Add, subtract and multiply polynomials and explain why the results are always polynomials

Learning goal: Add, subtract and multiply polynomials correctly, and understand that polynomials form a system analogous to the integers: the sum, difference and product of two polynomials is always another polynomial, which is what it means to say the system is closed under addition, subtraction and multiplication.

What a polynomial is. A polynomial in x is a sum of terms, each of which is a number (the coefficient) multiplied by a whole-number power of x. So 3x² − 5x + 2 is a polynomial with three terms; 7 is a polynomial with one term (7x⁰); and 4x + 7 is a polynomial of degree 1. The degree of a polynomial is the largest exponent that appears with a nonzero coefficient. Two things are not allowed: a variable in a denominator, as in 1/x, and a variable under a root or with a fractional exponent, as in √x. Those expressions are useful, but they are not polynomials, and the reason the distinction matters is the theme of this lesson.

The analogy with the integers. The integers are the numbers …, −3, −2, −1, 0, 1, 2, 3, …. Add two integers and you get an integer. Subtract two integers and you get an integer. Multiply two integers and you get an integer. Mathematicians say the integers are closed under addition, subtraction and multiplication: you never leave the set by doing those three operations. Division is different. 6 ÷ 3 = 2 is an integer, but 7 ÷ 2 = 3.5 is not, so the integers are not closed under division. Polynomials behave in exactly the same way. Add, subtract or multiply two polynomials and the result is a polynomial; divide one by another and the result is often not a polynomial, as (x² + 1)/x shows, because it contains 1/x. That is why the standard says polynomials form a system analogous to the integers.

Why closure holds. Adding two polynomials means collecting like terms, terms with the same power of x. Each collected coefficient is a sum of numbers, so it is a number, and each power of x is still a whole number. The result is therefore a polynomial. Subtraction is addition of the opposite, so the same reasoning applies. Multiplication uses the distributive property: every term of the first polynomial multiplies every term of the second, and a term axᵐ times a term bxⁿ is (ab)xᵐ⁺ⁿ, a number times a whole-number power of x. Collecting like terms then gives a polynomial. Nothing in any of these steps can create a variable in a denominator or under a root, which is the whole argument.

Addition and subtraction. Take (3x² − 5x + 2) + (x² + 4x − 7). Collect the x² terms: 3x² + x² = 4x². Collect the x terms: −5x + 4x = −x. Collect the constants: 2 + (−7) = −5. The sum is 4x² − x − 5. For the difference (3x² − 5x + 2) − (x² + 4x − 7), the subtraction sign applies to every term of the second polynomial, so it becomes 3x² − 5x + 2 − x² − 4x + 7, which collects to 2x² − 9x + 9. The step students skip is changing all three signs, not just the first one; subtracting only the x² term and adding the rest gives a wrong answer of 2x² − x − 5.

Multiplication. To multiply (x + 3)(2x − 5), distribute each term of the first factor across the second: x·2x = 2x², x·(−5) = −5x, 3·2x = 6x, 3·(−5) = −15. Collect like terms: −5x + 6x = x. The product is 2x² + x − 15. The grid in the diagram is the same computation laid out like an area model: the sides are x + 3 and 2x − 5, the four regions are the four partial products, and the product is their sum. Because one side piece is −5, it is a signed multiplication grid rather than a literal picture of areas: regions such as −5x and −15 carry negative values.

A fully worked model with a cubic result. Multiply (2x − 1)(x² + 3x − 4). The first factor has two terms and the second has three, so there are 2 × 3 = 6 partial products. From 2x: 2x·x² = 2x³, 2x·3x = 6x², 2x·(−4) = −8x. From −1: −1·x² = −x², −1·3x = −3x, −1·(−4) = 4. Collect: 2x³ + (6x² − x²) + (−8x − 3x) + 4 = 2x³ + 5x² − 11x + 4. Check with a value: at x = 2, the original is (4 − 1)(4 + 6 − 4) = 3 × 6 = 18, and the product is 2·8 + 5·4 − 11·2 + 4 = 16 + 20 − 22 + 4 = 18. A single check does not prove an expansion, but it catches most sign and collecting errors. The product has degree 3, the sum of the degrees 1 and 2, which is always true for a product of nonzero polynomials.

Expressions that simplify to linear or quadratic forms. Many polynomial expressions look complicated but collapse. (x + 2)² − (x − 2)² expands to (x² + 4x + 4) − (x² − 4x + 4) = 8x, a linear expression, because the x² terms and the constants cancel. The sum (4x + 7) + (−4x + 2) is simply 9, a polynomial of degree 0. The degree of a sum can be lower than the degree of either addend; the degree of a product of nonzero polynomials never is.

Beyond quadratics. The same rules extend to any degree. (x² − 2x + 3)(x² + 2x + 3) has nine partial products, but a grouping shortcut helps: write it as ((x² + 3) − 2x)((x² + 3) + 2x), which is a difference of squares, (x² + 3)² − (2x)² = x⁴ + 6x² + 9 − 4x² = x⁴ + 2x² + 9. Expanding all nine products and collecting gives the same answer, which is a good exercise in bookkeeping.

The misconception to remove. Students sometimes claim that (x + 3)² = x² + 9, "squaring each term". Test it at x = 1: the left side is 16 and the right side is 10. Squaring is multiplying (x + 3)(x + 3), which has four partial products, x² + 3x + 3x + 9 = x² + 6x + 9. The middle term is not optional.

Worked addition with cancelling terms: Add 4x + 7 and −4x + 2. Collect the x terms: 4x + (−4x) = 0x, which vanishes. Collect the constants: 7 + 2 = 9. The sum is the polynomial 9, of degree 0. This shows that the sum of two degree-1 polynomials need not have degree 1; it is still a polynomial, so closure holds, but the degree dropped because the leading terms cancelled.

Worked subtraction of squares: Simplify (x + 2)² − (x − 2)². Expand each square with all four partial products: (x + 2)² = x² + 2x + 2x + 4 = x² + 4x + 4, and (x − 2)² = x² − 2x − 2x + 4 = x² − 4x + 4. Subtract every term of the second: x² + 4x + 4 − x² + 4x − 4 = 8x. Check at x = 3: (5)² − (1)² = 25 − 1 = 24 and 8 × 3 = 24. The expression simplifies to a linear form even though it began with squares.

Worked product of two quadratics: Multiply (x² − 2x + 3)(x² + 2x + 3). From x²: x⁴ + 2x³ + 3x². From −2x: −2x³ − 4x² − 6x. From 3: 3x² + 6x + 9. Collect by degree: x⁴; 2x³ − 2x³ = 0; 3x² − 4x² + 3x² = 2x²; −6x + 6x = 0; 9. The product is x⁴ + 2x² + 9, a polynomial of degree 4 = 2 + 2. Check at x = 1: (1 − 2 + 3)(1 + 2 + 3) = 2 × 6 = 12 and 1 + 2 + 9 = 12.

Worked closure argument: Explain why the product of any two polynomials is a polynomial. Each term of the first is a number times a whole-number power of x, and so is each term of the second. Any single partial product is (a·b)·x^(m+n), a number times a whole-number power, because whole numbers are closed under addition. The product is the sum of finitely many such terms, and collecting like terms adds numbers, so the result is a polynomial. The same sentence with "sum" in place of "product" proves closure under addition.

Representation task: Draw an area model for (2x + 1)(x + 4): a rectangle split into four regions, sides labeled 2x, 1 along one edge and x, 4 along the other. Write the partial product inside each region, then record the four products in a list, collect like terms and state the final polynomial. Below the model, write a two-column table headed "Integers" and "Polynomials" with one row for each of addition, subtraction, multiplication and division, and in each cell give one example showing whether the result stays in the system. The division row must contain an example that leaves each system.

Independent performance: Choose two polynomials of your own, one of degree 2 and one of degree 3, each with at least three terms and at least one negative coefficient. Compute their sum, their difference in both orders, and their product. For each result, state its degree and explain in one sentence why the degree came out as it did. Check the product by substituting x = 2 into the original factors and into your expansion. Success criteria: every partial product is listed before collecting; the two differences are opposites of each other; the product has degree 5 and the substitution check matches; and one sentence explains why none of the four results could have failed to be a polynomial.