Every polynomial multiplication, no matter the size, is the same idea done more or less times: every term of one polynomial multiplies every term of the other, and the resulting products are combined with the same like-terms process from Adding Polynomials. This lesson works through that idea at increasing sizes — one term times many, two binomials, and larger polynomials — plus the handful of products worth recognizing on sight.
Multiplying a Monomial by a Polynomial
Distribute the monomial across every term inside the parentheses, multiplying coefficients and applying the product rule for exponents to each variable.
$$ 3x^{2}(4x^{3}-5x+2) = 3x^{2}\cdot 4x^{3} - 3x^{2}\cdot 5x + 3x^{2}\cdot 2 = 12x^{5} - 15x^{3} + 6x^{2} $$
Each term’s exponent adds to the monomial’s exponent independently — this is exactly the product rule from Exponent Rules, applied once per term.
Multiplying Two Binomials: FOIL
FOIL — First, Outer, Inner, Last — names the four multiplications needed to multiply two binomials, in a fixed order that guarantees nothing gets missed.
$$ (x+3)(x+5) $$
| Step | Multiply | Result |
|---|---|---|
| First | \(x \cdot x\) | \(x^{2}\) |
| Outer | \(x \cdot 5\) | \(5x\) |
| Inner | \(3 \cdot x\) | \(3x\) |
| Last | \(3 \cdot 5\) | \(15\) |
Adding all four results and combining the two like terms (\(5x\) and \(3x\)) finishes the multiplication:
$$ x^{2} + 5x + 3x + 15 = x^{2} + 8x + 15 $$
Multiplying Larger Polynomials: The Box Method
Once one factor has three or more terms, FOIL’s four letters no longer cover every needed multiplication. The box method lays every product out in a grid — one row per term of the first polynomial, one column per term of the second — so nothing is skipped.
$$ (x+4)(x^{2}-3x+2) $$
| \(\times\) | \(x^{2}\) | \(-3x\) | \(2\) |
|---|---|---|---|
| \(x\) | \(x^{3}\) | \(-3x^{2}\) | \(2x\) |
| \(4\) | \(4x^{2}\) | \(-12x\) | \(8\) |
Reading every cell and combining like terms:
$$ x^{3} + (-3x^{2}+4x^{2}) + (2x-12x) + 8 = x^{3} + x^{2} - 10x + 8 $$
Special Product Patterns Worth Memorizing
Three products come up often enough in later factoring work that recognizing the pattern on sight is faster than FOILing from scratch every time.
| Pattern | Expansion | Example |
|---|---|---|
| Difference of squares | \((a+b)(a-b) = a^{2}-b^{2}\) | \((x+6)(x-6) = x^{2}-36\) |
| Perfect square (sum) | \((a+b)^{2} = a^{2}+2ab+b^{2}\) | \((x+5)^{2} = x^{2}+10x+25\) |
| Perfect square (difference) | \((a-b)^{2} = a^{2}-2ab+b^{2}\) | \((x-4)^{2} = x^{2}-8x+16\) |
The middle term in both perfect-square patterns is genuinely there — \((a+b)^{2}\) is not \(a^{2}+b^{2}\), a shortcut that skips a real term. The difference-of-squares pattern is the one case where the middle term is guaranteed to cancel to \(0\), which is exactly what makes it worth recognizing.
Worked Example A: A Monomial Times a Trinomial
Multiply \(-2x^{3}(5x^{2}-3x+4)\).
Step 1 — distribute across every term:
$$ -2x^{3}\cdot 5x^{2} - (-2x^{3})\cdot 3x + (-2x^{3})\cdot 4 $$
Step 2 — multiply coefficients and add exponents in each term:
$$ -10x^{5} + 6x^{4} - 8x^{3} $$
Answer: \(-10x^{5}+6x^{4}-8x^{3}\)
Worked Example B: Two Binomials with FOIL
Multiply \((2x-3)(x+7)\).
Step 1 — apply FOIL:
$$ (2x)(x) + (2x)(7) + (-3)(x) + (-3)(7) $$
Step 2 — evaluate each product:
$$ 2x^{2} + 14x - 3x - 21 $$
Step 3 — combine the like terms:
$$ 2x^{2} + 11x - 21 $$
Answer: \(2x^{2}+11x-21\)
Worked Example C: A Binomial Times a Trinomial with the Box Method
Multiply \((x-2)(3x^{2}+x-5)\).
Step 1 — build the grid:
| \(\times\) | \(3x^{2}\) | \(x\) | \(-5\) |
|---|---|---|---|
| \(x\) | \(3x^{3}\) | \(x^{2}\) | \(-5x\) |
| \(-2\) | \(-6x^{2}\) | \(-2x\) | \(10\) |
Step 2 — combine like terms across the grid:
$$ 3x^{3} + (x^{2}-6x^{2}) + (-5x-2x) + 10 = 3x^{3} - 5x^{2} - 7x + 10 $$
Answer: \(3x^{3}-5x^{2}-7x+10\)
Worked Example D: Recognizing a Special Product
Multiply \((3x+4)^{2}\) using the perfect-square pattern.
Step 1 — identify \(a=3x\) and \(b=4\) in the pattern \((a+b)^{2}=a^{2}+2ab+b^{2}\):
$$ (3x)^{2} + 2(3x)(4) + (4)^{2} $$
Step 2 — evaluate each piece:
$$ 9x^{2} + 24x + 16 $$
Answer: \(9x^{2}+24x+16\). FOILing \((3x+4)(3x+4)\) directly gives the identical answer — the pattern is a shortcut, not a different result.
Common Mistakes to Avoid
- Forgetting the middle term in a perfect-square product. \((a+b)^{2} = a^{2}+2ab+b^{2}\), not \(a^{2}+b^{2}\) — the middle term is real and never disappears.
- Using FOIL on more than two binomials, or on a trinomial factor. FOIL only covers exactly four multiplications; anything larger needs the box method (or repeated distribution) to avoid missing a product.
- Adding exponents incorrectly during distribution. \(x \cdot x^{2} = x^{3}\) (exponents add), not \(x^{2}\) — this is the product rule from Exponent Rules, still in effect during multiplication.
- Dropping a sign when a factor is negative. In \((x-2)(3x^{2}+x-5)\), every product from the \(-2\) row needs its sign tracked carefully — \((-2)(x) = -2x\), not \(2x\).
- Combining terms that aren’t actually alike. After filling out a box or grid, only combine cells whose variable part genuinely matches — a \(x^{2}\) cell and an \(x\) cell stay separate.
- Assuming difference of squares applies to a sum on both sides. \((a+b)(a+b)\) is a perfect-square product, not a difference of squares; the pattern only applies when one factor adds and the other subtracts the exact same two terms.
Where This Shows Up Later
- Factoring Polynomials and Factoring Quadratics. Factoring is multiplication in reverse; recognizing that \(x^{2}-36\) came from \((x+6)(x-6)\) depends on already knowing the difference-of-squares pattern from this lesson.
- Solving quadratic equations. Many quadratic equations arrive already in factored form and need to be expanded (or the reverse) to match a particular solving method.
- Rational Expressions. Multiplying two rational expressions multiplies their numerators and denominators separately, using this exact polynomial-multiplication technique on each.
- Functions. \((fg)(x)\) is defined as the product of two functions’ outputs, computed with this same term-by-term multiplication applied to two function rules.
Practice Problems
Work each problem yourself before opening the answer.
Problem 1. Multiply \(4x(3x^{2}-2x+5)\).
Show answer
$$ 4x \cdot 3x^{2} - 4x \cdot 2x + 4x \cdot 5 = 12x^{3} - 8x^{2} + 20x $$
Answer: \(12x^{3}-8x^{2}+20x\)
Problem 2. Multiply \((x+2)(x+9)\) using FOIL.
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$$ x\cdot x + x\cdot 9 + 2\cdot x + 2\cdot 9 = x^{2}+9x+2x+18 = x^{2}+11x+18 $$
Answer: \(x^{2}+11x+18\)
Problem 3. Multiply \((3x-2)(2x-5)\) using FOIL.
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$$ 3x\cdot 2x + 3x\cdot(-5) + (-2)\cdot 2x + (-2)\cdot(-5) = 6x^{2}-15x-4x+10 = 6x^{2}-19x+10 $$
Answer: \(6x^{2}-19x+10\)
Problem 4. Multiply \((x+8)(x-8)\) using the difference of squares pattern.
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$$ x^{2} - 8^{2} = x^{2} - 64 $$
Answer: \(x^{2}-64\)
Problem 5. Multiply \((x-7)^{2}\) using the perfect-square pattern.
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$$ x^{2} - 2(x)(7) + 7^{2} = x^{2} - 14x + 49 $$
Answer: \(x^{2}-14x+49\)
Problem 6. Multiply \((2x+1)(x^{2}-3x+4)\) using the box method.
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| \(\times\) | \(x^{2}\) | \(-3x\) | \(4\) |
|---|---|---|---|
| \(2x\) | \(2x^{3}\) | \(-6x^{2}\) | \(8x\) |
| \(1\) | \(x^{2}\) | \(-3x\) | \(4\) |
$$ 2x^{3} + (-6x^{2}+x^{2}) + (8x-3x) + 4 = 2x^{3} - 5x^{2} + 5x + 4 $$
Answer: \(2x^{3}-5x^{2}+5x+4\)
Problem 7. Multiply \(-3x^{2}(x^{2}+4x-6)\).
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$$ -3x^{2}\cdot x^{2} - 3x^{2}\cdot 4x - (-3x^{2})\cdot 6 = -3x^{4} - 12x^{3} + 18x^{2} $$
Answer: \(-3x^{4}-12x^{3}+18x^{2}\)
Problem 8. Multiply \((5x+2y)(5x-2y)\) using the difference of squares pattern.
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$$ (5x)^{2} - (2y)^{2} = 25x^{2} - 4y^{2} $$
Answer: \(25x^{2}-4y^{2}\)
Problem 9. Multiply \((x-3)(x^{2}+3x+9)\) using the box method.
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| \(\times\) | \(x^{2}\) | \(3x\) | \(9\) |
|---|---|---|---|
| \(x\) | \(x^{3}\) | \(3x^{2}\) | \(9x\) |
| \(-3\) | \(-3x^{2}\) | \(-9x\) | \(-27\) |
$$ x^{3} + (3x^{2}-3x^{2}) + (9x-9x) + (-27) = x^{3} - 27 $$
Answer: \(x^{3}-27\)
Problem 10. Multiply \((x+1)(x+2)(x+3)\) by multiplying two factors first, then multiplying the result by the third.
Show answer
Step 1 — multiply the first two factors with FOIL:
$$ (x+1)(x+2) = x^{2}+3x+2 $$
Step 2 — multiply that trinomial-in-progress by the third factor using the box method:
| \(\times\) | \(x^{2}\) | \(3x\) | \(2\) |
|---|---|---|---|
| \(x\) | \(x^{3}\) | \(3x^{2}\) | \(2x\) |
| \(3\) | \(3x^{2}\) | \(9x\) | \(6\) |
$$ x^{3} + (3x^{2}+3x^{2}) + (2x+9x) + 6 = x^{3} + 6x^{2} + 11x + 6 $$
Answer: \(x^{3}+6x^{2}+11x+6\)
Quick Reference
| Situation | Move |
|---|---|
| Monomial times polynomial | Distribute across every term; multiply coefficients, add exponents |
| Two binomials | FOIL: First, Outer, Inner, Last, then combine the two middle terms |
| Three or more terms in either factor | Use the box method — one row per term, one column per term |
| \((a+b)(a-b)\) | Difference of squares: \(a^{2}-b^{2}\), middle term always cancels |
| \((a+b)^{2}\) or \((a-b)^{2}\) | Perfect square: \(a^{2}\pm 2ab+b^{2}\) — the middle term is real |
| Degree of the product | Always the sum of the two factors’ degrees |
Once expanding a product feels automatic, Factoring Polynomials and Factoring Quadratics run this exact process in reverse. Revisit Adding Polynomials for the like-terms step every multiplication ends with, or Polynomials for the full overview these lessons split apart. Build speed with the Multiplying Polynomials Generator, or browse the rest of the Algebra lessons as new ones publish.