Distributive property

You have used the distributive property with numbers before, perhaps without naming it. To find $7 \times 23$ mentally, split $23$ into $20 + 3$ and multiply each piece:

That mental trick is not a coincidence — it is a law that holds for every number and every algebraic expression, whenever a sum (or difference) sits inside a bracket being multiplied by something outside it.

where: $a$ is the term outside the bracket (the multiplier), $b$ and $c$ are the terms inside it, and the rule says each of $b$ and $c$ gets multiplied by $a$ on its own before the results are combined with the same $+$ or $-$ sign that joined them inside the bracket. A rectangle makes the rule visible instead of abstract. Split a rectangle of width $a$ into two parts of length $b$ and $c$: the whole area is $a(b+c)$, but it is also exactly the sum of the two smaller rectangles' areas, $ab$ and $ac$. Multiplying a sum and adding two products are the same operation seen two ways.

The rule extends when the outside factor is itself a monomial rather than a single number: multiply the numerical coefficients together and combine the powers of a repeated letter using $x^m \times x^n = x^{m+n}$. So $(3x^2)(4x^3) = 12x^5$ — coefficients multiply, exponents add, never the reverse. And distribution is not limited to two-term brackets: for a trinomial, $a(b+c+d) = ab+ac+ad$, distributing to every single term inside, with no exceptions. The same idea keeps working when the outside factor itself has more than one term, which is how binomial-by-binomial products such as $(a+b)(c+d)$ are eventually built: treat $(c+d)$ as a single block and distribute it over both $a$ and $b$ first, giving $a(c+d) + b(c+d)$, then distribute again inside each piece to get $ac+ad+bc+bd$ — four terms, one for every pairing of a term from the first bracket with a term from the second. Class 8 usually meets that two-layer expansion soon after this topic, so recognising it now as "distribute twice" rather than a brand-new rule saves a lot of memorising later.

The goal is to show, step by step and with full sign-tracking, exactly how a monomial multiplies through a bracket containing several terms with mixed signs — the calculation that Common traps below shows students most often get wrong.

Step 1. Write the general product with the outside multiplier as a monomial $m$ and the bracket holding three terms, $m(p+q-r)$. The distributive law says every term inside the bracket is multiplied by $m$ independently, keeping the sign that precedes it.

Step 2. Multiply term by term: $m\times p = mp$ (sign unchanged, since $p$ has a $+$), $m\times q = mq$ (sign unchanged), and $m\times(-r) = -mr$ (the minus sign travels with $r$ into the product).

Step 3. Recombine with the original signs between terms: $m(p+q-r) = mp+mq-mr$. Nothing is added or dropped — the number of terms inside the bracket equals the number of terms in the answer.

Step 4. Check for like terms in the result. If any two of $mp$, $mq$, $mr$ share the same letters raised to the same powers, they must be combined by adding their coefficients; only then is the expansion fully simplified. In this general case with distinct letters $p$, $q$, $r$, no further combining is possible.

Step 5. Confirm the rule with a quick numeric substitution, which catches sign slips instantly. Let $m=2$, $p=3$, $q=1$, $r=4$. Left side: $2(3+1-4) = 2(0) = 0$. Right side: $2(3)+2(1)-2(4) = 6+2-8=0$. Both sides match, so the general expansion $m(p+q-r)=mp+mq-mr$ is verified — the same check you should run on any expansion where you are unsure of a sign.

Expand and simplify $5x(2x+3y-4)$, then use the result to check a numeric case.

Step 1. Identify the multiplier outside the bracket, $5x$, and the three terms inside: $2x$, $3y$, and $-4$. Every one of the three must be multiplied by $5x$ separately.

Step 2. Multiply term by term. $5x\times2x=10x^2$. $5x\times3y=15xy$. $5x\times(-4)=-20x$.

Step 3. Recombine with the original signs: $5x(2x+3y-4) = 10x^2+15xy-20x$. None of these three terms share the same combination of letters and powers, so this is fully simplified.

Step 4. Check by substituting $x=1$, $y=2$. Left side: $5(1)(2(1)+3(2)-4) = 5(2+6-4)=5(4)=20$. Right side: $10(1)^2+15(1)(2)-20(1) = 10+30-20=20$. Both sides agree, confirming the expansion. Final answer: $5x(2x+3y-4) = 10x^2+15xy-20x$.

Competitive stretch

A rectangular garden has length $(x+7)$ metres and width $4$ metres; a flower-bed of length $(x+7)$ and width $x$ is cut from inside it. Garden area: $4(x+7)=4x+28$. Bed area: $x(x+7)=x^2+7x$. Subtracting carefully — the minus sign applies to every term of the bed's area, not just the first — gives the remaining area $= (4x+28)-(x^2+7x) = 4x+28-x^2-7x = -x^2-3x+28$ square metres, valid for $x<4$ so the cut-out actually fits inside the garden. Substituting a specific value, say $x=2$, gives $-4-6+28=18$ square metres of garden left over — a positive, sensible answer, which is a useful check that the algebra behaved.

• Forgetting to distribute to every term. In $3x(2x+5)$, multiplying $3x$ only by $2x$ gives the wrong $6x^2$ instead of the correct $6x^2+15x$.
• Sign errors with a minus inside the bracket. In $a(b-c)$ the second product is $-ac$, not $+ac$; this compounds when the whole expression is negative, as in $-5x(2x-3)=-10x^2+15x$, where both signs flip.
• Mixing multiplication of coefficients with addition of powers. For $3x^2 \times 4x^3$, some students add $3+4$ instead of multiplying, or multiply the exponents instead of adding them. The rule is coefficients multiply ($3\times4=12$), exponents add ($2+3=5$).
• Declaring an expansion finished before checking for like terms. After distributing, always scan the resulting terms for matching letters and powers before calling the expression simplified.
• Distributing only part of a two-term outside factor. When both the multiplier and the bracket have several terms, such as in a garden-area problem, every term of the shorter expression must still reach every term of the other — skipping this is the most common reason a "simplified" answer turns out wrong on a recheck.

Ask the tutor: if a sign flip after expanding a bracket does not make sense, or you want to practice turning a word problem about areas into an algebraic expression, ask the AI tutor to walk through it step by step.