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Algebraic Expressions

Grade 10
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Everything covered in this lesson, in one place - useful for revision or printing.

\( (a+b)(a^2-ab+b^2) = a^3+b^3 \)

Algebra is the language of the whole Grade 10 year, so this lesson builds the tools you will use in every other topic.

By the end you will be able to:
• sort numbers into rational, irrational and non-real
• round off and place a surd between two integers
• expand products, including the special products and cubes
• factorise using common factors, grouping, difference of squares, trinomials and cubes
• simplify algebraic fractions, stating the restrictions

Use Next to move through the steps. Try each quick check before you open the worked answer.

1. Number system

\( \mathbb{N} \subset \mathbb{N}_0 \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \)

Every number you meet in Grade 10 belongs to a family.

• Rational \( (\mathbb{Q}) \): can be written as \( \dfrac{a}{b} \) with \( a, b \) integers and \( b \neq 0 \). Terminating and recurring decimals are rational.
• Irrational \( (\mathbb{Q}') \): real, but cannot be written as a fraction of integers, e.g. \( \pi \), \( \sqrt{2} \). Their decimals never end and never repeat.
• Non-real: the even root of a negative number, e.g. \( \sqrt{-4} \). No real number squared gives a negative.

Rational and irrational numbers together make up the real numbers \( \mathbb{R} \).

\( 0{,}75 = \dfrac{75}{100} = \dfrac{3}{4} \)

Show that \( 0{,}75 \) is rational.

Write the decimal over a power of 10, then simplify:

\( 0{,}75 = \dfrac{75}{100} = \dfrac{3}{4} \)

Because \( 0{,}75 = \dfrac{3}{4} \), a ratio of integers (with a non-zero denominator), it is rational.
Quick check
Write \( 0{,}45 \) as a common fraction in simplest form.
Show worked answer
\( 0{,}45 = \dfrac{45}{100} \). Divide top and bottom by 5: \( \dfrac{9}{20} \).
Answer: \( \dfrac{9}{20} \)

\( \sqrt{25} \qquad \sqrt{-7} \qquad \pi \)

Classify each number as rational, irrational or non-real.

Always simplify first — a root sign does not automatically mean irrational.

Quick check
Classify \( \sqrt{25} \), \( \sqrt{-7} \) and \( \pi \).
Show worked answer
• \( \sqrt{25} = 5 \), an integer, so it is rational.
• \( \sqrt{-7} \): square root of a negative, so it is non-real.
• \( \pi \): its decimal never ends or repeats, so it is irrational.

\( \text{rational} \;|\; \text{irrational} \;|\; \text{non-real} \)

Number system in a nutshell
• Rational = fraction of integers (terminating or recurring decimals)
• Irrational = real, but no fraction form (e.g. \( \pi \), \( \sqrt{2} \))
• Non-real = even root of a negative
• Simplify before you classify: \( \sqrt{25} = 5 \) is rational.

2. Rounding & surds

\( 34{,}4678 \approx 34{,}47 \)

Round \( 34{,}4678 \) to two decimal places.

Look at the digit just after the place you are rounding to:

\( 34{,}46\underline{7}8 \): the third decimal is 7, which is 5 or more, so round the 6 up.

Answer: \( 34{,}47 \)

Rule: 0 to 4 leaves the digit as it is; 5 to 9 rounds it up by one.

\( 5 < \sqrt{33} < 6 \)

Between which two consecutive integers does \( \sqrt{33} \) lie?

Find the perfect squares on either side of 33:

\( 25 < 33 < 36 \)
\( \sqrt{25} < \sqrt{33} < \sqrt{36} \)
\( 5 < \sqrt{33} < 6 \)

Answer: 5 and 6. (A calculator gives \( \sqrt{33} \approx 5{,}74 \), which agrees.)

\( 5 < \;?\; < 6 \)

Which of these is an irrational number between 5 and 6?
A. \( \sqrt[3]{28} \)    B. \( \sqrt{20+8} \)    C. \( \sqrt{20 \times 8} \)    D. \( 2\pi \)

Quick check
Estimate each value and choose.
Show worked answer
• \( \sqrt[3]{28} \approx 3{,}04 \) (since \( 3^3 = 27 \)) — too small
• \( \sqrt{20+8} = \sqrt{28} \approx 5{,}29 \) (since \( 25 < 28 < 36 \)) — between 5 and 6
• \( \sqrt{160} \approx 12{,}65 \) — too big
• \( 2\pi \approx 6{,}28 \) — too big
Answer: B, \( \sqrt{28} \)

\( a^2 < n < b^2 \Rightarrow a < \sqrt{n} < b \)

Key ideas
• Rounding: look one digit past the required place; 5 or more rounds up.
• To place \( \sqrt{n} \): trap \( n \) between two perfect squares.
• For cube roots use perfect cubes: \( 27 < 28 < 64 \Rightarrow 3 < \sqrt[3]{28} < 4 \).

3. Products

\( -3x^2y(5xy^2 + xy) \)

Multiply the term outside by every term inside. Multiply coefficients, add exponents of like bases.

\( -3x^2y \cdot 5xy^2 = -15x^3y^3 \)
\( -3x^2y \cdot xy = -3x^3y^2 \)

Answer: \( -15x^3y^3 - 3x^3y^2 \)
Quick check
Expand and simplify \( 3x(2x - 4xy) \).
Show worked answer
\( 3x \cdot 2x = 6x^2 \) and \( 3x \cdot (-4xy) = -12x^2y \).
Answer: \( 6x^2 - 12x^2y \)

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

Each term in the first bracket multiplies each term in the second (four products), then collect like terms.

\( (2x+3)(5-x) = 10x - 2x^2 + 15 - 3x \)
\( = -2x^2 + 7x + 15 \)

Write the answer in descending powers of \( x \).
Quick check
Expand \( (x-3)^2 \).
Show worked answer
\( (x-3)^2 = (x-3)(x-3) \)
\( = x^2 - 3x - 3x + 9 \)
Answer: \( x^2 - 6x + 9 \)
Square of a binomial: square the first, double the product, square the last. Never just \( x^2 + 9 \).

\( (2x-1)(x^2-3x+1) \)

Multiply each term of the binomial by all three terms of the trinomial (six products), then collect like terms.

\( 2x(x^2-3x+1) = 2x^3 - 6x^2 + 2x \)
\( -1(x^2-3x+1) = -x^2 + 3x - 1 \)
\( = 2x^3 - 7x^2 + 5x - 1 \)

\( (2r-p)(3r^2-4rp+p^2) \)

Two variables work exactly the same way. Keep the letters in the same order in every term so like terms are easy to spot.

Quick check
Expand and simplify \( (2r-p)(3r^2-4rp+p^2) \).
Show worked answer
\( 2r(3r^2-4rp+p^2) = 6r^3 - 8r^2p + 2rp^2 \)
\( -p(3r^2-4rp+p^2) = -3r^2p + 4rp^2 - p^3 \)
Answer: \( 6r^3 - 11r^2p + 6rp^2 - p^3 \)

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

Products checklist
• Every term in one bracket times every term in the other
• Coefficients multiply, exponents of like bases add
• Watch the signs, then collect like terms
• \( (a-b)^2 = a^2 - 2ab + b^2 \), never \( a^2 - b^2 \)

4. Special products

\( (a+b)(a^2-ab+b^2) = a^3+b^3 \)

Two products collapse to just two terms because the middle terms cancel:

\( (a+b)(a^2-ab+b^2) = a^3 + b^3 \)
\( (a-b)(a^2+ab+b^2) = a^3 - b^3 \)

Pattern for the trinomial: square the first, opposite sign times the product, square the last. If the trinomial does not follow this pattern, the terms do not all cancel — expand it normally.
Quick check
Expand \( (a+2)(a^2-2a+8) \). Does it give a sum of cubes?
Show worked answer
\( a(a^2-2a+8) = a^3 - 2a^2 + 8a \)
\( 2(a^2-2a+8) = 2a^2 - 4a + 16 \)
Answer: \( a^3 + 4a + 16 \)
Not a sum of cubes: the last term is 8, not \( 2^2 = 4 \), so the \( a \) terms do not cancel.

\( (xy^3-3)(x^2y^6+3xy^3+9) \)

Let \( a = xy^3 \) and \( b = 3 \). Check the trinomial:
\( a^2 = x^2y^6 \), \( ab = 3xy^3 \), \( b^2 = 9 \). It matches \( (a-b)(a^2+ab+b^2) \).

\( = a^3 - b^3 = (xy^3)^3 - 3^3 = x^3y^9 - 27 \)
Quick check
\( a^3 + 27 = (a+3)(a^2 + ma + 9) \). Find \( m \).
Show worked answer
Sum of cubes with \( b = 3 \): \( a^3 + 27 = (a+3)(a^2 - 3a + 9) \).
Answer: \( m = -3 \)

\( \text{Area of picture} = \;? \)

A picture is framed using four rectangular pieces of wood of width \( x \). Read the lengths from the diagram and find the area of the picture in simplest form.

picturex7x + 37x + 3x5x + 2xx5x + 2
The frame is made of four pieces of width \( x \). Each \( 7x+3 \) length covers the picture plus one frame width, and so does each \( 5x+2 \) length.
Quick check
Find the area of the picture.
Show worked answer
Picture length \( = 7x + 3 - x = 6x + 3 \)
Picture breadth \( = 5x + 2 - x = 4x + 2 \)
\( A = (6x+3)(4x+2) \)
\( = 24x^2 + 12x + 12x + 6 \)
Answer: \( 24x^2 + 24x + 6 \)

\( a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2) \)

Special products
• \( (a+b)(a-b) = a^2 - b^2 \)
• \( (a \pm b)^2 = a^2 \pm 2ab + b^2 \)
• \( (a+b)(a^2-ab+b^2) = a^3 + b^3 \) and \( (a-b)(a^2+ab+b^2) = a^3 - b^3 \)
• Word problems: read lengths carefully from the diagram before multiplying.

5. Factorising I

\( xy^2 + 3x^2y = xy(y + 3x) \)

Factorising is expanding in reverse. Always look for a highest common factor (HCF) first.

HCF of \( xy^2 \) and \( 3x^2y \) is \( xy \).
\( xy^2 + 3x^2y = xy(y + 3x) \)

Check by expanding: \( xy \cdot y + xy \cdot 3x = xy^2 + 3x^2y \).
Quick check
Factorise fully: \( 2x^2 - 8 \).
Show worked answer
Common factor 2: \( 2(x^2 - 4) \). Then difference of squares: \( x^2 - 4 = (x-2)(x+2) \).
Answer: \( 2(x-2)(x+2) \)

\( x^2 - 7x - 18 \)

For \( x^2 + bx + c \), find two numbers that multiply to \( c \) and add to \( b \).

Multiply to \( -18 \), add to \( -7 \): \( -9 \) and \( 2 \).
\( x^2 - 7x - 18 = (x-9)(x+2) \)
Quick check
Factorise fully: \( x^2 - 4x + 3 \).
Show worked answer
Multiply to \( 3 \), add to \( -4 \): \( -3 \) and \( -1 \).
Answer: \( (x-3)(x-1) \)

\( 6x^2 + 7x - 20 \)

When the \( x^2 \) coefficient is not 1, try factor pairs of the first and last terms until the inner and outer products add to the middle term.

\( 6x^2 \): try \( 2x \) and \( 3x \). \( -20 \): try \( +5 \) and \( -4 \).
\( (2x+5)(3x-4) \): outer \( -8x \), inner \( 15x \), total \( 7x \)

Answer: \( (2x+5)(3x-4) \)
Quick check
Factorise fully: \( 4x^2 - 3x - 27 \).
Show worked answer
Try \( (4x + 9)(x - 3) \): outer \( -12x \), inner \( 9x \), total \( -3x \).
Answer: \( (x-3)(4x+9) \)

\( \text{HCF} \rightarrow a^2-b^2 \rightarrow \text{trinomial} \)

Order of attack
• 1. Take out the HCF
• 2. Two terms, both squares, minus sign: \( a^2 - b^2 = (a-b)(a+b) \)
• 3. Three terms: find the pair that multiplies to the last and adds to the middle
• Always check by expanding.

6. Factorising II

\( 2px + 3qx - 2py - 3qy \)

Four terms usually means grouping: pair the terms so each pair has a common factor, and the brackets left over match.

\( = x(2p + 3q) - y(2p + 3q) \)
\( = (2p + 3q)(x - y) \)
Quick check
Factorise fully: \( f + e - 1 - ef \).
Show worked answer
Regroup: \( (f - 1) + (e - ef) = (f-1) - e(f - 1) \)
Answer: \( (1-e)(f-1) \)

\( x^2y - 16 + 4y - 4x^2 \)

Rearrange so the pairs share a factor:

\( x^2y - 4x^2 + 4y - 16 \)
\( = x^2(y - 4) + 4(y - 4) \)
\( = (y-4)(x^2 + 4) \)

\( x^2 + 4 \) is a sum of squares and does not factorise further.
Answer: \( (x^2+4)(y-4) \)
Quick check
Factorise fully: \( x^3 + x^2 - x - 1 \).
Show worked answer
\( x^2(x + 1) - 1(x + 1) = (x+1)(x^2 - 1) \)
and \( x^2 - 1 = (x-1)(x+1) \), so
Answer: \( (x-1)(x+1)^2 \)

\( x^3 - y^9 \)

Write each term as a cube: \( x^3 = (x)^3 \) and \( y^9 = (y^3)^3 \). So \( a = x \), \( b = y^3 \).

\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)
\( x^3 - y^9 = (x - y^3)(x^2 + xy^3 + y^6) \)

The trinomial bracket from a sum or difference of cubes does not factorise further.

\( a^3 - b^3 = (a-b)(a^2+ab+b^2) \)

More tools
• Four terms: group into pairs with a common bracket
• Cubes: first bracket copies the sign; trinomial has the opposite middle sign, last sign positive
• \( x^2 + 4 \) (sum of squares) and the cube trinomial do not factorise further
• Keep going until no bracket can be factorised again.

7. Algebraic fractions

\( \dfrac{x^2 - 25}{25x - 125} \)

You may only cancel factors, never separate terms. So factorise the numerator and denominator first.

The denominator may never be zero, so state the restriction.

\( \dfrac{(x-5)(x+5)}{25(x-5)} = \dfrac{x+5}{25}, \quad x \neq 5 \)

Answer: \( \dfrac{x+5}{25} \) (equivalently \( \dfrac{x}{25} + \dfrac{1}{5} \)), \( x \neq 5 \)

\( \dfrac{27x^3 - 8}{27x^2 + 18x + 12} \)

Numerator: difference of cubes with \( a = 3x \), \( b = 2 \). Denominator: common factor 3.

\( \dfrac{(3x-2)(9x^2+6x+4)}{3(9x^2+6x+4)} = \dfrac{3x-2}{3} \)

Answer: \( \dfrac{3x-2}{3} \), which is \( x - \dfrac{2}{3} \). The bracket \( 9x^2 + 6x + 4 \) is never zero for real \( x \), so there is no real restriction here.

\( \dfrac{y^2-y-2}{y^2-4} \times \dfrac{y^2+2y}{y^2+y} \)

Factorise every numerator and denominator, then cancel common factors across the whole product.

\( \dfrac{(y-2)(y+1)}{(y-2)(y+2)} \times \dfrac{y(y+2)}{y(y+1)} = 1 \)

Restrictions (from every denominator): \( y \neq 2,\; -2,\; 0,\; -1 \).
Answer: \( 1 \)

\( \dfrac{x^2-1}{(x+2) + x(x+2)} \div \dfrac{x-1}{2x+4} \)

To divide, multiply by the reciprocal of the second fraction. Then factorise and cancel.

Quick check
Simplify fully: \( \dfrac{x^2-1}{(x+2) + x(x+2)} \div \dfrac{x-1}{2x+4} \).
Show worked answer
Denominator: \( (x+2) + x(x+2) = (x+2)(1+x) \).
\( \dfrac{(x-1)(x+1)}{(x+2)(x+1)} \times \dfrac{2(x+2)}{x-1} = 2 \)
Restrictions: \( x \neq -2,\; -1,\; 1 \).
Answer: \( 2 \)

\( \dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c} \)

Simplifying fractions
• Factorise every numerator and denominator fully
• Cancel factors only, never terms
• Divide = multiply by the reciprocal
• State restrictions: any value that makes a denominator (or a divisor) zero is excluded.

\( \text{expand} \;\rightleftarrows\; \text{factorise} \)

You have covered the whole Grade 10 algebraic expressions topic.

• Rational, irrational and non-real numbers; rounding; surds between integers
• Products of binomials and trinomials, squares, sum and difference of cubes
• Factorising: HCF, difference of squares, trinomials, grouping, cubes
• Algebraic fractions: factorise, cancel factors, multiply and divide, state restrictions

Expanding and factorising are opposite processes — check every factorisation by expanding it again.