Lesson 2 ~35 min Algebraic Techniques C · Path +85 XP

Expanding Binomial Products

One rule expands every product of brackets: multiply each term in the first by each term in the second. The special products are not exceptions to it, they are the cases worth recognising before you start.

Today's hook: Expand $(x + 3)(x + 5)(x - 2)$. There is no identity for three brackets, and no shortcut. But there is a method that never fails, works for any number of brackets, and makes the arithmetic hard to get wrong. It is the same method you already use for two.

0/5QUESTS

Think First

warm-up

Expand $(x + 2)(x + 6)$ by any method you know. Now, without expanding, predict the constant term and the coefficient of $x$ in $(x + 4)(x + 9)$. Where did each of your two predictions come from, and would the same reasoning still work for $(x - 4)(x + 9)$?

Record your answer in your workbook.

The Big Idea

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Expanding brackets is the distributive law applied more than once. Every term in the first bracket multiplies every term in the second, and then like terms are collected. Nothing else is involved, which is why the method scales to three brackets or more without any new rule.

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

Two terms times two terms gives four products, before any collecting. If you finish an expansion with fewer than four products written down, you have missed one. The special products from Lesson 1 are simply the cases where two of those four collect or cancel.

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

Count the products

Two brackets of two terms always give four products before collecting.

Signs travel

A minus belongs to the term after it. Carry it into every product that term makes.

Check a value

Substitute $x = 1$ into both forms. They must agree.

What You'll Master

objectives

Know

That expanding $(a+b)(c+d)$ produces four products before collecting
That the special products are particular cases of the same expansion

Understand

Why a negative sign must be carried into every product its term makes
Why three brackets need no new rule, only the method applied twice

Can Do

Expand and simplify any product of two binomials
Expand products of three brackets by working two at a time
Expand and simplify expressions combining several products, such as $(x+3)^2 - (x+1)(x-4)$

Words You Need

vocabulary

Distributive law$a(b + c) = ab + ac$. The rule behind every bracket expansion.

Binomial productThe product of two two-term expressions, such as $(x+3)(x-7)$.

Like termsTerms with identical variable parts, such as $5x$ and $-2x$, which can be collected.

TrinomialAn expression with three terms, the usual result of expanding a binomial product.

CoefficientThe number multiplying a variable term, such as the $-4$ in $-4x$.

Constant termThe term with no variable, produced by multiplying the two numbers.

The Four Products

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Work along the first bracket, and for each term sweep the whole of the second.

$$(x + 3)(x + 5)$$

The $x$ gives $x \times x = x^2$ and $x \times 5 = 5x$. The $3$ gives $3 \times x = 3x$ and $3 \times 5 = 15$. Four products:

$$x^2 + 5x + 3x + 15$$

Collect the like terms:

$$x^2 + 8x + 15$$

Notice where the trinomial's parts came from. The constant $15$ is the product of the two numbers, and the coefficient $8$ is their sum. That observation is what makes factorising possible in Lesson 4, so it is worth noticing now.

Carrying the Signs

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A minus sign belongs to the term it precedes, and travels into every product that term makes.

$$(x - 4)(x + 7)$$

The term is $-4$, not $4$. So the four products are $x^2$, $+7x$, $-4x$ and $-28$:

$$x^2 + 7x - 4x - 28 = x^2 + 3x - 28$$

With both signs negative, the constant turns positive:

$$(x - 4)(x - 7) = x^2 - 7x - 4x + 28 = x^2 - 11x + 28$$

Quick check

Substitute $x = 1$ into both the original and your answer. For the last one, $(1-4)(1-7) = 18$ and $1 - 11 + 28 = 18$. Agreement is strong evidence you have not dropped a sign.

When a Coefficient Is in Front

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Nothing changes except the arithmetic. Multiply the whole terms.

$$(2x + 3)(4x - 5)$$

The four products are $2x \times 4x = 8x^2$, then $2x \times (-5) = -10x$, then $3 \times 4x = 12x$, then $3 \times (-5) = -15$:

$$8x^2 - 10x + 12x - 15 = 8x^2 + 2x - 15$$

The two middle products no longer collect neatly into an obvious sum, which is exactly why non-monic expressions are harder to factorise. Lesson 4 deals with that; here it only means the arithmetic needs care.

Three Brackets, and Combined Expressions

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For three brackets, expand any two first, then multiply the resulting trinomial by the remaining bracket.

$$(x+1)(x+2)(x+3) = \big(x^2 + 3x + 2\big)(x+3)$$

Now sweep: $x^3 + 3x^2 + 3x^2 + 9x + 2x + 6$, which collects to

$$x^3 + 6x^2 + 11x + 6$$

For an expression combining several products, expand each part completely before combining, and watch the sign in front of a subtracted bracket:

$$(x+3)^2 - (x+1)(x-4) = \big(x^2 + 6x + 9\big) - \big(x^2 - 3x - 4\big)$$

The minus applies to the whole second bracket, so every term inside it changes sign:

$$= x^2 + 6x + 9 - x^2 + 3x + 4 = 9x + 13$$

Common Pitfalls

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Multiplying only the first terms and the last terms: writing $(x+3)(x+5) = x^2 + 15$.

Fix: count them. Two terms times two terms is four products. Two are missing here, and they are the ones that make the $8x$.

Subtracting a bracket without distributing the minus: writing $-(x^2 - 3x - 4)$ as $-x^2 - 3x - 4$.

Fix: the minus applies to every term inside, giving $-x^2 + 3x + 4$. Put the bracket in first, then remove it deliberately.

Collecting unlike terms, such as adding $8x^2$ and $2x$.

Fix: like terms need identical variable parts. $x^2$ and $x$ are different and stay apart.

Watch Me Solve It · 3 examples

Watch Me Solve It · A binomial product with signs

+15 XP per step

PROBLEM

Expand and simplify $(3x - 4)(2x + 5)$.

1

List the four products

$3x \times 2x, \quad 3x \times 5, \quad -4 \times 2x, \quad -4 \times 5$

The $-4$ carries its sign into both of its products.
2

Evaluate each

$6x^2, \quad 15x, \quad -8x, \quad -20$
3

Collect like terms

$6x^2 + 15x - 8x - 20 = 6x^2 + 7x - 20$

Only the two $x$ terms are alike.
4

Check at x = 1

$(3-4)(2+5) = -7$

$6 + 7 - 20 = -7 \;\checkmark$

Both forms agree, so no sign was dropped.

Answer$6x^2 + 7x - 20$

Watch Me Solve It · Three brackets

+15 XP per step

PROBLEM

Expand and simplify $(x - 2)(x + 1)(x + 4)$.

1

Expand the first two

$(x-2)(x+1) = x^2 + x - 2x - 2 = x^2 - x - 2$

Any two may be chosen; take the pair that looks easiest.
2

Multiply the trinomial by the third bracket

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

Three terms times two terms will give six products.
3

Sweep and list

$x^3 + 4x^2 - x^2 - 4x - 2x - 8$
4

Collect

$x^3 + 3x^2 - 6x - 8$

The $x^2$ terms collect, and so do the $x$ terms.

Answer$x^3 + 3x^2 - 6x - 8$

Watch Me Solve It · Combining two expansions

+15 XP per step

PROBLEM

Expand and simplify $(2x + 1)^2 - (x - 3)(x + 3)$.

1

Expand the perfect square

$(2x+1)^2 = 4x^2 + 4x + 1$

Special product from Lesson 1, with $a = 2x$ and $b = 1$.
2

Expand the difference of two squares

$(x-3)(x+3) = x^2 - 9$

The brackets differ only in sign, so the middle terms cancel.
3

Subtract the WHOLE second expression

$\big(4x^2 + 4x + 1\big) - \big(x^2 - 9\big)$

Keeping the brackets is what stops the sign error.
4

Distribute the minus and collect

$4x^2 + 4x + 1 - x^2 + 9 = 3x^2 + 4x + 10$

$-(-9)$ becomes $+9$.

Answer$3x^2 + 4x + 10$

Brain Trainer · 4 problems

Brain Trainer · Expand and simplify

4 problems

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

1 Expand $(x + 6)(x - 2)$.

Four products: $x^2$, $-2x$, $6x$, $-12$.$x^2 + 4x - 12$
2 Expand $(4x - 1)(x - 5)$.

$4x^2 - 20x - x + 5$.$4x^2 - 21x + 5$
3 Expand $(x + 2)(x - 2)(x + 5)$.

The first two are a difference of two squares: $(x^2 - 4)(x + 5)$.$x^3 + 5x^2 - 4x - 20$
4 Simplify $(x + 4)^2 - (x + 1)^2$.

$(x^2 + 8x + 16) - (x^2 + 2x + 1)$.$6x + 15$

Complete in your workbook.

Multiple Choice · 5 questions

MC1

Counting the products

+10 XP

Expanding $(a + b)(c + d)$ before collecting gives how many products?

MC2

A product with signs

+10 XP

$(x - 3)(x + 8)$ equals:

MC3

Subtracting a bracket

+10 XP

$(x + 5)^2 - (x^2 - 4)$ simplifies to:

MC4

Where the constant comes from

+10 XP

In the expansion of $(x + p)(x + q)$, the constant term is:

MC5

Three brackets

+10 XP

The highest power of $x$ in the expansion of $(x+1)(x-2)(x+7)$ is:

Short Answer · 3 questions

Expand and simplify

+15 XP

SHORT ANSWER

Expand and simplify each of the following, showing your four products where relevant.
(a) $(2x + 7)(3x - 1)$
(b) $(x - 5)(x + 5) + (x + 2)^2$
(c) $(x + 1)(x + 2)(x - 3)$

Write your working in your book.

Find the error

+15 XP

SHORT ANSWER

A student writes:
$(3x - 2)(x + 4) = 3x^2 + 12x - 2x - 8 = 3x^2 + 10x - 8$
and then
$(x + 6)^2 - (x - 1)(x + 1) = x^2 + 36 - x^2 - 1 = 35$.
(a) Is the first line correct? Justify by substituting a value.
(b) Identify every error in the second line and give the correct simplification.

Write your working in your book.

Reason about the structure

+15 XP

SHORT ANSWER

(a) Without expanding fully, state the coefficient of $x$ and the constant term in $(x - 9)(x + 4)$, and explain where each comes from.
(b) Two numbers have a sum of $-2$ and a product of $-15$. Write down a binomial product that expands to $x^2 - 2x - 15$, and verify it.
(c) Explain why this reasoning would not directly work for $2x^2 - 2x - 15$.

Write your working in your book.

Stretch Challenge · A cube, and a pattern

+25 XP

CHALLENGE

(a) Expand $(a + b)^3$ by writing it as $(a+b)(a+b)^2$ and simplifying fully.
(b) Write down the coefficients of your answer, then the coefficients of $(a+b)^2$ and of $(a+b)^1$. What pattern do the three rows form?
(c) Predict the expansion of $(a+b)^4$ from the pattern, and check your prediction by multiplying your answer to (a) by $(a+b)$.

Quick Review

recap

The rule

Every term times every term, then collect

Count

Two by two gives four products

Signs

A minus travels into both of its products

Structure

$(x+p)(x+q) = x^2 + (p+q)x + pq$

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