Here is a technique that looks like magic the first time you see it, and then becomes the most reliable tool in your algebra kit. Problems that look like they need clever guessing collapse into routine linear systems once you see the trick.
The setup is always the same. Somebody writes down an identity — an equation in x that is supposed to hold for every value of x — and leaves some unknown constants (usually called A, B, C, k, and so on) sitting inside it. You have to find those constants.
The technique: compare coefficients. The constant term on the left must equal the constant term on the right. The coefficient of x on the left must equal the coefficient of x on the right. The coefficient of x² on the left must equal the coefficient of x² on the right. And so on, power by power.
One equation-in-x becomes several equations in the unknown constants. Linear system. Solve. Done.
Why coefficient comparison works
This is not a convention or a trick of notation — it is a theorem. Here is why it is true.
Suppose p(x) and q(x) are polynomials, and suppose p(x) = q(x) for every real number x. Consider their difference, r(x) = p(x) - q(x). Since p and q agree everywhere, r(x) = 0 for every x.
Now, the only polynomial that equals zero for every x is the zero polynomial — the one where every single coefficient is 0. (A non-zero polynomial of degree n can have at most n roots; it cannot vanish at infinitely many points.)
So every coefficient of r = p - q must be zero. But the coefficient of x^k in p - q is just (coefficient of x^k in p) - (coefficient of x^k in q). If their difference is zero, they are equal.
Conclusion: if p(x) = q(x) for all x, then coefficient-by-coefficient, p and q match. That is your license. Whenever you see "for all x," you may compare coefficients.
The partial-fraction decomposition problem
Here is the classic application. You will meet this when you learn integration.
Problem: find constants A and B such that
1 / ((x-1)(x+2)) = A/(x-1) + B/(x+2).
Multiply both sides by (x-1)(x+2) to clear denominators:
1 = A(x+2) + B(x-1).
Expand the right-hand side:
1 = Ax + 2A + Bx - B = (A + B)x + (2A - B).
The left-hand side is just the constant 1 — you can read it as 0·x + 1. Now compare coefficients:
- x-term: A + B = 0.
- Constant term: 2A - B = 1.
This is a simple two-variable linear system. Adding the equations: 3A = 1, so A = 1/3. And then B = -A = -1/3.
Two unknowns found, cleanly, with no guessing.
A cleaner example
Find A, B, C such that x² + 1 = A(x-1)² + B(x-1) + C for all x.
Expand the right-hand side:
A(x² - 2x + 1) + B(x - 1) + C = Ax² + (-2A + B)x + (A - B + C).
Compare coefficients:
- x² coefficient: A = 1.
- x coefficient: -2A + B = 0, so B = 2A = 2.
- Constant: A - B + C = 1, so 1 - 2 + C = 1, giving C = 2.
Answer: A = 1, B = 2, C = 2. Three unknowns, three equations, all determined.
Alternative — substitute specific values
There is a sister technique that you should also know: plug in clever values of x.
Take the same identity x² + 1 = A(x-1)² + B(x-1) + C.
Set x = 1. The (x-1) factors kill the A and B terms:
1 + 1 = 0 + 0 + C, so C = 2.
One unknown, one substitution. This is lightning-fast when the identity contains factors with clean zeros that knock out most of the right-hand side.
When coefficient comparison beats substitution
- When the equation has mixed terms you cannot separate by choosing a clever x. No single x value makes things vanish.
- When you want all unknowns at once, as a single clean linear system.
- When the identity is a structural claim about the polynomial shape rather than a statement about specific roots.
When substitution beats coefficient comparison
- When one root trivially isolates one unknown — plug in that root, watch the other terms vanish, read off the answer.
- When you want a quick first pass. Often you use substitution to nail one or two constants fast, then use coefficient comparison for whatever remains.
The two techniques are not rivals; they are partners. A strong student carries both and picks whichever is shorter for the problem at hand.
Applications
Coefficient comparison is not a party trick. You will use it constantly:
- Partial fractions for integration in calculus.
- Undetermined coefficients when solving differential equations with a polynomial ansatz — you guess the form, then match coefficients to fix the constants.
- Polynomial identity proofs — showing that two expressions are equal for every x.
- Binomial theorem verification — confirming that an expansion matches the claimed form.
- Generating functions in combinatorics, where equalities between power series become equalities between term coefficients.
Worked example — partial fractions
Decompose (3x + 5) / ((x+1)(x+2)) into simpler fractions:
(3x + 5) / ((x+1)(x+2)) = A/(x+1) + B/(x+2).
Multiply both sides by (x+1)(x+2):
3x + 5 = A(x+2) + B(x+1).
Expand the right-hand side:
3x + 5 = (A + B)x + (2A + B).
Compare coefficients:
- x-term: A + B = 3.
- Constant: 2A + B = 5.
Subtract the first from the second: A = 2. Then B = 3 - A = 1.
So (3x + 5) / ((x+1)(x+2)) = 2/(x+1) + 1/(x+2). You can check by combining the right side — the numerator rebuilds to 3x + 5. It works.
Why the technique is reliable
Here is the guarantee. If you expand the right-hand side fully, you get a polynomial with coefficients that are linear combinations of your unknown constants. You get one equation for each distinct power of x appearing anywhere.
If you have k unknowns and the expanded polynomial has degree at least k - 1, you get at least k equations — enough to pin down all the unknowns (provided the system is not degenerate, which well-posed textbook problems will not be).
Coefficient comparison never asks you to guess. It does not require hunting for lucky substitution values. It is a mechanical procedure: expand, group by powers, match, solve.
Recognition drill
For each identity, name what you would compare:
2x + 3 = Ax + B. Compare x-coefficient (A = 2) and constant (B = 3). Done in one line.x² + x + 1 = A(x-1)² + B(x-1) + C. Three equations: one from x² coefficient, one from x coefficient, one from the constant term.(x+1)² = Ax² + Bx + C. Expand LHS to x² + 2x + 1, read off A = 1, B = 2, C = 1.x³ + 3x + 2 = A(x-1) + B(x+1) + C(x-1)(x+1) + D(x² + 1). Four unknowns; expand RHS, collect by powers of x, get four equations from the coefficients of x³, x², x, and the constant.
When you see the phrase "for all x" attached to a polynomial equation with unknown constants inside, that is your cue. Translate it instantly: "coefficient comparison is legal here."
Closing
"Equal for all x" is not just a piece of Latin decoration. It is a license. It lets you replace one polynomial identity by n linear equations in the unknown constants — one equation per power of x.
Every single time you see an identity with letters floating inside it, ask: expand, then match. The mess of x's disappears and you are left with a system of numbers. Linear algebra finishes the job.
That is the whole trick. Simple to state, endlessly useful, and a permanent resident of your algebra toolkit.