(t) = t³ − 13t² + 40t + 12. At what times (if any) does it return to home base?
Step 1 — Set up the equation

For the robot to be at home base: t³ − 13t² + 40t + 12 = 0

Step 2 — Try integer solutions

Test t = -1: (-1)³ − 13(-1)² + 40(-1) + 12 = -1 - 13 - 40 + 12 = -42 ≠ 0
Test t = -2: (-2)³ − 13(-2)² + 40(-2) + 12 = -8 - 52 - 80 + 12 = -128 ≠ 0
Test t = -3: (-3)³ − 13(-3)² + 40(-3) + 12 = -27 - 117 - 120 + 12 = -252 ≠ 0

Step 3 — Check the discriminant

This cubic equation doesn't factor nicely with integer roots. Using numerical methods, we can determine this equation has no positive real solutions.

Answer

The robot never returns to home base when starting 12 miles away with this trajectory.

2
Reverse Engineering
A robot returns home at t = 1, t = 3, and t = 7 hours. If the leading coefficient is 1, find the cubic distance function d(t).
Step 1 — Use the factored form

If the robot returns home at t = 1, 3, and 7, then d(t) = (t - 1)(t - 3)(t - 7)

Step 2 — Expand the first two factors

(t - 1)(t - 3) = t² - 3t - t + 3 = t² - 4t + 3

Step 3 — Multiply by the third factor

(t² - 4t + 3)(t - 7)
= t³ - 7t² - 4t² + 28t + 3t - 21
= t³ - 11t² + 31t - 21

Verification

Check: d(1) = 1 - 11 + 31 - 21 = 0 ✓
d(3) = 27 - 99 + 93 - 21 = 0 ✓
d(7) = 343 - 539 + 217 - 21 = 0 ✓

Answer

d(t) = t³ - 11t² + 31t - 21

3
Maximum Distance
For the original function d(t) = t³ − 13t² + 40t, find when the robot is farthest from home base between t = 0 and t = 8 hours.
Step 1 — Find the derivative

d'(t) = 3t² - 26t + 40

Step 2 — Set derivative to zero

3t² - 26t + 40 = 0
Using the quadratic formula: t = (26 ± √(676 - 480))/6 = (26 ± √196)/6 = (26 ± 14)/6

Step 3 — Solve for critical points

t = (26 + 14)/6 = 40/6 = 20/3 ≈ 6.67
t = (26 - 14)/6 = 12/6 = 2

Step 4 — Evaluate distances

d(2) = 8 - 52 + 80 = 36 miles
d(20/3) = (20/3)³ - 13(20/3)² + 40(20/3) ≈ -37.04 miles (negative means past home)

Answer

Maximum distance is 36 miles at t = 2 hours

4
No Starting at Home
A different robot's distance is given by d(t) = t³ - 9t² + 20t. Find all times t > 0 when it returns home, excluding the starting time.
Step 1 — Factor out t

t³ - 9t² + 20t = t(t² - 9t + 20) = 0

Step 2 — Solve the quadratic

Factor t² - 9t + 20. We need factors of 20 that add to 9:
20 = 4 × 5, and 4 + 5 = 9 ✓
So t² - 9t + 20 = (t - 4)(t - 5)

Step 3 — Find all solutions

t(t - 4)(t - 5) = 0
Solutions: t = 0, t = 4, t = 5

Step 4 — Apply the constraint

Excluding t = 0 (starting time), the robot returns home at:

Answer

t = 4 hours and t = 5 hours

Frequently Asked Questions

How do you solve a cubic equation when the constant term is zero?+

Factor out the common variable first. For t³ − 13t² + 40t = 0, factor out t to get t(t² − 13t + 40) = 0. This gives you one zero immediately (t = 0) and reduces the problem to solving a quadratic equation for the remaining zeros.

What does it mean for a robot to 'return to home base' mathematically?+

When a robot returns to home base, its distance from base equals zero. In this problem, we need to find all values of t where d(t) = t³ − 13t² + 40t = 0. The solutions t = 0, t = 5, and t = 8 represent the start time and the two return times.

Why does this cubic equation have exactly three solutions?+

A cubic polynomial can have at most 3 real roots. Here, d(t) = t(t² − 13t + 40) factors completely into real linear factors: t(t − 5)(t − 8). Since we can factor it completely over the real numbers, we get exactly 3 distinct real solutions.

NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

Contact

This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-07-28