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Year 8 Maths Algebra

Expanding Expressions and Simplifying

20 practice questions 2 video lessons Theory + worked examples
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Practice questions

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Theory

The distributive law: \(a(b + c) = a \times b + a \times c = ab + ac\). To expand, multiply every term inside the bracket by the term outside, then collect like terms: \(3 + 2(x + 5) = 3 + 2x + 10 = 2x + 13\). This Year 8 maths page shows how.

The distributive law. \(a(b + c) = ab + ac\) and \(a(b - c) = ab - ac\). Removing the brackets this way is called expanding.

The outside term keeps its sign. A negative outside multiplies every term: \(-3(x + 4) = -3x - 12\) and \(-5(p - 2) = -5p + 10\), because a negative times a negative is positive.

A term outside. Multiply the numbers, then the pronumerals: \(2x(x + 5) = 2x^2 + 10x\).

Subtracting a bracket. The minus belongs to the term in front of the bracket: \(9 - 2(x - 3) = 9 - 2x + 6 = 15 - 2x\). A bare minus means \(-1\): \(-(x - 4) = -x + 4\), and \(-x(4x + 3) = -4x^2 - 3x\).

Then collect like terms, squared term first, then the pronumeral term, then the number. Check by substituting the same number into the question and the answer.

Area model for 2x(x + 5) A rectangle of height 2x split into two parts of widths x and 5. The parts have areas 2x squared and 10x. x2x² 510x 2x 2x(x + 5) = 2x² + 10x
The two parts of the rectangle add to the whole area, which is exactly what expanding does: \(2x \times x + 2x \times 5\).

The distributive law:

\[ a(b + c) = ab + ac \]
a ( b + c )= ab + ac
\[ a(b - c) = ab - ac \]
a ( b − c )= ab − ac
\[ -a(b - c) = -ab + ac \]
− a ( b − c )=− ab + ac
Every term, not just the first. \(5(2n + 3) = 10n + 15\), not \(10n + 3\).
Expand before combining. \(20 - 3(x + 2) = 20 - 3x - 6 = 14 - 3x\), not \(17(x + 2)\).

How to expand and simplify

  1. Multiply every term inside each bracket by the term in front (with its sign).
  2. Collect the like terms.
  3. Check by substituting a number.
Example 1 — Expand, then add
Simplify \(3 + 2(x + 5)\).
Solution
\(3 + 2(x + 5)\)\(=\)\(3 + 2x + 2 \times 5\)
\(\)\(=\)\(3 + 2x + 10\)
\(\)\(=\)\(2x + 13\)
2 x +13

\(2x + 13\)

Example 2 — Subtract a term times a bracket
Simplify \(2(x - 4) - x(x + 2)\).
Solution
\(\)\(=\)\(2x - 8 - x^2 - 2x\)
\(\)\(=\)\(-x^2 - 8\)
− x −8

\(-x^2 - 8\) (the \(2x\) terms cancel)

Example 3 — Expand a bracket
Expand
(i) \(4(3p - 5)\)
(ii) \(-2(a + 7b)\)
Solution
\(4(3p - 5)\)\(=\)\(12p - 20\)
\(-2(a + 7b)\)\(=\)\(-2a - 14b\)
12 p −20−2 a −14 b

(i) \(12p - 20\)
(ii) \(-2a - 14b\)

Example 4 — Expand and collect
Expand and simplify \(3k(k - 2) + 5k\).
Solution
\(3k(k - 2)\)\(=\)\(3k^2 - 6k\)
\(3k^2 - 6k + 5k\)\(=\)\(3k^2 - k\)
3 k − k

\(3k^2 - k\). Check with \(k = 4\): \(12 \times 2 + 20 = 44\) and \(48 - 4 = 44\).

Example 5 — Perimeter
A rectangle is \((3x - 1)\) cm long and \((x + 4)\) cm wide. Find its perimeter, expanded and simplified.
Solution
\(P\)\(=\)\(2(3x - 1) + 2(x + 4)\)
\(\)\(=\)\(6x - 2 + 2x + 8\)
\(\)\(=\)\(8x + 6\)
(8 x +6) cm

\((8x + 6)\) cm. Check with \(x = 3\): sides \(8\) and \(7\), so \(P = 30\), and \(8 \times 3 + 6 = 30\).

Example 6 — Subtract a bracket
Expand and simplify \(4c(c + 3) - 2c(c - 5)\).
Solution
\(4c(c + 3)\)\(=\)\(4c^2 + 12c\)
\(-2c(c - 5)\)\(=\)\(-2c^2 + 10c\)
\(\)\(=\)\(2c^2 + 22c\)
2 c +22 c

\(2c^2 + 22c\). Check with \(c = 3\): \(12 \times 6 - 6 \times (-2) = 84\) and \(18 + 66 = 84\).

Common pitfalls

Multiplying only the first term. \(3(x + 4) = 3x + 12\).
Losing the second sign. \(-4(y - 6) = -4y + 24\).
Forgetting the square. \(x(x + 3) = x^2 + 3x\), not \(2x + 3x\).
Adding before expanding. \(5 + 2(x + 1)\) is \(2x + 7\), not \(7x + 7\).

Frequently asked questions

What is the distributive law?

\(a(b + c) = ab + ac\): the term outside multiplies every term inside.

How do you expand a negative times a bracket?

Multiply each term by the negative: \(-2(x - 3) = -2x + 6\).

How do you expand x(x + 4)?

\(x \times x + x \times 4 = x^2 + 4x\).

What do you do after expanding?

Collect like terms, writing the answer in descending powers.

How can I check an expansion?

Substitute a number into the original and your answer; both must give the same value.