A4 - Simplifying, expanding and common factors

The same expression can be written as a sum, a product or a shorter expression with powers. Learn to collect like terms, expand a single bracket, extract common factors and apply integer index laws without changing the value.

Collecting like terms

Do $3x+2x$ and $(3x)(2x)$ mean the same thing? At $x=2$ , the first gives $10$ , but the second gives $24$ .

Addition counts copies; multiplication combines factors. In $4x+7x$ , both terms count copies of $x$ , so $4x+7x=(4+7)x=11x$ . The letter part stays unchanged.

You already know

A term includes its sign. In $5a-2b-3a$ , the terms are $5a$ , $-2b$ and $-3a$ . A coefficient is the numerical multiplier.

Before collecting terms, compare their entire letter parts.

Like terms

Like terms have the same letter factors with the same powers. Thus $3xy$ and $-5yx$ are like terms because $xy=yx$ . The terms $x$ and $x^2$ , or $x$ and $y$ , are not like terms.

For instance, $x$ and $x^2$ give $3$ and $9$ at $x=3$ . They cannot be collected just because they share a letter.

Worked example 1: collecting signed terms

Question. Simplify $5a-2b-3a+6b-4$ .

Plan. Group like terms with their signs, then add their coefficients.

\begin{aligned} 5a-2b-3a+6b-4&=(5-3)a+(-2+6)b-4 &&\text{group matching letter parts}\\ &=2a+4b-4 &&\text{add the coefficients} \end{aligned}

Why this step? The constant $-4$ has no letter part, so it stays separate.

Answer. $2a+4b-4$ .

Check. At $a=2$ , $b=1$ , the original gives $10-2-6+6-4=4$ ; the answer gives $4+4-4=4$ .

Change only the coefficient of the second $a$ term. The collected $a$ coefficient will now be negative.

Worked example 2: a negative collected coefficient

Question. Simplify $5a-2b-7a+6b-4$ .

Plan. Collect the same groups, keeping the subtraction with $7a$ .

\begin{aligned} 5a-2b-7a+6b-4&=(5-7)a+(-2+6)b-4 &&\text{group signed terms}\\ &=-2a+4b-4 &&\text{add each group} \end{aligned}

Why this step? $5-7=-2$ , so a negative coefficient is the correct outcome.

Answer. $-2a+4b-4$ .

Check. At $a=2$ , $b=1$ , both forms give $-4$ .

Reordering factors does not change their product: $3xy-5yx+2x^2=-2xy+2x^2$ . The squared term stays separate. The same counting idea works with an identical supplied root: $3\sqrt2+5\sqrt2=8\sqrt2$ . Both terms count the same unevaluated quantity $\sqrt2$ ; there is no need to find its decimal value or change the root.

Watch out

Wrong: $3x+2x=5x^2$ . Right: $3x+2x=5x$ . The product is different: $(3x)(2x)=6x^2$ .

Change only the $6b$ term in Worked example 2.

Distributing a factor across a bracket

Multiplying a bracket means multiplying every term inside it by the outside factor.

Three groups of $2x+5$ contain six copies of $x$ and fifteen ones: $3(2x+5)=6x+15$ . Changing the inside addition to subtraction gives $3(2x-5)=6x-15$ .

Distributive law

a(b+c)=ab+ac,\qquad a(b-c)=ab-ac

These identities hold for real values of $a$ , $b$ and $c$ .

Not on the formula sheet: learn it.

An outside letter is a factor too: $4x(2x-3)=8x^2-12x$ . There are two factors of $x$ in the first product and one in the second.

Worked example 3: expanding a negative factor

Question. Expand $-2(6x-8)$ .

Plan. Multiply each signed term by $-2$ .

\begin{aligned} -2(6x-8)&=(-2)(6x)+(-2)(-8) &&\text{multiply both terms}\\ &=-12x+16 &&\text{two negatives multiply positively} \end{aligned}

Why this step? The $-8$ is part of the bracket, so it must also be multiplied.

Answer. $-12x+16$ .

Check. At $x=1$ , the bracket gives $-2$ , so the original is $4$ ; $-12+16=4$ .

Change only the outside factor from $-2$ to $-\frac12$ . Fractional factors distribute in exactly the same way.

Worked example 4: expanding a fractional factor

Question. Expand $-\frac12(6x-8)$ .

Plan. Multiply each term by the signed half.

\begin{aligned} -\frac12(6x-8)&=-\frac12(6x)+\left(-\frac12\right)(-8) &&\text{distribute the factor}\\ &=-3x+4 &&\text{halve each magnitude} \end{aligned}

Why this step? The factor is negative, so the two products have opposite signs to the original terms.

Answer. $-3x+4$ .

Check. At $x=1$ , both forms give $1$ . This is one quarter of Worked example 3's value, matching the changed outside factor.

The positive half gives $\frac12(6y-8)=3y-4$ . A plain minus before a bracket means multiplication by $-1$ , so $7x-(2x-5)=7x-2x+5=5x+5$ . With several brackets, expand each before collecting: $3(2x-5)-2(x+4)=6x-15-2x-8=4x-23$ .

Watch out

Wrong: $-2(3x-4)=-6x-8$ . The second product is $(-2)(-4)=8$ , so the correct result is $-6x+8$ .

Change the outside factor again, keeping the bracket fixed.

Taking out common factors

Factorising reverses expansion: it rewrites a sum as a product.

Since $6x+15=3(2x+5)$ , the terms have a common factor of $3$ . Divide each term by the extracted factor to find what belongs inside the bracket.

Factorising fully

To factorise fully by common factors, take out the highest common numerical factor and every letter factor shared by all terms. For letter powers, use the smallest power present in every term.

Taking out a smaller common factor can give a correct product that is not yet fully factorised.

Worked example 5: a number and a letter in common

Question. Factorise fully $12a^2-18a$ .

Plan. Find the numerical highest common factor, then the shared letter factor.

\begin{aligned} 12a^2-18a&=6a(2a)+6a(-3) &&\text{both terms contain }6a\\ &=6a(2a-3) &&\text{take out the shared factor} \end{aligned}

Why this step? $18a$ contains only one $a$ , so $a^2$ cannot be extracted from both terms.

Answer. $6a(2a-3)$ .

Check. Expanding gives $12a^2-18a$ . Taking out only $3a$ leaves a further factor of $2$ inside.

Now multiply only the first term by a new letter $b$ . That letter will stay inside because the second term does not contain it.

Worked example 6: a letter that is not shared

Question. Factorise fully $12a^2b-18a$ .

Plan. Extract $6a$ , the factor common to both complete terms.

\begin{aligned} 12a^2b-18a&=6a(2ab)+6a(-3) &&\text{divide each term by }6a\\ &=6a(2ab-3) &&\text{write a single product} \end{aligned}

Why this step? Extracting $b$ would require it to be a factor of $-18a$ too.

Answer. $6a(2ab-3)$ .

Check. $6a\times2ab=12a^2b$ and $6a\times(-3)=-18a$ .

When both terms contain two letters, keep both shared factors: $8x^2y+12xy^2=4xy(2x+3y)$ . If a whole term is extracted, it leaves $1$ , not zero: $5x+5=5(x+1)$ . Negative outside factors also work: $-3x+6=-3(x-2)=3(2-x)$ .

Factorisation is an identity obtained by multiplication, so it remains valid when an extracted letter is zero. Dividing an equation by that letter is a different operation and could lose a zero solution.

Watch out

Wrong: $5x+5=5(x+0)$ . Dividing the constant $5$ by $5$ leaves $1$ ; the correct bracket is $x+1$ .

Change only the coefficient of the second term in Worked example 6.

Integer index laws in expressions

Index laws shorten products and powers by counting repeated factors.

In $x^3x^2$ , there are five factors of $x$ , so the result is $x^5$ . In $(x^2)^3$ , there are three groups of two, giving $x^6$ . A power of a product acts on every factor: $(2x^3)^2=4x^6$ .

Integer indices

x^mx^n=x^{m+n},\qquad \frac{x^m}{x^n}=x^{m-n},\qquad (x^m)^n=x^{mn}

The product, quotient and power laws above use integer indices. For non-zero $x$ , $x^0=1$ and $x^{-n}=\frac1{x^n}$ , where $n$ is a positive integer. Quotients and negative powers require their original denominators to be non-zero.

Not on the formula sheet: learn it.

Every non-zero number, however large or small, has zeroth power $1$ : the quotient $x^m/x^m$ is $1$ , while subtracting indices gives $x^0$ . This reasoning does not define $0^0$ , because it would divide by zero.

Worked example 7: powers followed by division

Question. Simplify $\frac{(3x^2)^2}{9x^3}$ , where $x\ne0$ .

Plan. Apply the outer power to every factor, then divide matching bases.

\begin{aligned} \frac{(3x^2)^2}{9x^3}&=\frac{9x^4}{9x^3} &&\text{square the coefficient and power}\\ &=x^{4-3}=x &&\text{divide coefficients and subtract indices} \end{aligned}

Why this step? The outer square applies to $3$ as well as $x^2$ .

Answer. $x$ , for $x\ne0$ .

Check. At $x=2$ , the original is $144/72=2$ , matching the answer.

Change only the outer power from $2$ to $3$ . Both the coefficient and letter power change.

Worked example 8: cubing the whole numerator

Question. Simplify $\frac{(3x^2)^3}{9x^3}$ , where $x\ne0$ .

Plan. Cube both factors before simplifying the quotient.

\begin{aligned} \frac{(3x^2)^3}{9x^3}&=\frac{27x^6}{9x^3} &&\text{cube }3\text{ and triple the index}\\ &=3x^{6-3}=3x^3 &&\text{divide matching factors} \end{aligned}

Why this step? $(3x^2)^3$ means three copies of the whole product, not just three copies of $x^2$ .

Answer. $3x^3$ , for $x\ne0$ .

Check. At $x=2$ , $1728/72=24$ , and $3(2^3)=24$ .

The same laws give $\frac{12a^4b}{3ab}=4a^3$ for $a,b\ne0$ , and $2x^3\times3x^{-1}=6x^2$ for $x\ne0$ . The cancelled $b$ restriction still belongs to the original fraction. A negative index means reciprocal, not a negative value.

Watch out

Wrong: $x^3+x^3=x^6$ . A sum collects coefficients: $x^3+x^3=2x^3$ . Indices add in a product. Nor does the product law give $(x+2)^2=x^2+4$ : a sum inside a bracket needs expansion.

The opening expressions are now clear: $3x+2x=5x$ , while $(3x)(2x)=6x^2$ . In your turn, change only the denominator's index from Worked example 8.

Quick Recap

Collect coefficients only when the whole letter part matches.
Expand by multiplying every term in a bracket by the outside factor.
Factorise by extracting the greatest shared numerical and letter factors; expand to check.
Index laws count factors in products, quotients and powers. Keep original non-zero restrictions.

The opening sum and product are different: at $x=2$ , $5x=10$ whereas $6x^2=24$ .

Last Accessed

Papers 1, 2 and 3

A4 - Simplifying, expanding and common factors

Collecting like terms

Distributing a factor across a bracket

Taking out common factors

Integer index laws in expressions

Quick Recap

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