Lesson 1 ~35 min Algebraic Techniques C · Path +85 XP

The Special Products

Three expansions turn up so often that recognising them is faster than doing the work. This lesson proves all three, algebraically and geometrically, so you can trust them and rebuild them.

Today's hook: Work out $51^2$ in your head. Most people reach for a calculator. But $51 = 50 + 1$, and there is an identity that turns $(50 + 1)^2$ into $2500 + 100 + 1$ in one line. The same identity is how you will expand $(2x + 3)^2$ next year, and the year after.

0/5QUESTS

Think First

warm-up

A student writes $(a + b)^2 = a^2 + b^2$. Test it with $a = 3$ and $b = 4$: what does each side come to? Now draw a square of side $a + b$ and cut it into a square of side $a$, a square of side $b$, and whatever is left over. What are the leftover pieces, and what does that tell you the student is missing?

Record your answer in your workbook.

The Big Idea

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The three special products are ordinary expansions that occur often enough to be worth knowing by sight. Each one is provable in two lines by expanding the brackets, and the first is also visible in a picture: a square of side $a + b$ splits into four pieces, and their areas are the four terms.

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

The picture is the proof of the first. The whole square has area $(a+b)^2$, and the four pieces have areas $a^2$, $ab$, $ab$ and $b^2$. Two of them are identical, which is exactly where the middle term $2ab$ comes from, and why $(a+b)^2 = a^2 + b^2$ is wrong by precisely that amount.

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

The middle term

$2ab$ is the piece students drop. It is two identical rectangles, not one.

Only the sign changes

$(a-b)^2$ is the same three terms with the middle one negative.

The middle vanishes

In $(a+b)(a-b)$ the two middle terms cancel, leaving just $a^2 - b^2$.

What You'll Master

objectives

Know

The perfect square identities $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$
The difference of two squares $(a+b)(a-b) = a^2 - b^2$

Understand

Why the middle term of a perfect square is $2ab$ rather than $ab$
Why the middle terms cancel in the difference of two squares

Can Do

Prove each special product by expansion, and the first one geometrically
Apply the identities to expressions such as $(3x - 5)^2$ and $(2y + 7)(2y - 7)$
Use them to evaluate numerical squares and products mentally

Words You Need

vocabulary

IdentityAn equation true for every value of the variables, such as $(a+b)^2 = a^2 + 2ab + b^2$.

Perfect squareAn expression of the form $(a \pm b)^2$, or its expansion.

Difference of two squaresAn expression of the form $a^2 - b^2$, which factorises as $(a+b)(a-b)$.

BinomialAn expression with exactly two terms, such as $3x - 5$.

ExpandTo remove brackets by multiplying out.

Middle termThe $2ab$ in a perfect square, produced by two identical products.

Proving the Perfect Square

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Nothing here is new machinery. Write the square as a product and expand it term by term.

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

Multiply each term of the first bracket by each term of the second:

$$= a \times a + a \times b + b \times a + b \times b = a^2 + ab + ba + b^2$$

Since $ab$ and $ba$ are the same thing, they collect:

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

The proof of $(a-b)^2$ is identical with $-b$ in place of $b$. The two cross terms become $-ab$ each, so the middle term is $-2ab$, and $(-b)^2 = b^2$ keeps the last term positive.

Notice

The last term is $+b^2$ in BOTH identities. Only the middle term carries the sign.

Proving the Difference of Two Squares

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The same expansion, with one sign changed, produces a very different result.

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

Now $-ab$ and $+ba$ are opposites, so they cancel completely:

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

That cancellation is the whole point of the identity. Two binomials multiply to give only two terms, which almost never happens, and it is why the pattern is worth recognising the moment you see brackets differing only in a sign.

Applying Them to Algebra

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The letters $a$ and $b$ stand for whole terms, not just single letters. Identify what plays each role first, then substitute.

For $(3x + 4)^2$, take $a = 3x$ and $b = 4$:

$$(3x)^2 + 2(3x)(4) + 4^2 = 9x^2 + 24x + 16$$

For $(5y - 2)^2$, take $a = 5y$ and $b = 2$:

$$(5y)^2 - 2(5y)(2) + 2^2 = 25y^2 - 20y + 4$$

For $(2m + 7)(2m - 7)$, take $a = 2m$ and $b = 7$:

$$(2m)^2 - 7^2 = 4m^2 - 49$$

HSC Tip

Square the whole term, coefficient included. $(3x)^2$ is $9x^2$, never $3x^2$.

Using Them on Numbers

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The identities work on numbers too, and that is often the fastest mental route.

$51^2$. Write $51 = 50 + 1$, so $51^2 = 50^2 + 2(50)(1) + 1^2 = 2500 + 100 + 1 = 2601$.

$98^2$. Write $98 = 100 - 2$, so $98^2 = 10\,000 - 400 + 4 = 9604$.

$43 \times 37$. These are $40 + 3$ and $40 - 3$, a difference of two squares: $40^2 - 3^2 = 1600 - 9 = 1591$.

The last one is the most striking. A two-digit multiplication becomes two squares and a subtraction, provided the numbers sit symmetrically either side of a round number.

Common Pitfalls

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Writing $(a+b)^2 = a^2 + b^2$, dropping the middle term.

Fix: test it once with numbers. $(3+4)^2 = 49$ but $3^2 + 4^2 = 25$. The missing $24$ is exactly $2ab$, the two rectangles in the diagram.

Forgetting to square the coefficient: writing $(3x)^2 = 3x^2$.

Fix: the whole term is squared, so $(3x)^2 = 3^2 \times x^2 = 9x^2$.

Making the last term negative in $(a-b)^2$, writing $a^2 - 2ab - b^2$.

Fix: $(-b)^2 = +b^2$, since a negative squared is positive. Only the middle term is negative.

Watch Me Solve It · 3 examples

Watch Me Solve It · A perfect square with a coefficient

+15 XP per step

PROBLEM

Expand $(4x - 3)^2$.

1

Identify a and b

$a = 4x, \qquad b = 3$

Both are whole terms; the coefficient travels with the letter.
2

Write the identity

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

The subtraction means the middle term is negative.
3

Substitute

$(4x)^2 - 2(4x)(3) + 3^2$
4

Evaluate each term

$16x^2 - 24x + 9$

$(4x)^2 = 16x^2$, not $4x^2$. The last term is positive.

Answer$16x^2 - 24x + 9$

Watch Me Solve It · Spotting a difference of two squares

+15 XP per step

PROBLEM

Expand $(5p + 6q)(5p - 6q)$.

1

Recognise the pattern

$\text{brackets differ only in the middle sign}$

That is the signature of a difference of two squares.
2

Identify a and b

$a = 5p, \qquad b = 6q$
3

Apply the identity

$a^2 - b^2 = (5p)^2 - (6q)^2$

The two cross terms cancel, so no middle term survives.
4

Evaluate

$25p^2 - 36q^2$

Two terms only, which is the whole saving.

Answer$25p^2 - 36q^2$

Watch Me Solve It · Mental arithmetic with an identity

+15 XP per step

PROBLEM

Without a calculator, evaluate $103^2$ and $67 \times 73$.

1

Rewrite 103 around a round number

$103 = 100 + 3$

Round numbers make the squares easy.
2

Apply the perfect square

$103^2 = 100^2 + 2(100)(3) + 3^2$

$= 10\,000 + 600 + 9 = 10\,609$
3

Check 67 and 73 for symmetry

$67 = 70 - 3, \qquad 73 = 70 + 3$

Equal distances either side of 70, so a difference of two squares applies.
4

Apply it

$70^2 - 3^2 = 4900 - 9 = 4891$

A two-digit multiplication reduced to a subtraction.

Answer$103^2 = 10\,609$ and $67 \times 73 = 4891$

Brain Trainer · 4 problems

Brain Trainer · Expand each one

4 problems

Four quick problems. Work each one, then reveal the answer.

1 Expand $(x + 5)^2$.

$a = x$, $b = 5$, middle term $2(x)(5) = 10x$.$x^2 + 10x + 25$
2 Expand $(2y - 7)^2$.

$(2y)^2 = 4y^2$, middle term $-2(2y)(7) = -28y$, last term $+49$.$4y^2 - 28y + 49$
3 Expand $(3m + 8)(3m - 8)$.

Difference of two squares: $(3m)^2 - 8^2$.$9m^2 - 64$
4 Evaluate $99^2$ without a calculator.

$99 = 100 - 1$, so $10\,000 - 200 + 1$.$9801$

Complete in your workbook.

Multiple Choice · 5 questions

MC1

The perfect square

+10 XP

$(a + b)^2$ expands to:

MC2

Squaring a coefficient

+10 XP

$(5x - 2)^2$ equals:

MC3

Recognising the pattern

+10 XP

Which product expands to only TWO terms?

MC4

Mental arithmetic

+10 XP

Using a special product, $58 \times 62$ equals:

MC5

Testing a claim

+10 XP

A student claims $(x - 4)^2 = x^2 - 16$. Substituting $x = 5$ shows the claim is:

Short Answer · 3 questions

Prove and apply

+15 XP

SHORT ANSWER

(a) Prove algebraically that $(a - b)^2 = a^2 - 2ab + b^2$.
(b) Use the result to expand $(7x - 2y)^2$.
(c) Explain why the final term is positive even though the bracket contains a subtraction.

Write your working in your book.

Choose the efficient method

+15 XP

SHORT ANSWER

Evaluate each without a calculator, stating which special product you used and why it applies.
(a) $102^2$
(b) $95^2$
(c) $48 \times 52$
(d) $199 \times 201$

Write your working in your book.

The geometric argument

+15 XP

SHORT ANSWER

A square of side $a + b$ is cut into four rectangles by one horizontal and one vertical line, positioned so that one corner piece is a square of side $a$.
(a) State the area of each of the four pieces.
(b) Use the diagram to explain why $(a+b)^2 = a^2 + 2ab + b^2$.
(c) Explain what the diagram shows about the claim $(a+b)^2 = a^2 + b^2$.

Write your working in your book.

Stretch Challenge · A chain of squares

+25 XP

CHALLENGE

(a) Show that $(a+b)^2 - (a-b)^2 = 4ab$.
(b) Use that result to evaluate $87^2 - 83^2$ mentally, without squaring either number.
(c) Hence explain how to compute the product of any two numbers from the squares of their sum and difference, and demonstrate it on $14 \times 6$.

Quick Review

recap

Perfect square

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

Middle term

$2ab$, from two identical rectangles

Two squares

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

Last term

Always positive, even when the bracket subtracts

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