Multiplying algebraic expressions

Addition and subtraction of algebraic expressions only ever combined like terms. Multiplication is different: every term of one expression must meet every term of the other, and the exponent rule $x^m \times x^n = x^{m+n}$ decides what happens when two powers of the same letter collide. This article builds that skill in three stages — monomial times monomial, monomial times polynomial, and binomial times binomial — the exact ladder the ICSE Class 8 chapter Algebraic Expressions climbs after addition and subtraction are secure. Two habits carry the whole topic. First, the distributive law $a(b+c) = ab+ac$ is applied as many times as there are terms to distribute over — miss one term and the product is wrong. Second, after every term has been multiplied out, like terms in the result must be collected, exactly as before, to reach the simplest form. Get the exponent rule and the distributive law working together and any product of two simple algebraic expressions in this chapter becomes routine.

The core insight

Multiplying algebraic expressions means applying the distributive law so every term of one expression multiplies every term of the other, using $x^m \times x^n = x^{m+n}$ for matching letters, then collecting like terms in the result.

The core idea

Monomial times monomial. Multiply the numerical coefficients as ordinary numbers, and for each letter that appears in either monomial, add its exponents (a missing letter counts as exponent $1$ , or contributes nothing if absent from that factor).

5x^2y \times 3xy^3 = (5\times3)\,x^{2+1}y^{1+3} = 15x^3y^4

where: the coefficients $5$ and $3$ multiply to $15$ , the $x$ -exponents $2$ and $1$ add to $3$ , and the $y$ -exponents $1$ and $3$ add to $4$ .

Monomial times polynomial. Distribute the monomial over every term of the polynomial in turn, keeping each term's sign.

a(b+c-d) = ab+ac-ad

where: $a$ is the monomial multiplier, and $b$ , $c$ , $-d$ are the three terms of the polynomial it is distributed over.

Binomial times binomial. Apply the distributive law twice: every term of the first binomial multiplies every term of the second, giving four products in general.

(a+b)(c+d) = ac+ad+bc+bd

where: $ac$ and $bd$ are the "outer-pair" and "inner-pair" style products of a and c, and b and d, while $ad$ and $bc$ are the cross products — this four-product pattern is often remembered as First, Outer, Inner, Last (FOIL). A quick sanity check uses degree: the degree of a monomial is the sum of its variables' exponents (so $x^2y$ has degree $3$ ), and the degree of a product of two monomials equals the sum of the two factors' degrees. Multiply a degree-2 monomial by a degree-3 monomial and every term of the answer must be degree 5 — if it is not, a term was multiplied wrong. The same check extends to a monomial times a polynomial: each resulting term's degree equals the monomial's degree plus that particular polynomial term's degree, so different terms of the answer can end up with different degrees whenever the polynomial itself mixes terms of different degrees. It also helps to remember what multiplication of algebraic expressions is not. Addition only combines terms that already match, so $2x + 3x = 5x$ , but $2 + x$ cannot be simplified further because $2$ and $x$ are unlike terms. Multiplication has no such restriction on matching — any monomial can multiply any other monomial, whether or not their letters match — but the classic curriculum trap is confusing the two operations: $2+x \neq 2x$ , and $7x+y \neq 7xy$ . Addition of unlike terms simply cannot be shortened into a single term, while multiplication of the very same unlike terms is perfectly legal and always produces one combined term with every letter's exponent added in. A binomial product can also be pictured as an area. A rectangle of length $(a+b)$ and breadth $(c+d)$ , split by one vertical and one horizontal line, breaks into four smaller rectangles of areas $ac$ , $ad$ , $bc$ , $bd$ ; their total area is $(a+b)(c+d)$ , which is exactly the four terms of the FOIL expansion drawn as a picture.

Figure

A rectangle of length a plus b and breadth c plus d split into four smaller rectangles labelled ac ad bc and bd whose total area equals the product a plus b times c plus d

Derivation AMultiplying two binomials, term by term

The goal is to show, by careful repeated use of the distributive law alone, why the product of two binomials $(a+b)$ and $(c+d)$ always expands to the four-term sum $ac+ad+bc+bd$ , so that the FOIL pattern used above is not a rule to memorise blindly but a direct consequence of distributing twice, once over each bracket, and it is this same reasoning that explains why a like-terms step is sometimes needed afterward and sometimes not.

Step 1. Treat the whole second bracket $(c+d)$ as a single block, and distribute the first bracket over it exactly as in the monomial-times-polynomial rule above, temporarily replacing $a+b$ by a single letter would give $(\text{block})\times c + (\text{block})\times d$ ; written out in full this is $(a+b)\times c + (a+b)\times d$ .

Step 2. Distribute each of those two pieces separately. $(a+b)\times c = ac+bc$ by the ordinary distributive law, and $(a+b)\times d = ad+bd$ by the same law applied again. So the first term of Step 1 becomes $ac+bc$ and the second becomes $ad+bd$ .

Step 3. Add the two results from Step 2 together: $(ac+bc) + (ad+bd) = ac+bc+ad+bd$ , which is the same four terms as $ac+ad+bc+bd$ just written in a different order — addition of terms does not depend on the order they are listed in.

Step 4. Read off the general result, then collect like terms only if the specific letters allow it. $(a+b)(c+d) = ac+ad+bc+bd$ always holds as an identity with four named products; whether the numerical answer to a particular problem has two, three, or four terms after that depends only on whether any of $ac$ , $ad$ , $bc$ , $bd$ turn out to be like terms once actual expressions are substituted for $a$ , $b$ , $c$ , $d$ .

Worked example

Multiply the binomials $(2x+3y)$ and $(x-4y)$ , then simplify, applying the same four-product-then-collect pattern proved in the derivation above.

Step 1. Multiply each term of the first bracket by each term of the second: $2x \times x = 2x^2$ , $2x \times (-4y) = -8xy$ , $3y \times x = 3xy$ , $3y \times (-4y) = -12y^2$ .

Step 2. Add all four products in the order found.

2x^2 - 8xy + 3xy - 12y^2

Step 3. Collect the like terms $-8xy$ and $3xy$ : they combine to $-5xy$ .

(2x+3y)(x-4y) = \mathbf{2x^2 - 5xy - 12y^2}

Competitive stretch

Multiply $3x^2y$ by $(2x^3 - 5xy^2 + 4y)$ — a monomial times a trinomial, distributed the same way. Distributing gives $3x^2y \times 2x^3 = 6x^5y$ , $3x^2y \times (-5xy^2) = -15x^3y^3$ , and $3x^2y \times 4y = 12x^2y^2$ . None of these three terms share the same combination of exponents, so no further collecting is possible, and the answer is $\mathbf{6x^5y - 15x^3y^3 + 12x^2y^2}$ .

Common traps

Forgetting to add exponents when multiplying like variables — writing $x^2 \times x^3 = x^6$ instead of the correct $x^5$ .
Sign errors when a bracket has a negative term: a term like $-5xy^2$ must carry its negative sign through the whole multiplication, not get attached only at the end.
Only multiplying "first with first, last with last" in a binomial product and skipping the two cross terms, giving an incomplete expansion of just two terms instead of four.
Leaving the final answer without collecting like terms, e.g. stopping at $2x^2-8xy+3xy-12y^2$ instead of simplifying to $2x^2-5xy-12y^2$ .
Treating unlike terms as if multiplication merges them the way addition sometimes tempts students to, e.g. writing $2x \times 3y = 6xy^2$ by miscounting which variable gets the extra power, instead of the correct $6xy$ .

Ask the tutor: Want to work through a few more binomial products, including one where two terms cancel out completely, step by step with the AI tutor?

Why do we add exponents instead of multiplying them when multiplying powers of the same letter?

Because $x^2$ means $x$ used as a factor twice, and $x^3$ means $x$ used as a factor three times; multiplying them together means $x$ appears as a factor $2+3=5$ times in total, which is $x^5$ . Multiplying the exponents instead would answer a different question — that rule belongs to raising a power to a power, like $(x^2)^3 = x^6$ , not to multiplying two separate powers of $x$ .

How do I know when a binomial times binomial product will simplify to fewer than four terms?

Look at the two cross products from Steps 1 and 2 of the expansion. If they involve exactly the same combination of letters and exponents, they are like terms and can be added or subtracted into one term, taking the answer from four terms down to three. Occasionally the two cross terms are exact opposites, such as in $(a+b)(a-b)$ , and they cancel completely, leaving only two terms. If the two cross products use different letter combinations, all four terms of the expansion stay separate.

Do I always have to distribute both terms of the first bracket, or can I shortcut it for easy numbers?

Always distribute both terms — there is no safe shortcut, even when the numbers look simple. Multiplication of algebraic expressions works by the distributive law, and the law does not care how easy the coefficients look: every term of the first expression must meet every term of the second. Skipping a term, even out of habit, is the single most common way marks are lost on this topic, so writing out all the individual products before adding them is worth the extra line.

Multiplying algebraic expressions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads