What is a polynomial equation?
A polynomial equation is p(x) = 0, where p(x) is a polynomial. The degree is the highest power of x.
- Degree 3: cubic, e.g. x³ − 6x² + 11x − 6 = 0.
- Degree 4: quartic, e.g. x⁴ − 5x² + 4 = 0.
A solution (or root) is a value of x that makes the equation true. On the graph y = p(x), each real solution is a point where the curve meets the x-axis.
A polynomial of degree n has at most n real roots. A cubic always has at least one real root, because its two ends go in opposite directions, so the curve must cross the axis somewhere.
Solving cubics and quartics with the factor theorem
The factor theorem says: if p(a) = 0, then (x − a) is a factor. We use it in three moves.
- Guess: list the divisors of the constant term (for 6: ±1, ±2, ±3, ±6). Test them until p(a) = 0.
- Divide: divide p(x) by (x − a), using synthetic or long division. The degree drops by 1.
- Finish: solve the smaller polynomial. A quadratic can be factored or solved with the quadratic formula.
If the leading coefficient is not 1, rational roots p/q are also possible, where p divides the constant term and q divides the leading coefficient.
Grouping
Sometimes terms pair up nicely: x³ + 2x² − 9x − 18 = x²(x + 2) − 9(x + 2) = (x + 2)(x² − 9) = (x + 2)(x − 3)(x + 3).
Common factor first
For 2x⁴ − 8x² = 0, take out 2x²: 2x²(x² − 4) = 0, so x = 0 (a double root), 2 or −2.
Equations that turn into quadratics
A biquadratic equation has only even powers: ax⁴ + bx² + c = 0. Put t = x² (so x⁴ = t²) and you get a quadratic in t.
Example: x⁴ − 5x² + 4 = 0 → t² − 5t + 4 = 0 → (t − 1)(t − 4) = 0 → t = 1 or t = 4. Then x² = 1 gives x = ±1, and x² = 4 gives x = ±2.
Careful: t = x² can never be negative. If you get t = −3, it gives no real x.
The same trick works for other repeated parts, e.g. (x² + x)² − 8(x² + x) + 12 = 0 with t = x² + x.
Polynomial inequalities with a sign chart
To solve p(x) > 0 or p(x) < 0:
- Factor p(x) and find its roots.
- Mark the roots on a number line. They cut it into intervals.
- Check the sign of p(x) in each interval (pick a test value).
- Choose the intervals you need.
Example: (x − 1)(x − 2)(x − 3) > 0. Roots 1, 2, 3. For x > 3 all brackets are positive, so p > 0. The sign flips at each single root: + for x > 3, − for 2 < x < 3, + for 1 < x < 2, − for x < 1. Answer: 1 < x < 2 or x > 3.
At a double root (like (x − 2)²) the sign does not flip; the curve only touches the axis.
Rational equations and inequalities
A rational expression is a fraction of two polynomials, P(x)/Q(x). We can never divide by 0, so first write the domain: all x with Q(x) ≠ 0.
P(x)/Q(x) = 0 means P(x) = 0 and Q(x) ≠ 0.
Example: (x² − 4)/(x − 2) = 0. The numerator is 0 at x = ±2, but x = 2 makes the denominator 0. So the only solution is x = −2.
Equations with fractions on both sides
Multiply both sides by the common denominator, solve the polynomial equation, then check every answer against the domain.
Example: 3/(x − 1) = x + 1. Domain x ≠ 1. Multiply: 3 = (x + 1)(x − 1) = x² − 1, so x² = 4, x = ±2. Both are allowed.
Rational inequalities
Do not multiply by an expression whose sign you do not know. Instead move everything to one side, write one fraction, and use a sign chart with the zeros of both the top and the bottom. Zeros of the bottom are never included.
Example: (x − 3)/(x + 1) ≥ 0 gives x < −1 or x ≥ 3.
Try it: a practical
In the 3D: in the last step, move the slider to shift the cubic up or down. Predict first: will it have 1 or 3 real solutions? Then check the counter. Find a value of k where two roots join into one (the curve just touches the axis).
At home: take an A4 sheet (about 30 cm × 21 cm). Cut equal squares of side x from the corners and fold a box. Measure the volume for x = 2, 3, 4, 5 cm. Which x gives the largest box? You are exploring the cubic V = x(30 − 2x)(21 − 2x).
Key formulas and definitions
- Factor theorem: p(a) = 0 ⇔ (x − a) is a factor
- Integer roots divide the constant term; rational roots p/q: p | constant, q | leading coefficient
- Degree n ⇒ at most n real roots; a cubic has at least 1
- Biquadratic ax⁴ + bx² + c = 0: put t = x², need t ≥ 0
- P(x)/Q(x) = 0 ⇔ P(x) = 0 and Q(x) ≠ 0
- Quadratic formula: x = (−b ± √(b² − 4ac)) / 2a
Worked examples
1. Solve x³ − 6x² + 11x − 6 = 0.
Test x = 1: 1 − 6 + 11 − 6 = 0. Divide by (x − 1): quotient x² − 5x + 6 = (x − 2)(x − 3). Solutions x = 1, 2, 3.
2. Solve x³ + 2x² − 9x − 18 = 0.
Group: x²(x + 2) − 9(x + 2) = (x + 2)(x² − 9) = (x + 2)(x − 3)(x + 3). Solutions x = −2, 3, −3.
3. Solve x⁴ − 5x² + 4 = 0.
Let t = x²: t² − 5t + 4 = 0, (t − 1)(t − 4) = 0, t = 1 or 4. x² = 1 → x = ±1; x² = 4 → x = ±2.
4. Solve x⁴ + x² − 12 = 0 over the real numbers.
t = x²: t² + t − 12 = (t + 4)(t − 3) = 0, so t = −4 or 3. t = −4 is impossible (x² ≥ 0). x² = 3 gives x = ±√3.
5. Solve x³ − 3x² − 4x + 12 = 0, then solve x³ − 3x² − 4x + 12 < 0.
Group: x²(x − 3) − 4(x − 3) = (x − 3)(x − 2)(x + 2). Roots −2, 2, 3. Signs: − for x < −2, + for −2 < x < 2, − for 2 < x < 3, + for x > 3. So p < 0 when x < −2 or 2 < x < 3.
6. Solve (x² − 9)/(x + 3) = 0.
Domain x ≠ −3. Numerator zero at x = ±3, but −3 is not allowed. Solution x = 3.
7. Solve 2x³ − 3x² − 3x + 2 = 0.
Test x = 2: 16 − 12 − 6 + 2 = 0. Divide by (x − 2): 2x² + x − 1 = (2x − 1)(x + 1). Solutions x = 2, 1/2, −1.
8. Solve (x + 2)/(x − 1) ≤ 0.
Zeros: top at x = −2, bottom at x = 1. Sign chart: + for x < −2, − for −2 < x < 1, + for x > 1. Include −2 (top zero), exclude 1. Answer −2 ≤ x < 1.
Common mistakes
- Dividing both sides by x and losing the root x = 0. Factor out x instead.
- Keeping t = x² negative and writing x = √(−4). Negative t gives no real roots.
- Forgetting to check rational-equation answers against the domain (denominator 0).
- Multiplying a rational inequality by (x − 1) without knowing its sign, which can flip the inequality.