The Method of Undetermined Coefficients · Little Blue Book, Visualized

What this move does

Take the quartic $x^{4} + 2 x^{2} + x + 2$ . The sieve from the previous stop says it has no rational root (neither of the candidates $\pm 1, \pm 2$ works), so it has no linear factor. But it could still be a product of two quadratics:

x^{4} + 2 x^{2} + x + 2 = ? (x^{2} + a x + b) (x^{2} + c x + d)

We do not know the coefficients yet, so we let letters hold their places (they are undetermined), and let the equation tell us what they are.

Fix the shape: from the degree and the leading coefficient, write down what the factors look like, using letters for the unknown coefficients.
Multiply out and compare term by term: two polynomials are equal exactly when every coefficient matches. Each coefficient gives one equation.
Solve the system (usually starting with the integer ways to split the constant), and then check every equation.

Predict first

Does $x^{4} + 2 x^{2} + x + 2$ have a linear factor with rational coefficients?

YesNo

Choose an answer first; then you can see the result.

Work through one

Example: factor x⁴ + 2x² + x + 2

Step 1 / 6

Set up the shape
The leading coefficient is $1$ , so both quadratics get leading coefficient $1$ . Four undetermined letters; four lights for four equations.

$(x^{2} + a x + b) (x^{2} + c x + d)$
Start with the constant term
The cubic term
The linear term
One equation left
Write the result

Undetermined coefficients can also find a parameter. Say $x^{3} + 4 x^{2} + m x - 6$ is divisible by $x^{2} + x - 2$ . Call the quotient $x + k$ and multiply out: $(x^{2} + x - 2) (x + k) = x^{3} + (k + 1) x^{2} + (k - 2) x - 2 k$ . Comparing gives $k + 1 = 4$ and $- 2 k = - 6$ , which both give $k = 3$ , so $m = k - 2 = 1$ . The $r_{1}$ and $r_{2}$ of the double cross method at stop 6 used exactly the same idea.

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: It factors in the shape you set up

Write $x^{4} + 2 x^{2} + x + 2$ as $(x^{2} + a x + 1) (x^{2} + c x + 2)$ : the system has a solution.

Practice

Exercise 1Basic

Given $(x + a) (x^{2} + 2 x + 3) = x^{3} + 5 x^{2} + 9 x + 9$ , find $a$ .

Exercise 2Advanced

Given that $x^{3} + 4 x^{2} + m x - 6$ is divisible by $x^{2} + x - 2$ , find $m$ .

Exercise 3Advanced

Factor $x^{4} + x^{3} + 2 x^{2} + x + 1$ .

Exercise 4Challenge

Factor $x^{4} - x^{3} + 4 x^{2} + 3 x + 5$ .

Mistake Museum · Exhibits from this stop

Before you leave, look at the traps other people fell into here.See every exhibit in the museum →

Exhibit

Not using every equation

Write

x^{4} + 2 x^{2} + x + 2 = (x^{2} + a x - 1) (x^{2} + c x - 2)

. From

a + c = 0

and

- 2 a - c = 1

we get

a = - 1

c = 1

, so it factors as

(x^{2} - x - 1) (x^{2} + x - 2)

What went wrong? Think first, then open

Diagnosis One equation was never checked: the coefficient of $x^{2}$ , $a c - 3$ , should equal $2$ , but $(- 1) (1) - 3 = - 4$ . There are more equations than unknowns, and every one of them must hold.

These constants (

- 1

and

- 2

) do not work; switch to

1

and

2

a = 1

c = - 1

, all three equations hold, giving

(x^{2} + x + 1) (x^{2} - x + 2)

“Once you have solved, substitute into the equations you did not use as well”

Exhibit

Thinking only of positive factors of the constant

The constant term of

x^{4} - 3 x^{2} + 1

1

, so it must be

(x^{2} + a x + 1) (x^{2} + c x + 1)

; there is no integer solution, so it cannot be factored.

What went wrong? Think first, then open

Diagnosis $1 = (- 1) \times (- 1)$ is another way to split it. Try $(x^{2} + a x - 1) (x^{2} + c x - 1)$ : $a + c = 0$ and $a c - 2 = - 3$ give $a = 1$ , $c = - 1$ .

x^{4} - 3 x^{2} + 1 = (x^{2} + x - 1) (x^{2} - x - 1)

. When you set up the shape, consider both the positive and the negative splits.

“When splitting the constant, remember that minus times minus is plus”