C. Polynomial and Rational Functions · Polynomial Functions and Equations

Solving Polynomial Equations

C3.4 C3.3About 75 minutes

Learning goals

I can explain the connection between the real roots of P(x) = 0 and the x-intercepts of y = P(x).
I can solve polynomial equations of degree up to 4 by factoring, using the factor theorem and special patterns.
I can use the quadratic formula on a quadratic factor and recognize when it has no real roots.
I can estimate a real root that cannot be found exactly, using a graph or a table with a calculator.

Words to know

root (of an equation)
A value of x that makes the equation true. The real roots of P(x) = 0 are the zeros of P and the x-intercepts of y = P(x).

zero product property
If a product is 0, at least one factor is 0.

real root / non-real root
Some polynomial equations have roots that are not real numbers (for example x² = −4). They do not appear as x-intercepts. You will meet them in university as complex numbers.

repeated (double) root
A root that comes from a repeated factor, such as x = 3 in (x − 3)²(x + 1) = 0.

Three words for one idea

For a polynomial P(x):

the roots of the equation P(x) = 0,
the zeros of the function P,
the x-intercepts of the graph y = P(x)

are the same real numbers. So you can solve an equation by factoring (algebra) or by finding x-intercepts (a graph, by hand or with technology).

A polynomial equation of degree n has at most n real roots. Some equations have fewer, because some of their roots are not real numbers. For example, x² + 4 = 0 has no real roots: x² is never negative. Its graph never meets the x-axis.

Method for solving P(x) = 0 algebraically:

Move every term to one side so the other side is 0. (Never divide both sides by x; you would lose the root x = 0.)
Factor fully: common factor, patterns, grouping, factor theorem.
Set each factor equal to 0 (zero product property). Use the quadratic formula on a quadratic factor that will not factor.
Check, or compare with a graph.

Worked example

Example 1: A cubic with three integer roots

Solve x³ − 2x² − 5x + 6 = 0.

From lesson 11: x³ − 2x² − 5x + 6 = (x − 1)(x − 3)(x + 2).
Set each factor to 0: x − 1 = 0, x − 3 = 0 or x + 2 = 0.

x = 1, x = 3 or x = −2

Worked example

Example 2: Only one real root

Solve 2x³ − 3x² + 8x − 12 = 0.

Group: x²(2x − 3) + 4(2x − 3) = (2x − 3)(x² + 4) = 0.
2x − 3 = 0 gives x = 3/2.
x² + 4 = 0 gives x² = −4, which has no real solution.

The only real root is x = 3/2. The graph crosses the x-axis only once.

y = 2x³ − 3x² + 8x − 12 meets the x-axis only at x = 1.5. The factor x² + 4 never equals 0 for a real x, so it adds no x-intercepts.

Worked example

Example 3: A quartic in quadratic form

Solve x⁴ − 5x² + 4 = 0.

Let m = x²: m² − 5m + 4 = (m − 1)(m − 4) = 0, so m = 1 or m = 4.
x² = 1 gives x = ±1. x² = 4 gives x = ±2.

x = −2, −1, 1, 2

Worked example

Example 4: Irrational roots

Solve x³ − 3x² − 2x + 4 = 0. Give exact roots and decimal approximations.

P(1) = 1 − 3 − 2 + 4 = 0, so x − 1 is a factor. Dividing gives x² − 2x − 4.
(x − 1)(x² − 2x − 4) = 0, so x = 1 or x² − 2x − 4 = 0.
Quadratic formula: x = [2 ± √(4 + 16)]/2 = (2 ± √20)/2 = (2 ± 2√5)/2 = 1 ± √5.

x = 1, x = 1 + √5 ≈ 3.24, x = 1 − √5 ≈ −1.24

Worked example

Example 5: Do not divide by x

Solve x⁴ = 3x³ + 4x².

Move everything to one side: x⁴ − 3x³ − 4x² = 0.
Common factor: x²(x² − 3x − 4) = 0, so x²(x − 4)(x + 1) = 0.
x² = 0 gives the double root x = 0. The other factors give x = 4 and x = −1.
If you had divided both sides by x², you would have lost x = 0.

x = 0 (double root), x = 4, x = −1

Worked example

Example 6: A root you can only estimate

Find the real root of x³ − 2x − 5 = 0 to two decimal places.

The rational root candidates ±1, ±5 do not work: P(1) = −6, P(−1) = −4, P(5) = 110, P(−5) = −120. So there is no rational root.
P(2) = −1 and P(3) = 16: the sign changes, so a root lies between 2 and 3.
Narrow it down with a calculator: P(2.1) ≈ 0.061 (positive), so the root is between 2 and 2.1. P(2.09) ≈ −0.051 (negative), so it is between 2.09 and 2.1. P(2.095) ≈ 0.005 (positive), so it is between 2.09 and 2.095.
Every number between 2.09 and 2.095 rounds to 2.09. A graphing calculator's zero feature gives about 2.0946.

x ≈ 2.09

Left: y = x³ − 2x − 5 crosses the x-axis once, between 2 and 3. Right: zooming in on the dashed box shows the sign change between x = 2.09 and x = 2.1.

Using technology well

Graphing technology (a graphing calculator, Desmos or GeoGebra) shows how many real roots to expect and roughly where they are. Use it to check factored answers and to estimate roots that do not factor nicely. The method of narrowing an interval where the sign changes, used in Example 6, is the same idea computers use to find roots.

Just for fun

In 16th-century Italy, mathematicians held public problem-solving duels. In 1535, Niccolò Tartaglia won a famous contest because he had worked out how to solve cubic equations; Gerolamo Cardano published a cubic method in his 1545 book Ars Magna. Today you can solve many cubics with the factor theorem, no duel required.

Hands-on activity

Hot or cold: hunting a root

You need: a calculator and a partner.

The equation x³ + x − 3 = 0 has one real root, and it is not a nice number.

Your partner keeps a secret: just the rule P(x) = x³ + x − 3. You name a value of x; your partner calculates P(x) and says "positive" or "negative" (not the number).
Find two whole numbers where the answer switches from negative to positive.
Keep halving the interval (try the midpoint each time) until you know the root to one decimal place, then to two.
Count how many guesses you needed. Swap roles with P(x) = x³ − 2x − 5.

What to look for: P(1) = −1 < 0 and P(2) = 7 > 0, so the root is between 1 and 2. Continuing: P(1.2) ≈ −0.07, P(1.25) ≈ 0.20, P(1.21) ≈ −0.02, P(1.22) ≈ 0.04, so the root is about 1.21 (more precisely 1.2134). For x³ − 2x − 5 it is about 2.09.

Your turn

Solve (x − 2)(x + 3)(2x − 1) = 0.

Solve x³ − 7x + 6 = 0.

Solve x³ + 2x² − 9x − 18 = 0.

Solve x⁴ − 13x² + 36 = 0.

Find all real roots of x³ − 8 = 0.

Solve 2x³ − 5x² − 4x + 3 = 0.

Solve x³ = 4x.

More practice: warm-up

Solve x(x − 5) = 0.

What is the greatest number of real roots a cubic equation can have?

Solve (x + 1)²(x − 4) = 0.

Solve x⁴ = 16 over the real numbers.

How many real roots does x² + 9 = 0 have?

More practice: on your own

Solve x⁴ − 2x³ − 13x² + 14x + 24 = 0.

Solve x³ − 4x² + x + 2 = 0 exactly.

Solve 3x³ + x² − 8x + 4 = 0.

Find all real roots of x⁴ + 3x² − 4 = 0.

Solve x³ + x² = 12x.

Challenge

Solve x⁴ − 10x² + 16 = 0 exactly.

x = 2 is a root of x³ + kx² − 4x − 12 = 0. Find k and the other roots.

Explain why every cubic equation with real coefficients has at least one real root, but a quartic equation might have none.

Check your understanding

Quiz

What are the real roots of x³ − x² − 6x = 0?

How many real roots does (x − 1)(x² + 1) = 0 have?

What are the real roots of x⁴ − 9x² = 0?

Which statement about the graph of y = (x + 2)²(x − 5) is true?

What are the exact roots of x³ − 2x² − 2x = 0?

P(1.4) = −0.3 and P(1.5) = 0.2 for a polynomial P. What can you conclude?

For parents and teachers

Big idea (C3.3, C3.4): the real roots of P(x) = 0 are the x-intercepts of y = P(x). To solve, move everything to one side, factor fully, and set each factor equal to zero. Quadratic factors may need the quadratic formula, and some have no real roots.

How to help: After your teen solves an equation, graph the polynomial together (Desmos is free) and check that the x-intercepts match. If a root is missing or extra, look for a sign error or a lost factor.

Common difficulties:

Dividing both sides by x (or x²) and losing the root 0.
Forgetting the negative root when solving x² = 4.
Writing "x = ±2i" style answers; in this course, say the factor has no real roots.
Not rearranging to = 0 before factoring.