What Does It Mean To Multiply Two Binomials?
Multiplying two binomials means finding the product of two two-term expressions by multiplying each term of the first by each term of the second, then combining like terms. For $(a+b)(c+d)$, that produces four products: $ac$, $ad$, $bc$, and $bd$. Add them, simplify, and you have the answer.
This is the same idea as the area of a rectangle whose sides are each split into two pieces. If one side measures $a+b$ and the other measures $c+d$, the rectangle breaks into four smaller rectangles with areas $ac$, $ad$, $bc$, and $bd$. The total area is the sum. The algebra and the geometry are the same fact wearing different clothes.
The three methods, and when each one helps
There is more than one route to the same product, and you should know they all exist even if you settle on a favourite.
The distributive property is the foundation. Distribute the first binomial across the second, then distribute again. It works for any polynomials, not just binomials, so it never lets you down.
FOIL is a memory aid for the distributive property applied to exactly two binomials. The letters stand for First, Outer, Inner, Last — the four products in the order you write them. FOIL is the quickest method for two binomials, but it only works for binomials.
The box method (also called the grid or area method) organises the four products in a 2-by-2 table. The grid that comes from the rusher forgetting a term is much harder to forget when every product has its own cell.
How Do You Multiply Binomials Using FOIL?
FOIL gives you the four products in a fixed order so none goes missing. Take $(x + 3)(x + 5)$:
First — multiply the first terms: $x \cdot x = x^2$
Outer — multiply the outer terms: $x \cdot 5 = 5x$
Inner — multiply the inner terms: $3 \cdot x = 3x$
Last — multiply the last terms: $3 \cdot 5 = 15$
Now collect the four pieces and combine the like terms $5x$ and $3x$:
$$x^2 + 5x + 3x + 15 = x^2 + 8x + 15$$
The two middle terms almost always combine, which is why a multiplied pair of binomials so often lands as a three-term trinomial.
The special products worth recognising on sight
A few binomial products show up so often that recognising the pattern beats running all four FOIL steps. They are still ordinary binomial multiplication; you are just spotting the shortcut.
Square of a binomial: $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$. The Outer and Inner products are identical here, so they combine into the single middle term $2ab$.
Difference of squares: $(a+b)(a-b) = a^2 - b^2$. The Outer and Inner products cancel, so the middle term vanishes and you are left with just two terms.
Both are worth memorising, but neither replaces the underlying rule: every term in the first bracket still meets every term in the second.
Examples of Multiplying Binomial
Example 1
Multiply $(x + 2)(x + 6)$.
First $x \cdot x = x^2$.
Outer $x \cdot 6 = 6x$.
Inner $2 \cdot x = 2x$.
Last $2 \cdot 6 = 12$.
Combine the like terms:
$$x^2 + 6x + 2x + 12 = x^2 + 8x + 12$$
Final answer: $x^2 + 8x + 12$.
Example 2
Multiply $(x - 4)(x + 7)$.
A common first instinct is to treat the subtraction as if it disappears once you start multiplying, and write the constant term as $4 \cdot 7 = 28$. Watch where that goes wrong.
Wrong attempt. A student keeps the $-4$ in the binomial but writes the last product as $+28$, getting $x^2 + 7x - 4x + 28 = x^2 + 3x + 28$. Check it against the structure: the two constants are $-4$ and $+7$, and a negative times a positive is negative. The last product cannot be positive.
Correct. Carry the sign through every product:
First $x \cdot x = x^2$.
Outer $x \cdot 7 = 7x$.
Inner $(-4) \cdot x = -4x$.
Last $(-4) \cdot 7 = -28$.
$$x^2 + 7x - 4x - 28 = x^2 + 3x - 28$$
Final answer: $x^2 + 3x - 28$.
Example 3
Multiply $(2x + 3)(x - 5)$.
First $2x \cdot x = 2x^2$.
Outer $2x \cdot (-5) = -10x$.
Inner $3 \cdot x = 3x$.
Last $3 \cdot (-5) = -15$.
$$2x^2 - 10x + 3x - 15 = 2x^2 - 7x - 15$$
Final answer: $2x^2 - 7x - 15$.
Example 4
Multiply $(3a - 2b)(4a + b)$ — two variables.
First $3a \cdot 4a = 12a^2$.
Outer $3a \cdot b = 3ab$.
Inner $(-2b) \cdot 4a = -8ab$.
Last $(-2b) \cdot b = -2b^2$.
$$12a^2 + 3ab - 8ab - 2b^2 = 12a^2 - 5ab - 2b^2$$
Final answer: $12a^2 - 5ab - 2b^2$.
Example 5
Multiply $\left(x + \tfrac{1}{2}\right)\left(x - \tfrac{1}{3}\right)$ — with fractions.
First $x \cdot x = x^2$.
Outer $x \cdot \left(-\tfrac{1}{3}\right) = -\tfrac{1}{3}x$.
Inner $\tfrac{1}{2} \cdot x = \tfrac{1}{2}x$.
Last $\tfrac{1}{2} \cdot \left(-\tfrac{1}{3}\right) = -\tfrac{1}{6}$.
Combine the middle terms over a common denominator of 6: $-\tfrac{2}{6}x + \tfrac{3}{6}x = \tfrac{1}{6}x$.
$$x^2 + \tfrac{1}{6}x - \tfrac{1}{6}$$
Final answer: $x^2 + \tfrac{1}{6}x - \tfrac{1}{6}$.
Example 6
Multiply three binomials: $(x + 1)(x + 2)(x + 3)$.
Multiply two of them first, then multiply the result by the third. Start with $(x+1)(x+2)$:
$$(x+1)(x+2) = x^2 + 3x + 2$$
Now multiply $(x^2 + 3x + 2)$ by $(x + 3)$, distributing each term:
$$x^2 \cdot (x+3) = x^3 + 3x^2$$
$$3x \cdot (x+3) = 3x^2 + 9x$$
$$2 \cdot (x+3) = 2x + 6$$
Add and combine like terms:
$$x^3 + 3x^2 + 3x^2 + 9x + 2x + 6 = x^3 + 6x^2 + 11x + 6$$
Final answer: $x^3 + 6x^2 + 11x + 6$.
Where The FOIL Shortcut Quietly Stops Working
The honest caution about FOIL: it is not a multiplication rule, it is a counting trick for one specific case. When students learn FOIL as the way to multiply, the method breaks the moment a problem has more than two terms.
The first instinct when multiplying $(x + 3)$ by a trinomial like $(x^2 + 2x + 5)$ is to reach for FOIL out of habit — but there is no clean First-Outer-Inner-Last with three terms in the second bracket. The distributive property is the rule underneath; FOIL is just the distributive property with the four steps named. Keep the distributive property as your real tool and FOIL as the fast lane for the two-binomial case.
This connects forward to the binomial theorem, which generalises the whole idea to any power, and back to multiplying a monomial across a bracket, which is the single-distribution version of the same move.
Where Students Trip Up On Multiplying Binomial
Mistake 1: Dropping a sign on the inner or last product
Where it slips in: When one or both binomials contain a subtraction, students carry the minus into the first product or two but forget it on the last.
Don't do this: Writing $(x - 4)(x + 7) = x^2 + 3x + 28$ — the $+28$ should be $-28$, because $(-4)(+7)$ is negative.
The correct way: Treat each term as carrying its own sign, including the minus, and apply the sign rules on every one of the four products. A negative times a positive is always negative.
Mistake 2: Forgetting to combine the two middle terms
Where it slips in: The memorizer who can recite "First, Outer, Inner, Last" sometimes stops after writing the four products and never simplifies.
Don't do this: Leaving the answer as $x^2 + 5x + 3x + 15$ and calling it final.
The correct way: The Outer and Inner products are usually like terms — combine them. $x^2 + 5x + 3x + 15$ simplifies to $x^2 + 8x + 15$. The skipped-prerequisite here is combining like terms; the first-instinct error students reach for is leaving the middle un-simplified, and it is what makes a perfectly correct expansion look unfinished on a test.
Mistake 3: Multiplying only the first terms and the last terms
Where it slips in: The rusher pattern-matches binomial multiplication onto adding fractions and multiplies "straight across" — first-with-first, last-with-last — skipping the two cross-products entirely.
Don't do this: Writing $(x + 3)(x + 5) = x^2 + 15$, which loses the $8x$ middle term.
The correct way: Every term in the first binomial must meet every term in the second — four products, not two. The box method makes the dropped terms impossible to ignore, because the empty cells stare back at you.
Practice Questions on Multiplying Binomial
Work each of these with whichever method you prefer, then combine like terms. Answers follow so you can check your sign and middle-term work.
Multiply $(x + 4)(x + 6)$.
Multiply $(x - 5)(x + 2)$.
Multiply $(2x + 1)(3x - 4)$.
Multiply $(a - 3b)(2a + b)$.
Expand $(x + 7)(x - 7)$ using the difference-of-squares pattern.
Multiply $(x + 1)(x + 2)(x - 3)$.
Answers
Answer to Question 1: $x^2 + 10x + 24$.
Answer to Question 2: $x^2 - 3x - 10$.
Answer to Question 3: $6x^2 - 5x - 4$.
Answer to Question 4: $2a^2 - 5ab - 3b^2$.
Answer to Question 5: $x^2 - 49$.
Answer to Question 6: $x^3 - 7x - 6$.
Key Takeaways
Multiplying a binomial means multiplying every term of the first by every term of the second, so $(a+b)(c+d) = ac + ad + bc + bd$.
The distributive property is the real rule; FOIL is a memory aid for the two-binomial case, and the box method organises the same four products in a grid.
Two binomials usually produce a trinomial because the Outer and Inner products combine into one middle term.
The most common mistake is dropping a sign on a product or skipping the two cross-terms entirely.
This rule generalises to the binomial theorem and underlies polynomial multiplication of every size.
A Practical Next Step
Work through these three problems to solidify your understanding, using whichever method you prefer:
Multiply $(x + 5)(x - 2)$.
Multiply $(3x - 1)(2x + 4)$.
Multiply $(x + 2)(x - 2)(x + 3)$.
If you get stuck on the sign of a product, come back to Mistake 1 and check each of the four products separately. At Bhanzu, our trainers teach multiplying binomials through the area model first, so students see why there are four products before they memorise any shortcut. Want a live trainer to walk through more binomial problems? Book a free demo class.
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