Expanding Algebraic Expressions Using Identities

Hello. Last lesson established a crucial habit: algebraic rules depend on the operation and structure actually present. For surds, you learned not to distribute a square root across addition. The same discipline now applies to brackets: expanding means multiplying every required term, preserving signs, and only then combining like terms.

In this lesson, you will build the general multiplication method behind expansion, then use standard identities to expand common expressions much faster. These identities are especially useful in entrance-exam questions, where recognising structure saves time and reduces careless errors.

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Expansion is repeated distribution

To expand an expression is to remove brackets by carrying out multiplication. The governing rule is the distributive property:

$$a(b+c)=ab+ac$$

For example,

$$3(x-4)=3x-12$$

When two brackets are multiplied, every term in one bracket must multiply every term in the other. Consider:

$$(3x-2)(2x+5)$$

Distribute $3x$ across the second bracket, then distribute $-2$ across it:

$$(3x-2)(2x+5)$$ $$=3x(2x+5)-2(2x+5)$$ $$=6x^2+15x-4x-10$$

Now combine like terms:

$$=6x^2+11x-10$$

For two binomials, this four-product routine is often remembered as FOIL:

• First: multiply the first terms.
• Outer: multiply the outside terms.
• Inner: multiply the inside terms.
• Last: multiply the last terms.

FOIL is useful shorthand, but distribution is the real principle. That matters because FOIL works only for a binomial multiplied by a binomial, whereas distribution works for every polynomial product you will meet later.

A reliable written layout under time pressure is:

$$(3x-2)(2x+5)$$ $$=3x(2x)+3x(5)-2(2x)-2(5)$$

Writing the signs explicitly makes the common error $(-2)(5)=+10$ much less likely.

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Squaring a binomial: why the middle term is doubled

A particularly frequent structure is a binomial squared:

$$(a+b{)}^2$$

The exponent applies to the whole bracket, so this means:

$$(a+b{)}^2=(a+b)(a+b)$$

Expanding gives four terms:

$$(a+b)(a+b)$$ $$=a^2+ab+ab+b^2$$

The two middle terms are like terms:

$$ab+ab=2ab$$

Therefore,

$$\boxed{(a+b{)}^2=a^2+2ab+b^2}$$

Read this identity as:

1. Square the first term.
2. Add twice the product of the two terms.
3. Add the square of the second term.

The supplied square-area diagram gives a geometric reason for the pattern.

The two blue rectangles each have area $ab$. That is why the middle term is $2ab$, not merely $ab$.

Example: a positive binomial

Expand:

$$(2x+3{)}^2$$

Here the two terms are $2x$ and $3$, not merely $x$ and $3$. Apply the identity carefully:

$$(2x+3{)}^2=(2x{)}^2+2(2x)(3)+3^2$$ $$=4x^2+12x+9$$

A common wrong answer is:

$$4x^2+6x+9$$

The error is forgetting that the product $2x\cdot 3$ must be doubled.

The subtraction version

Now consider:

$$(a-b{)}^2$$

Treat the second term as $-b$:

$$(a-b{)}^2=(a+(-b){)}^2$$

Using the same square identity gives:

$$\boxed{(a-b{)}^2=a^2-2ab+b^2}$$

Notice the sign pattern:

• the middle term is negative;
• the last term is positive, because $(-b{)}^2=b^2$.

For example,

$$(2x-3y{)}^2$$ $$=(2x{)}^2-2(2x)(3y)+(3y{)}^2$$ $$=4x^2-12xy+9y^2$$

Do not write $4x^2-9y^2$. That would incorrectly omit the middle product.

The two binomial-square identities are worth knowing together:

$$\boxed{(a+b{)}^2=a^2+2ab+b^2}$$ $$\boxed{(a-b{)}^2=a^2-2ab+b^2}$$

A quick memory check: squaring any real quantity cannot produce a negative final square term.

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Conjugates and the difference of squares

Now compare these two brackets:

$$(a+b)(a-b)$$

They have the same terms but opposite signs between them. Such a pair is called a pair of conjugates.

Expand once to see the pattern:

$$(a+b)(a-b)$$ $$=a^2-ab+ab-b^2$$

The middle terms cancel:

$$-ab+ab=0$$

So:

$$\boxed{(a+b)(a-b)=a^2-b^2}$$

This is called the difference of squares identity.

For example,

$$(5x+4)(5x-4)$$ $$=(5x{)}^2-4^2$$ $$=25x^2-16$$

The identity is also very useful when radicals occur. Using the surd rules from the previous lesson:

$$(3+\sqrt{2})(3-\sqrt{2})$$ $$=3^2-(\sqrt{2}{)}^2$$ $$=9-2$$ $$=7$$

This cancellation is the idea behind rationalising denominators that contain expressions such as $a+\sqrt{b}$, which you will be able to handle systematically once factorisation is secure.

Do not confuse these three structures

Expression	Recognise it as	Expansion
$(a+b{)}^2$	A sum squared	$a^2+2ab+b^2$
$(a-b{)}^2$	A difference squared	$a^2-2ab+b^2$
$(a+b)(a-b)$	Conjugates	$a^2-b^2$

The brackets determine the identity. For instance:

$$(x+4{)}^2=x^2+8x+16$$

but

$$(x+4)(x-4)=x^2-16$$

The first expression has a middle term; the second does not.

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Two further standard expansions

The three identities above are the highest-priority set. Two related forms also occur often enough to recognise.

Squaring three terms

For three terms, every individual term is squared, and every pairwise product appears twice:

$$\boxed{(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca}$$

For example,

$$(x+y-2{)}^2$$ $$=x^2+y^2+(-2{)}^2+2xy+2(y)(-2)+2(-2)(x)$$ $$=x^2+y^2+4+2xy-4y-4x$$

In a standard descending order:

$$\boxed{x^2+2xy+y^2-4x-4y+4}$$

A safe method is to regard $a+b+c$ as $(a+b)+c$, use the two-term square identity, and then simplify. This avoids trying to memorise signs mechanically.

Cubing a binomial

For a binomial cube, the standard identities are:

$$\boxed{(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3}$$ $$\boxed{(a-b{)}^3=a^3-3a^{2}b+3a{b}^2-b^3}$$

For example:

$$(x-2{)}^3$$ $$=x^3-3x^2(2)+3x(2^2)-2^3$$ $$=x^3-6x^2+12x-8$$

The signs in a difference cube alternate:

$$+,-,+,-$$

For now, use these identities only when the expression exactly has the required form. In the later binomial-expansion unit, you will develop a general method for higher powers rather than relying on a growing list of formulas.

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Recognition and checking under exam conditions

Before expanding, take two seconds to classify the structure.

1. If you see a binomial multiplied by a different binomial, distribute each term.
2. If the same binomial is squared, use a binomial-square identity.
3. If the brackets are identical except for one internal sign, use the difference of squares identity.
4. If the structure is not exact, do not force an identity. Use ordinary distribution.

For instance:

$$(x+7)(x+2)$$

is neither a square nor a conjugate pair. Expand directly:

$$(x+7)(x+2)$$ $$=x^2+2x+7x+14$$ $$=x^2+9x+14$$

There is also a useful shortcut for this special-looking form:

$$\boxed{(x+p)(x+q)=x^2+(p+q)x+pq}$$

Here $p=7$ and $q=2$, so the middle coefficient is $7+2=9$, while the constant is $7\cdot 2=14$.

A fast numerical check

After expanding, substitute a convenient value for the variable into both the original expression and your result. If they differ, the expansion is definitely wrong.

Take the earlier result:

$$(3x-2)(2x+5)=6x^2+11x-10$$

Set $x=1$. The original expression gives:

$$(3(1)-2)(2(1)+5)$$ $$=(1)(7)$$ $$=7$$

The expanded expression gives:

$$6(1{)}^2+11(1)-10$$ $$=7$$

The values agree. This does not prove an identity in full, but it is an efficient way to catch a sign error in a timed set.

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Key takeaways

Expansion is repeated distribution: every term in one bracket must multiply every term in the other bracket before like terms are combined.

The three essential identities are:

$$(a+b{)}^2=a^2+2ab+b^2$$ $$(a-b{)}^2=a^2-2ab+b^2$$ $$(a+b)(a-b)=a^2-b^2$$

The key distinction is structural: squares of binomials produce three terms because the two middle products combine, while conjugates produce no middle term because those products cancel.

Next, you will reverse this process through factorisation: recognising an expanded expression such as $x^2-16$ or $x^2+8x+16$ and rewriting it efficiently as a product of factors.