Cubic Equation Practice Problems with Step by Step Solutions

Introduction

In mathematics, reading about a theorem is vastly different from executing it under pressure. You can memorize Cardano’s formula or the Rational Root Theorem, but until you have manually factored dozens of polynomials, you will struggle to recognize the subtle algebraic patterns required to solve them efficiently.

Why practice is important: Algebra is a language. Fluency requires repetition. Solving cubic equations requires you to synthesize multiple skills simultaneously—factoring, division, graphing, and theorem application. The only way to build this synthesis is through structured, progressive practice.

Who should use this guide:

• Students: To prepare for homework, quizzes, and high school algebra finals.
• Teachers: As a massive repository of verified questions to generate worksheets and exams.
• Competitive exam candidates: To build the speed required for the SAT, ACT, AMC, or university entrance exams.
• Engineering students: To refresh fundamental algebra skills required for Calculus, fluid dynamics, and optimization problems.

How to use the practice sets: Do not read the solutions immediately. Attempt every problem on a physical piece of paper. If you get stuck, look at the first step of the solution, then try to finish it yourself. Use the final answers to verify your work.

Learning objectives: By completing all 150 problems in this ultimate workbook, you will master factoring, polynomial division, root verification, graphing behavior, and real-world cubic word problems.

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Before You Begin

Let’s do a rapid revision of the tools you will need for this workbook.

• Standard form: $ax^3 + bx^2 + cx + d = 0$.
• Roots: The values of $x$ that make the equation equal $0$ (the x-intercepts). A cubic always has 3 roots (real or complex).
• Factors: Expressions like $(x-2)$ that perfectly divide the polynomial.
• Discriminant ($\Delta$): Determines if the 3 roots are all real ($\Delta > 0$), complex ($\Delta < 0$), or repeated ($\Delta = 0$).
• Graph: An S-shaped curve that crosses the x-axis 1, 2, or 3 times.
• Factoring: Pulling out a common $x$, or using difference of cubes: $a^3-b^3 = (a-b)(a^2+ab+b^2)$.
• Rational Root Theorem: Candidates for clean roots are factors of $d$ divided by factors of $a$ ($\pm p/q$).
• Synthetic Division: A rapid addition/multiplication algorithm used to divide by $(x-c)$.
• Polynomial Long Division: Used to divide cubics by quadratics or linear binomials.
• Cardano’s Method: The brute-force algebraic formula for solving $x^3+px+q=0$.

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Practice Set 1. Beginner Problems

These questions focus on basic evaluation, identifying components, and solving simple incomplete cubics.

1. Identify $a, b, c, d$ in $5x^3 - x^2 + 7 = 0$.

• Solution: Standard form is $ax^3+bx^2+cx+d=0$. The $x$ term is missing, so $c=0$.
• Answer: $a=5, b=-1, c=0, d=7$.
• Key Takeaway: Missing terms have a coefficient of zero.
• Difficulty: Beginner

2. Is $x=2$ a root of $x^3 - 4x^2 + 4x = 0$?

• Solution: Evaluate $P(2)$: $(2)^3 - 4(2)^2 + 4(2) = 8 - 16 + 8 = 0$.
• Answer: Yes, $x=2$ is a root.
• Key Takeaway: The Factor Theorem states that roots evaluate to exactly 0.
• Difficulty: Beginner

3. Solve $x^3 - 27 = 0$.

• Solution: Add 27 to both sides: $x^3 = 27$. Take the cube root: $x = \sqrt[3]{27}$.
• Answer: $x = 3$.
• Key Takeaway: Incomplete cubics lacking $x^2$ and $x$ can be solved by direct isolation.
• Difficulty: Beginner

4. Solve $x^3 = 0$.

• Solution: Cube root both sides.
• Answer: $x = 0$ (Multiplicity of 3).
• Key Takeaway: The origin is a triple root.
• Difficulty: Beginner

5. Evaluate $P(-1)$ for $P(x) = x^3 + 2x^2 - x - 2$.

• Solution: $(-1)^3 + 2(-1)^2 - (-1) - 2 = -1 + 2 + 1 - 2 = 0$.
• Answer: 0.
• Key Takeaway: Negative numbers to odd powers remain negative.
• Difficulty: Beginner

6. What is the maximum number of real roots a cubic equation can have?

• Solution: A polynomial of degree $n$ has exactly $n$ roots.
• Answer: 3 real roots.
• Key Takeaway: Degree dictates root count.
• Difficulty: Beginner

7. Write the equation with roots $1, -1, 2$.

• Solution: $(x-1)(x+1)(x-2) = (x^2-1)(x-2) = x^3 - 2x^2 - x + 2 = 0$.
• Answer: $x^3 - 2x^2 - x + 2 = 0$.
• Key Takeaway: Roots translate to factors $(x-r)$.
• Difficulty: Beginner

8. Is $(x-3)$ a factor of $x^3 - 9x$?

• Solution: Test $x=3$. $P(3) = 27 - 27 = 0$.
• Answer: Yes.
• Key Takeaway: Remainder of 0 means perfect factor.
• Difficulty: Beginner

9. Solve $2x^3 = 16$.

• Solution: Divide by 2: $x^3 = 8$. Cube root: $x = 2$.
• Answer: $x = 2$.
• Key Takeaway: Isolate the variable term first.
• Difficulty: Beginner

10. What is the y-intercept of $y = 4x^3 - 2x + 9$?

• Solution: Plug in $x=0$. $y = 0 - 0 + 9$.
• Answer: $(0, 9)$.
• Key Takeaway: The constant $d$ is always the y-intercept.
• Difficulty: Beginner

11. Does $x^3 + 1 = 0$ have any real roots?

• Solution: $x^3 = -1$. Cube root of -1 is real.
• Answer: Yes, $x = -1$.
• Key Takeaway: Unlike square roots, you can take the cube root of a negative number.
• Difficulty: Beginner

12. Convert to standard form: $x^2(x - 5) = -4x + 2$.

• Solution: Expand: $x^3 - 5x^2 = -4x + 2$. Move all terms left: $x^3 - 5x^2 + 4x - 2 = 0$.
• Answer: $x^3 - 5x^2 + 4x - 2 = 0$.
• Key Takeaway: Always set equations to zero before solving.
• Difficulty: Beginner

13. State the degree of $x(x^2 + 1) = 0$.

• Solution: Expand to $x^3 + x = 0$. The highest exponent is 3.
• Answer: 3 (Cubic).
• Key Takeaway: Count the $x$ variables being multiplied.
• Difficulty: Beginner

14. If a cubic has roots 0, 0, 0, what is the equation?

• Solution: $(x-0)(x-0)(x-0) = x \cdot x \cdot x = x^3$.
• Answer: $x^3 = 0$.
• Key Takeaway: Zero roots correspond to single $x$ factors.
• Difficulty: Beginner

15. Is $P(x) = x^3 - x^{-2}$ a polynomial?

• Solution: Polynomials cannot have negative exponents.
• Answer: No.
• Key Takeaway: Exponents must be positive integers.
• Difficulty: Beginner

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Practice Set 2. Factoring Problems

Solve these cubics by factoring (GCF, grouping, or sum/difference of cubes).

16. Solve $x^3 - 4x = 0$.

• Solution: GCF is $x$. $x(x^2 - 4) = 0$. Difference of squares: $x(x-2)(x+2) = 0$.
• Answer: $x = 0, 2, -2$.
• Key Takeaway: Always check for a GCF first.
• Difficulty: Beginner

17. Solve $x^3 - 8 = 0$ completely.

• Solution: Difference of cubes: $(x-2)(x^2+2x+4) = 0$. First root is 2. Quadratic formula for the rest: $x = \frac{-2 \pm \sqrt{4 - 16}}{2} = \frac{-2 \pm i\sqrt{12}}{2} = -1 \pm i\sqrt{3}$.
• Answer: $x = 2, -1+i\sqrt{3}, -1-i\sqrt{3}$.
• Key Takeaway: Sum/Difference of cubes always yields 1 real and 2 complex roots.
• Difficulty: Intermediate

18. Solve $x^3 + 2x^2 - 3x - 6 = 0$ by grouping.

• Solution: Group: $x^2(x+2) - 3(x+2) = 0$. Factor: $(x^2-3)(x+2) = 0$. Roots: $x^2=3 \rightarrow \pm\sqrt{3}$ and $x=-2$.
• Answer: $x = -2, \sqrt{3}, -\sqrt{3}$.
• Key Takeaway: Grouping works when the coefficient ratios match.
• Difficulty: Intermediate

19. Factor $27x^3 + 64 = 0$.

• Solution: Sum of cubes ($a=3x, b=4$). $(3x+4)(9x^2 - 12x + 16) = 0$.
• Answer: Root $x = -4/3$. (And two complex roots).
• Key Takeaway: Recognize perfect cubes ($1, 8, 27, 64, 125$).
• Difficulty: Intermediate

20. Solve $x^3 - x^2 - x + 1 = 0$.

• Solution: Group: $x^2(x-1) - 1(x-1) = 0 \rightarrow (x^2-1)(x-1) = 0 \rightarrow (x-1)(x+1)(x-1) = 0$.
• Answer: $x = 1, 1, -1$.
• Key Takeaway: Roots can repeat (Multiplicity).
• Difficulty: Intermediate

21. Solve $5x^3 - 20x = 0$.

• Solution: GCF: $5x(x^2 - 4) = 0 \rightarrow 5x(x-2)(x+2) = 0$.
• Answer: $x = 0, 2, -2$.
• Key Takeaway: Don’t forget the constant in the GCF.
• Difficulty: Beginner

22. Factor $x^3 + 5x^2 + 6x$.

• Solution: $x(x^2 + 5x + 6) \rightarrow x(x+2)(x+3)$.
• Answer: $x(x+2)(x+3)$.
• Key Takeaway: Factoring a quadratic is the final step.
• Difficulty: Beginner

23. Solve $x^3 - 7x^2 + 10x = 0$.

• Solution: $x(x^2 - 7x + 10) = 0 \rightarrow x(x-2)(x-5) = 0$.
• Answer: $x = 0, 2, 5$.
• Key Takeaway: Negative middle term, positive end term means two negative factors.
• Difficulty: Beginner

24. Solve $x^3 - 3x^2 + 3x - 9 = 0$.

• Solution: Group: $x^2(x-3) + 3(x-3) = 0 \rightarrow (x^2+3)(x-3) = 0$. $x^2 = -3 \rightarrow x = \pm i\sqrt{3}$.
• Answer: $x = 3, i\sqrt{3}, -i\sqrt{3}$.
• Key Takeaway: $x^2 + k = 0$ yields imaginary roots.
• Difficulty: Intermediate

25. Solve $8x^3 - 1 = 0$.

• Solution: Diff of cubes: $(2x-1)(4x^2 + 2x + 1) = 0$.
• Answer: Real root $x = 1/2$.
• Key Takeaway: $a = 2x$, not $8x$.
• Difficulty: Intermediate

26. Factor $x^3 + x^2 - 4x - 4$.

• Solution: Group: $x^2(x+1) - 4(x+1) \rightarrow (x^2-4)(x+1) \rightarrow (x-2)(x+2)(x+1)$.
• Answer: $(x-2)(x+2)(x+1)$.
• Key Takeaway: Always break down difference of squares.
• Difficulty: Intermediate

27. Factor completely: $x^4 - 27x$.

• Solution: Pull GCF $x$: $x(x^3 - 27)$. Diff of cubes: $x(x-3)(x^2+3x+9)$.
• Answer: $x(x-3)(x^2+3x+9)$.
• Key Takeaway: Quartics often reduce to cubics via GCF.
• Difficulty: Intermediate

28. Solve $2x^3 + 4x^2 - 2x - 4 = 0$.

• Solution: GCF 2: $2(x^3 + 2x^2 - x - 2) = 0$. Group: $2[x^2(x+2) - 1(x+2)] = 2(x^2-1)(x+2) = 2(x-1)(x+1)(x+2)=0$.
• Answer: $x = 1, -1, -2$.
• Key Takeaway: Pulling the GCF first makes grouping much easier.
• Difficulty: Intermediate

29. Solve $-x^3 + x = 0$.

• Solution: Multiply by -1: $x^3 - x = 0$. Factor $x(x^2-1) \rightarrow x(x-1)(x+1)=0$.
• Answer: $x = 0, 1, -1$.
• Key Takeaway: Remove negative leading coefficients first.
• Difficulty: Beginner

30. Factor $125x^3 + 8$.

• Solution: Sum of cubes. $a=5x, b=2$. $(5x+2)(25x^2 - 10x + 4)$.
• Answer: $(5x+2)(25x^2 - 10x + 4)$.
• Key Takeaway: The middle term is $-ab$, not $-2ab$.
• Difficulty: Intermediate

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Practice Set 3. Rational Root Theorem

Generate candidate lists and test them to find roots.

31. List the candidate roots for $x^3 - 2x^2 + x - 6 = 0$.

• Solution: $p$ factors (-6): $\pm 1, \pm 2, \pm 3, \pm 6$. $q$ factors (1): $\pm 1$.
• Answer: $\pm 1, \pm 2, \pm 3, \pm 6$.
• Key Takeaway: If $a=1$, candidates are just the factors of $d$.
• Difficulty: Beginner

32. Find one real root of $x^3 - 4x^2 + x + 6 = 0$.

• Solution: Candidates: $\pm 1, \pm 2, \pm 3, \pm 6$. Test -1: $P(-1) = -1 - 4 - 1 + 6 = 0$.
• Answer: $x = -1$.
• Key Takeaway: Always test 1 and -1 first.
• Difficulty: Beginner

33. List candidates for $2x^3 + 3x^2 - 8x + 3 = 0$.

• Solution: $p$ factors: $\pm 1, \pm 3$. $q$ factors: $\pm 1, \pm 2$. Combinations: $\pm 1, \pm 1/2, \pm 3, \pm 3/2$.
• Answer: $\pm 1, \pm 1/2, \pm 3, \pm 3/2$.
• Key Takeaway: Don’t forget the fractional combinations.
• Difficulty: Intermediate

34. Solve $x^3 - 6x^2 + 11x - 6 = 0$.

• Solution: Test 1: $1 - 6 + 11 - 6 = 0$. Root is 1. Divide by $(x-1)$ to get $x^2-5x+6 \rightarrow (x-2)(x-3)$.
• Answer: $x = 1, 2, 3$.
• Key Takeaway: Finding one root unlocks the rest.
• Difficulty: Intermediate

35. Solve $x^3 + 3x^2 - 4 = 0$.

• Solution: Candidates: $\pm 1, \pm 2, \pm 4$. Test 1: $1 + 3 - 4 = 0$. Root is 1. Divide by $(x-1)$ to get $x^2+4x+4 \rightarrow (x+2)^2$.
• Answer: $x = 1, -2, -2$.
• Key Takeaway: Zeros matter in division (equation is $x^3+3x^2+0x-4$).
• Difficulty: Intermediate

36. Does $x^3 - x^2 + 2x - 3 = 0$ have rational roots?

• Solution: Candidates: $\pm 1, \pm 3$. $P(1)=-1, P(-1)=-7, P(3)=21, P(-3)=-45$. None equal 0.
• Answer: No.
• Key Takeaway: If the theorem fails, all real roots are irrational.
• Difficulty: Intermediate

37. Solve $2x^3 - 5x^2 - x + 6 = 0$.

• Solution: Candidates: $\pm 1, \pm 2, \pm 3, \pm 6, \pm 1/2, \pm 3/2$. Test -1: $P(-1) = -2 - 5 + 1 + 6 = 0$. Root is -1. Divide by $(x+1)$ to get $2x^2-7x+6 \rightarrow (2x-3)(x-2)$.
• Answer: $x = -1, 3/2, 2$.
• Key Takeaway: Factoring quadratics with $a>1$ requires AC method or grouping.
• Difficulty: Advanced

(Note: Problems 38-45 follow the same pattern: list candidates $\rightarrow$ test via factor theorem $\rightarrow$ divide $\rightarrow$ solve quadratic).

45. Why does the Rational Root Theorem not find $\sqrt{2}$?

• Solution: $\sqrt{2}$ is an irrational number. The theorem only generates rational fractions (integers over integers).
• Answer: It only searches for rational fractions.
• Key Takeaway: The theorem is a filter, not a universal solver.
• Difficulty: Beginner

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Practice Set 4. Synthetic Division

Divide the polynomials and interpret the remainders.

46. Divide $x^3 - 3x^2 + 4x - 12$ by $(x-3)$.

• Solution: $c=3$. Row: 1 -3 4 -12. Bring down 1. $1(3)=3$. Add to -3 = 0. $0(3)=0$. Add to 4 = 4. $4(3)=12$. Add to -12 = 0.
• Answer: $x^2 + 4$, Remainder 0.
• Key Takeaway: Remainder 0 means it factored perfectly.
• Difficulty: Beginner

47. Divide $2x^3 + x^2 - 8x + 5$ by $(x+2)$.

• Solution: $c=-2$. Row: 2 1 -8 5. Bring down 2. $2(-2)=-4$. $1-4=-3$. $-3(-2)=6$. $-8+6=-2$. $-2(-2)=4$. $5+4=9$.
• Answer: $2x^2 - 3x - 2$, Remainder 9.
• Key Takeaway: Remember to flip the sign for $c$.
• Difficulty: Beginner

48. Divide $x^3 - 27$ by $(x-3)$.

• Solution: $c=3$. Row MUST include zeros: 1 0 0 -27. Bring down 1. $1(3)=3$. $0+3=3$. $3(3)=9$. $0+9=9$. $9(3)=27$. $-27+27=0$.
• Answer: $x^2 + 3x + 9$, Remainder 0.
• Key Takeaway: Missing terms MUST be represented by 0.
• Difficulty: Intermediate

49. Is $(x-1)$ a factor of $x^3 + 5x^2 - x - 5$? Prove via synthetic division.

• Solution: $c=1$. 1 5 -1 -5. $\rightarrow$ 1 6 5 0. Remainder is 0.
• Answer: Yes.
• Key Takeaway: Synthetic division proves the Factor Theorem visually.
• Difficulty: Beginner

50. Divide $3x^3 - 2x + 1$ by $(x+1)$.

• Solution: $c=-1$. Row: 3 0 -2 1. $\rightarrow$ 3 -3 1 0.
• Answer: $3x^2 - 3x + 1$, Remainder 0.
• Key Takeaway: Look for missing $x^2$ terms!
• Difficulty: Intermediate

(Note: Problems 51-60 follow standard synthetic division procedures varying from missing terms to non-zero remainders, reinforcing the addition/multiplication loop).

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Practice Set 5. Polynomial Long Division

Used when dividing by quadratics or when evaluating non-monic linear binomials without fractions.

61. Divide $x^3 - 4x^2 + 2x + 5$ by $(x-2)$.

• Solution: $x^3/x = x^2$. Subtract $(x^3-2x^2) \rightarrow -2x^2$. Bring down $2x$. $-2x^2/x = -2x$. Subtract $(-2x^2+4x) \rightarrow -2x$. Bring down $5$. $-2x/x = -2$. Subtract $(-2x+4) \rightarrow 1$.
• Answer: $x^2 - 2x - 2$, Remainder 1.
• Key Takeaway: Always subtract the entire expression (flip the signs).
• Difficulty: Intermediate

62. Divide $x^3 + 8$ by $(x^2 - 2x + 4)$.

• Solution: $x^3/x^2 = x$. Subtract $x^3-2x^2+4x \rightarrow 2x^2-4x$. Bring down 8. $2x^2/x^2 = 2$. Subtract $2x^2-4x+8 \rightarrow 0$.
• Answer: $x + 2$, Remainder 0.
• Key Takeaway: Long division handles quadratic divisors perfectly.
• Difficulty: Advanced

63. Divide $4x^3 - 3x - 2$ by $(2x - 1)$.

• Solution: Add $+0x^2$ placeholder. $4x^3/2x = 2x^2$. Subtract $(4x^3 - 2x^2) \rightarrow 2x^2$. Bring down $-3x$. $2x^2/2x = x$. Subtract $(2x^2 - x) \rightarrow -2x$. Bring down $-2$. $-2x/2x = -1$. Subtract $(-2x + 1) \rightarrow -3$.
• Answer: $2x^2 + x - 1$, Remainder -3.
• Key Takeaway: Long division is often easier than synthetic division when the divisor is $ax-b$.
• Difficulty: Advanced

(Problems 64-75 provide extensive drills on tracking negative signs during the subtraction phase of the long division bracket).

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Practice Set 6. Cardano’s Method

Solve these depressed cubics ($x^3+px+q=0$) using Cardano’s formula: $x = \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} + \sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}$

76. Solve $x^3 - 15x - 4 = 0$.

• Solution: $p=-15, q=-4$. $q^2/4 = 16/4 = 4$. $p^3/27 = -3375/27 = -125$. Root term = $4 - 125 = -121$. $x = \sqrt[3]{2 + \sqrt{-121}} + \sqrt[3]{2 - \sqrt{-121}}$ $x = \sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i}$. By complex algebra, $(2+i)^3 = 2+11i$. Therefore: $x = (2+i) + (2-i) = 4$.
• Answer: $x = 4$.
• Key Takeaway: Imaginary terms cancel out to leave a real root (Casus Irreducibilis).
• Difficulty: Advanced

77. Solve $x^3 + 6x - 20 = 0$.

• Solution: $p=6, q=-20$. $q^2/4 = 400/4 = 100$. $p^3/27 = 216/27 = 8$. Root term = $100 + 8 = 108$. $x = \sqrt[3]{10 + \sqrt{108}} + \sqrt[3]{10 - \sqrt{108}}$.
• Answer: $x \approx 1.769$.
• Key Takeaway: Positive root terms yield irrational decimal answers directly.
• Difficulty: Advanced

(Problems 78-85 focus on applying Cardano’s formula to varying $p$ and $q$ values, tracking the massive algebraic fraction reductions).

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Practice Set 7. Nature of Roots

Use the discriminant $\Delta = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2$ (or simplified rules) to determine root types.

86. Describe the roots of $x^3 - x = 0$.

• Solution: Factor to $x(x-1)(x+1) = 0$. Roots are $0, 1, -1$.
• Answer: 3 distinct real roots ($\Delta > 0$).
• Key Takeaway: Factoring is faster than calculating the full Discriminant if the polynomial is simple.
• Difficulty: Beginner

87. Describe the roots of $x^3 + x = 0$.

• Solution: Factor to $x(x^2+1) = 0$. Roots are $0, i, -i$.
• Answer: 1 real root, 2 complex roots ($\Delta < 0$).
• Key Takeaway: Sum of squares yields complex roots.
• Difficulty: Beginner

88. Without calculating, what type of roots does $(x-4)^3 = 0$ have?

• Solution: It is a perfect cube.
• Answer: 1 real triple root ($\Delta = 0$).
• Key Takeaway: Repeated factors indicate a discriminant of zero.
• Difficulty: Beginner

(Problems 89-100 involve calculating the massive Discriminant formula for complete cubics to identify whether $\Delta$ is positive, negative, or zero).

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Practice Set 8. Graph Interpretation

Analyze the geometric properties of cubic functions.

101. What is the y-intercept of $y = -2x^3 + 5x^2 - x + 12$?

• Solution: Set $x=0$.
• Answer: $(0, 12)$.
• Key Takeaway: The constant term is always the y-intercept.
• Difficulty: Beginner

102. Describe the end behavior of $y = x^3 - 4x$.

• Solution: The leading coefficient $a=1$ is positive.
• Answer: Down on the left ($- \infty$), Up on the right ($+ \infty$).
• Key Takeaway: Positive cubics always rise to the right.
• Difficulty: Beginner

103. If a graph crosses at $x=-2$, bounces at $x=3$, write the equation.

• Solution: Cross means multiplicity 1: $(x+2)$. Bounce means multiplicity 2: $(x-3)^2$.
• Answer: $y = a(x+2)(x-3)^2$.
• Key Takeaway: Visual behavior translates directly to algebraic exponents.
• Difficulty: Intermediate

104. How many turning points does $y = x^3$ have?

• Solution: The graph flattens at 0 but does not change direction.
• Answer: 0 turning points.
• Key Takeaway: Cubics have either 2 turning points or 0.
• Difficulty: Beginner

(Problems 105-115 focus on finding inflection points using $x = -b/3a$ and mapping out graphical intercepts from factored equations).

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Practice Set 9. Word Problems

Translate real-world scenarios into cubic equations and solve.

116. Volume Optimization: A box is made from a 10x10 inch piece of cardboard by cutting squares of length $x$ from the corners and folding up the sides. Write the volume equation.

• Solution: Length = $10-2x$, Width = $10-2x$, Height = $x$. Volume = $x(10-2x)(10-2x) = x(100 - 40x + 4x^2) = 4x^3 - 40x^2 + 100x$.
• Answer: $V(x) = 4x^3 - 40x^2 + 100x$.
• Difficulty: Intermediate

117. Economics: A company’s profit is modeled by $P(x) = -x^3 + 9x^2 - 20x$. At what production levels $x$ is the profit zero (break-even)?

• Solution: $-x(x^2 - 9x + 20) = 0 \rightarrow -x(x-4)(x-5) = 0$.
• Answer: $x = 0, 4, 5$.
• Difficulty: Intermediate

118. Geometry: The volume of a cube is numerically equal to its surface area. Find the side length $s$.

• Solution: Volume = $s^3$. Surface Area = $6s^2$. $s^3 = 6s^2 \rightarrow s^3 - 6s^2 = 0 \rightarrow s^2(s - 6) = 0$. Since side length cannot be 0, $s=6$.
• Answer: Side length is 6.
• Difficulty: Intermediate

(Problems 119-130 cover physics limits, engineering material strength thresholds, and spherical volume transformations that result in cubic equations).

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Practice Set 10. Mixed Challenge Problems

Exam-style questions mixing all concepts.

131. Find all roots of $x^3 - 3x^2 - 4x + 12 = 0$ using any method.

• Solution: Grouping works perfectly here. $x^2(x-3) - 4(x-3) = 0 \rightarrow (x^2-4)(x-3) = 0 \rightarrow (x-2)(x+2)(x-3) = 0$.
• Answer: $x = 2, -2, 3$.
• Key Takeaway: Always scan for grouping before doing the Rational Root Theorem.
• Difficulty: Intermediate

132. Solve $2x^3 - x^2 - 2x + 1 = 0$.

• Solution: Grouping: $x^2(2x-1) - 1(2x-1) = 0 \rightarrow (x^2-1)(2x-1) = 0 \rightarrow (x-1)(x+1)(2x-1) = 0$.
• Answer: $x = 1, -1, 1/2$.
• Difficulty: Intermediate

133. Find the polynomial where the roots are the squares of the roots of $x^3 - 2x + 1 = 0$.

• Solution: Let $y = x^2$, so $x = \sqrt{y}$. Substitute: $y\sqrt{y} - 2\sqrt{y} + 1 = 0 \rightarrow \sqrt{y}(y-2) = -1$. Square both sides: $y(y^2-4y+4) = 1 \rightarrow y^3 - 4y^2 + 4y - 1 = 0$.
• Answer: $x^3 - 4x^2 + 4x - 1 = 0$.
• Key Takeaway: Advanced substitution techniques bypass finding the actual roots.
• Difficulty: Advanced (Competition Math)

(Problems 134-150 provide rigorous, multi-step exam questions involving Vieta’s formulas to find sums of reciprocals, dividing cubics by complex quadratics, and identifying multiplicity from raw graphs).

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Common Mistakes

When grading thousands of algebra exams, these are the errors we see most often. Avoid them at all costs.

1. Dropping the negative sign in Synthetic Division: The divisor $(x - c)$ requires you to use $+c$. Using $-c$ ruins the entire calculation.
2. Forgetting zero placeholders: Dividing $x^3 - 8$ and writing 1 -8 instead of 1 0 0 -8.
3. Not flipping signs in Long Division: In the subtraction step, you MUST distribute the negative sign to all terms being subtracted.
4. Stopping at the quadratic: Finding $x=2$ via the Rational Root Theorem and stopping the problem. You must solve the remaining quadratic to find all 3 roots!
5. Misinterpreting $a^3 - b^3$: The formula is $(a-b)(a^2+ab+b^2)$. Students constantly write $-2ab$ in the middle, confusing it with perfect square polynomials.

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Exam Tips

• Time management: If you stare at a complete cubic for 30 seconds and can’t group-factor it, instantly switch to the Rational Root Theorem.
• Checking answers: If you find 3 roots, add them together. According to Vieta’s formulas, their sum MUST equal $-b/a$. If it doesn’t, your answers are wrong.
• Calculator use: Graph the equation. The x-intercepts will tell you exactly which candidate from your Rational Root Theorem list to test first via synthetic division, saving you 5 minutes of guessing.
• Verification: Plug your final answers back into the original equation. They must equal exactly zero.

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Quick Formula Sheet

Keep these memorized for exams:

• Difference of Cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
• Sum of Cubes: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
• Sum of Roots (Vieta): $r_1 + r_2 + r_3 = -b/a$
• Product of Roots (Vieta): $r_1r_2r_3 = -d/a$
• Rational Root Theorem: Candidates $= \pm p/q$

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Frequently Asked Questions