Adding, Subtracting, and Multiplying Polynomials

Polynomials are the building blocks of most expressions in this course. Being quick and accurate at adding, subtracting, and multiplying them makes everything that follows easier, from simplifying rational expressions to working with quadratic functions.

Key ideas

Terms and like terms

A polynomial is a sum of terms like $3x^2$, $-5x$, and $2$. Each term has a coefficient (the number) and a variable part.

Like terms have exactly the same variable part: $3x^2$ and $-7x^2$ are like terms, but $3x^2$ and $3x$ are not. You can only add or subtract like terms.

The degree of a polynomial is its highest exponent. $4x^3 - x + 6$ has degree $3$.

Adding and subtracting

• Adding: remove the brackets and combine like terms.
• Subtracting: the minus sign applies to every term in the second bracket. Change the sign of each term, then combine like terms.

$$(5x - 2) - (3x - 4) = 5x - 2 - 3x + 4 = 2x + 2$$

Multiplying

Multiply every term in the first bracket by every term in the second, then combine like terms. When you multiply powers with the same base, add the exponents: $x \cdot x^2 = x^3$.

$$(x + 3)(x - 5) = x^2 - 5x + 3x - 15 = x^2 - 2x - 15$$

Special products

These patterns come up constantly, so they’re worth memorizing:

$$\begin{aligned} (a + b)^2 &= a^2 + 2ab + b^2 \\ (a - b)^2 &= a^2 - 2ab + b^2 \\ (a + b)(a - b) &= a^2 - b^2 \end{aligned}$$

The middle term $2ab$ is the one students forget: $(x + 3)^2$ is not $x^2 + 9$.

Checking equivalence

Two expressions are equivalent if they’re equal for every value of the variable. A quick test: substitute a number (not $0$ or $1$, which can hide mistakes) into both. If the results differ, they aren’t equivalent. If they match, that’s good evidence, and simplifying both fully confirms it.

On the SAT

Desmos won’t expand or collect like terms, so on the SAT simplifying a polynomial is quickest by hand. To check, graph y = the original expression and y = your answer (or an answer choice): if the graphs lie exactly on top of each other, they’re equivalent. When a question asks for one coefficient, like “what is the coefficient of $x$?”, multiply out only the terms that make that power; it’s faster than expanding everything. See using Desmos on the SAT.

Worked examples

Example 1: Adding and subtracting

Let $P = 3x^2 - 5x + 2$ and $Q = x^2 + 7x - 9$. Find $P + Q$ and $P - Q$.

Solution.

$$P + Q = 3x^2 - 5x + 2 + x^2 + 7x - 9 = 4x^2 + 2x - 7$$ $$\begin{aligned} P - Q &= 3x^2 - 5x + 2 - x^2 - 7x + 9 && \text{change every sign in } Q \\ &= 2x^2 - 12x + 11 \end{aligned}$$

Example 2: Multiplying binomials

Expand and simplify $(2x - 3)(x + 4)$ and $(3x - 5)^2$.

Solution.

$$(2x - 3)(x + 4) = 2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12$$

For the square, use $(a - b)^2 = a^2 - 2ab + b^2$ with $a = 3x$ and $b = 5$:

$$(3x - 5)^2 = 9x^2 - 2(3x)(5) + 25 = 9x^2 - 30x + 25$$

Example 3: A binomial times a trinomial

Expand and simplify $(x + 2)(x^2 - 3x + 5)$.

Solution. Multiply each term of $x + 2$ by all three terms of the trinomial:

$$\begin{aligned} (x + 2)(x^2 - 3x + 5) &= x^3 - 3x^2 + 5x + 2x^2 - 6x + 10 \\ &= x^3 - x^2 - x + 10 \end{aligned}$$

Example 4: Several operations

Simplify $2(x - 3)^2 - (x + 1)(x - 1)$.

Solution. Expand each part first, keeping the second product in brackets so the minus sign reaches every term:

$$\begin{aligned} 2(x - 3)^2 - (x + 1)(x - 1) &= 2(x^2 - 6x + 9) - (x^2 - 1) \\ &= 2x^2 - 12x + 18 - x^2 + 1 \\ &= x^2 - 12x + 19 \end{aligned}$$

Check with $x = 1$: the original gives $2(4) - (2)(0) = 8$, and the answer gives $1 - 12 + 19 = 8$. ✓

Common mistakes

Subtracting only the first term. $(5x - 2) - (3x - 4)$ is $5x - 2 - 3x + 4$. The minus sign changes every sign in the bracket.

Squaring a binomial term by term. $(x + 3)^2 = x^2 + 6x + 9$, not $x^2 + 9$. Write it as $(x + 3)(x + 3)$ if you’re unsure.

Combining unlike terms. $x^2 + x$ can’t be simplified. It is not $x^3$ or $2x^2$.

Adding exponents when you should keep them. When multiplying, add exponents: $x^2 \cdot x^3 = x^5$. When adding like terms, keep them: $x^2 + x^2 = 2x^2$, not $x^4$.

Dropping a negative. In $(2x - 3)(x + 4)$, the product of $-3$ and $4$ is $-12$. Keep track of the sign on every term.

Practice

1. (Warm-up) Simplify $(4x^2 + 3x - 1) + (2x^2 - 5x + 6)$.

Solution

$$6x^2 - 2x + 5$$

2. (Warm-up) Simplify $(5a^2 - 2a + 3) - (a^2 + 4a - 7)$.

Solution

$$5a^2 - 2a + 3 - a^2 - 4a + 7 = 4a^2 - 6a + 10$$

3. (Warm-up) Expand $-3x(2x^2 - x + 4)$.

Solution

$$-6x^3 + 3x^2 - 12x$$

4. (Core) Expand and simplify $(3x + 2)(2x - 5)$.

Solution

$$6x^2 - 15x + 4x - 10 = 6x^2 - 11x - 10$$

5. (Core) Expand $(2y - 7)^2$.

Solution

$$(2y)^2 - 2(2y)(7) + 7^2 = 4y^2 - 28y + 49$$

6. (Core) Expand and simplify $(x - 3)(x^2 + 2x - 4)$.

Solution

$$\begin{aligned} (x - 3)(x^2 + 2x - 4) &= x^3 + 2x^2 - 4x - 3x^2 - 6x + 12 \\ &= x^3 - x^2 - 10x + 12 \end{aligned}$$

7. (Core) Simplify $3(x + 2)^2 - 2(x - 1)(x + 4)$.

Solution

$$\begin{aligned} 3(x + 2)^2 - 2(x - 1)(x + 4) &= 3(x^2 + 4x + 4) - 2(x^2 + 3x - 4) \\ &= 3x^2 + 12x + 12 - 2x^2 - 6x + 8 \\ &= x^2 + 6x + 20 \end{aligned}$$

8. (Core) Are $(x + 3)^2$ and $x^2 + 9$ equivalent? Explain.

Solution

No. Test $x = 1$: $(1 + 3)^2 = 16$, but $1^2 + 9 = 10$.

Expanding shows why: $(x + 3)^2 = x^2 + 6x + 9$, which has an extra $6x$.

9. (Challenge) A rectangle is $(2x + 3)$ m long and $(x + 1)$ m wide. A square with sides of $x$ m is cut out of one corner. Write a simplified expression for the area that’s left.

Solution

$$\begin{aligned} (2x + 3)(x + 1) - x^2 &= 2x^2 + 2x + 3x + 3 - x^2 \\ &= x^2 + 5x + 3 \end{aligned}$$

The remaining area is $(x^2 + 5x + 3)$ m².

10. (Challenge) Expand and simplify $(x + 1)^3$.

Solution

Write it as $(x + 1)^2(x + 1)$:

$$\begin{aligned} (x + 1)^3 &= (x^2 + 2x + 1)(x + 1) \\ &= x^3 + x^2 + 2x^2 + 2x + x + 1 \\ &= x^3 + 3x^2 + 3x + 1 \end{aligned}$$