Worked example 1 · Easy
Worked method
Simplify \(7p-3p+2p\).
\((7-3+2)p=\boxed{6p}\).
Chapter 20
Algebraic manipulation
Simplifying, expanding, factorising and completing the square. Original explanations and examination-focused practice based on the supplied textbook's topic sequence.
Simplifying algebraic expressionsSimplifying more complex algebraic expressionsExpanding a single bracketFactorising algebraic expressionsFactorising expressions of the form \(ax+bx+kay+kby\)Expanding a pair of bracketsExpanding more complex bracketsFactorising expressions of the form \(a^2x^2-b^2y^2\)Factorising expressions of the form \(x^2+bx+c\)Factorising expressions of the form \(a^2+2ab+b^2\)Factorising quadratic expressions of the form \(ax^2+bx+c\)Factorising expressions of the form \(ax^3+bx^2+cx\)Completing the square
By the end of this chapter, you should be able to
- Explain the key ideas and use correct mathematical terminology in each textbook subtopic.
- Use the appropriate methods and show enough working for examination marks.
- Check intermediate arithmetic, signs, units and the final answer.
1
Subtopic 1
Simplifying algebraic expressions
Only like terms can be added or subtracted: they have the same variables raised to the same powers. Coefficients are the numbers multiplying the variables.
When multiplying terms, multiply coefficients and add powers of the same letter. Do not confuse \(3x^2\) and \(3x\), or \(x+x=2x\) with \(x\times x=x^2\).
Rules and important ideas
- \(3a+5a-2a=6a\)
- \((2x)(3x^2)=6x^3\)
Worked example 2 · Medium
Worked method
Simplify \(4x+3y-7x+5y\).
\((4x-7x)+(3y+5y)=\boxed{-3x+8y}\).
Worked example 3 · Exam-style
Worked method
Simplify \((-3a^2b)(4ab^3)\).
\((-3)(4)a^{2+1}b^{1+3}=\boxed{-12a^3b^4}\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
2
Subtopic 2
Simplifying more complex algebraic expressions
With several types of term, group them by their letter combinations and indices. \(xy\) and \(yx\) are like terms, but \(x^2y\) and \(xy^2\) are not.
Watch signs before brackets and around negative terms. Multiplication and division of numerical coefficients and index laws still apply.
Rules and important ideas
- \(4a^2b-3ab^2+2a^2b=6a^2b-3ab^2\)
Worked example 4 · Easy
Worked method
Simplify \(3xy+5yx-2xy\).
Because \(xy=yx\), \((3+5-2)xy=\boxed{6xy}\).
Worked example 5 · Exam-style
Worked method
Simplify \(2x^2y-5xy^2+7xy^2-x^2y\).
\((2-1)x^2y+(-5+7)xy^2=\boxed{x^2y+2xy^2}\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
3
Subtopic 3
Expanding a single bracket
To expand one bracket, multiply every term inside it by the factor outside. For a negative factor, each product changes sign accordingly.
Expansion and factorisation are reverse operations. Check an expansion by substituting an easy number or by reverse-factorising it.
Rules and important ideas
- \(a(b+c)=ab+ac\)
- \(-2(x-3)=-2x+6\)
Worked example 6 · Easy
Worked method
Expand \(4(3x+2)\).
\(4(3x)+4(2)=\boxed{12x+8}\).
Worked example 7 · Exam-style
Worked method
Expand and simplify \(-3x(2x^2-4x+5)\).
\(-6x^3+12x^2-15x\). Thus \(\boxed{-6x^3+12x^2-15x}\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
4
Subtopic 4
Factorising algebraic expressions
To factorise, find the highest common factor of all terms, including numbers and letters, and place it outside a bracket.
Divide each term by the factor to form the expression inside the bracket. Factorise fully and always multiply back out to check.
Rules and important ideas
- \(6x^2+9x=3x(2x+3)\)
Worked example 8 · Easy
Worked method
Factorise \(8x+12\).
The highest common factor is 4. \(\boxed{4(2x+3)}\).
Worked example 9 · Exam-style
Worked method
Factorise \(15a^3b-10a^2b^2+5ab\).
Take out \(5ab\): \(\boxed{5ab(3a^2-2ab+1)}\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
5
Subtopic 5
Factorising expressions of the form \(ax+bx+kay+kby\)
Grouping is useful when four terms have no common factor all at once. Collect two pairs that each have a factor, then factorise the repeated bracket.
You may need to rearrange the terms or factor out a negative number. Your aim is to make the same bracket appear twice.
Rules and important ideas
- \(ax+bx+kay+kby=(a+b)(x+ky)\)
Worked example 10 · Easy
Worked method
Factorise \(ax+bx+3ay+3by\).
\(x(a+b)+3y(a+b)=\boxed{(a+b)(x+3y)}\).
Worked example 11 · Exam-style
Worked method
Factorise \(6xp-9x+4yp-6y\).
\(3x(2p-3)+2y(2p-3)=\boxed{(2p-3)(3x+2y)}\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
6
Subtopic 6
Expanding a pair of brackets
When multiplying two brackets, each term in the first bracket multiplies each term in the second. An area model, grid or distributive method can help you avoid missing products.
After multiplying, collect like terms. Negative signs matter: \((-3)(-4)=+12\). You can check by substitution.
Rules and important ideas
- \((x+a)(x+b)=x^2+(a+b)x+ab\)
Worked example 12 · Easy
Worked method
Expand \((x+3)(x+4)\).
\(x^2+4x+3x+12=\boxed{x^2+7x+12}\).
Worked example 13 · Exam-style
Worked method
Expand \((2x-5)(x+7)\).
\(2x^2+14x-5x-35=\boxed{2x^2+9x-35}\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
7
Subtopic 7
Expanding more complex brackets
The method is unchanged for algebraic coefficients or three brackets: multiply two brackets first, simplify, then multiply by the remaining bracket.
Keep intermediate working clearly arranged. Square brackets represent the same operation as ordinary brackets: multiply everything inside.
Rules and important ideas
- \((x+1)(x-1)=x^2-1\)
Worked example 14 · Medium
Worked method
Expand \((3x+2y)(2x-y)\).
\(6x^2-3xy+4xy-2y^2=\boxed{6x^2+xy-2y^2}\).
Worked example 15 · Exam-style
Worked method
Expand \((x+1)(x+2)(x-3)\).
First \((x+1)(x+2)=x^2+3x+2\). Then multiply by \(x-3\): \(x^3+3x^2+2x-3x^2-9x-6=\boxed{x^3-7x-6}\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
8
Subtopic 8
Factorising expressions of the form \(a^2x^2-b^2y^2\)
A difference of two squares occurs when one square term is subtracted from another. It factorises into two conjugate brackets.
If every term has a common factor, take it out first. Then factorise the remaining difference of squares.
Rules and important ideas
- \(A^2-B^2=(A-B)(A+B)\)
Worked example 16 · Easy
Worked method
Factorise \(x^2-49\).
\(49=7^2\), so \(\boxed{(x-7)(x+7)}\).
Worked example 17 · Exam-style
Worked method
Fully factorise \(3x^2-75y^2\).
Take out 3: \(3(x^2-25y^2)=\boxed{3(x-5y)(x+5y)}\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
9
Subtopic 9
Factorising expressions of the form \(x^2+bx+c\)
For a monic quadratic \(x^2+bx+c\), find two numbers with sum \(b\) and product \(c\). Place them into \((x+m)(x+n)\).
If \(c>0\), the signs match; if \(c<0\), the signs differ. Expand to verify both the middle and constant terms.
Rules and important ideas
- \(x^2+(m+n)x+mn=(x+m)(x+n)\)
Worked example 18 · Easy
Worked method
Factorise \(x^2+9x+20\).
Find 4 and 5: product 20, sum 9. \(\boxed{(x+4)(x+5)}\).
Worked example 19 · Exam-style
Worked method
Factorise \(x^2-3x-28\).
Find -7 and +4: product -28, sum -3. \(\boxed{(x-7)(x+4)}\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
10
Subtopic 10
Factorising expressions of the form \(a^2+2ab+b^2\)
A perfect-square trinomial has first and last terms that are squares and a middle term that is twice the product of their square roots.
The sign of the middle term tells you which bracket sign to use. When in doubt, expand your proposed answer to verify it.
Rules and important ideas
- \(A^2+2AB+B^2=(A+B)^2\)
- \(A^2-2AB+B^2=(A-B)^2\)
Worked example 20 · Easy
Worked method
Factorise \(x^2+10x+25\).
\(25=5^2\) and \(10x=2(x)(5)\). Thus \(\boxed{(x+5)^2}\).
Worked example 21 · Exam-style
Worked method
Factorise \(16a^2-24ab+9b^2\).
\((4a)^2-2(4a)(3b)+(3b)^2=\boxed{(4a-3b)^2}\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
11
Subtopic 11
Factorising quadratic expressions of the form \(ax^2+bx+c\)
For \(ax^2+bx+c\) with \(a\ne1\), seek two binomials whose leading coefficients multiply to \(a\), constant terms multiply to \(c\), and cross-products add to \(b\).
Always first take out any common factor. A reliable alternative is the split-middle-term method: find two numbers that multiply to \(ac\) and add to \(b\), split the middle term, then factorise by grouping.
Rules and important ideas
- \((px+q)(rx+s)=prx^2+(ps+qr)x+qs\)
Worked example 22 · Medium
Worked method
Factorise \(2x^2+7x+3\).
\(ac=6\); numbers 6 and 1 add to 7. \(2x^2+6x+x+3=2x(x+3)+(x+3)=\boxed{(2x+1)(x+3)}\).
Worked example 23 · Exam-style
Worked method
Fully factorise \(6x^2-15x+6\).
Take out 3: \(3(2x^2-5x+2)\). Split: \(2x^2-4x-x+2=(2x-1)(x-2)\). Answer \(\boxed{3(2x-1)(x-2)}\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
12
Subtopic 12
Factorising expressions of the form \(ax^3+bx^2+cx\)
Expressions of the form \(ax^3+bx^2+cx\) have at least one factor \(x\). First remove the highest common numerical factor and \(x\), then factorise the remaining quadratic if possible.
A full factorisation is a product of factors; stop only when no polynomial factor can be factorised further using the methods you know.
Rules and important ideas
- \(ax^3+bx^2+cx=x(ax^2+bx+c)\)
Worked example 24 · Medium
Worked method
Factorise \(x^3+5x^2+6x\).
\(x(x^2+5x+6)=\boxed{x(x+2)(x+3)}\).
Worked example 25 · Exam-style
Worked method
Fully factorise \(6x^3+3x^2-30x\).
\(3x(2x^2+x-10)=\boxed{3x(2x+5)(x-2)}\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
13
Subtopic 13
Completing the square
Completing the square rewrites a quadratic as a square plus or minus a constant. For \(x^2+bx+c\), take half the \(x\)-coefficient for the bracket, then adjust the constant.
For \(ax^2+bx+c\), first factor \(a\) from the two \(x\)-terms. Completing the square helps find turning points, minimum or maximum values and solve quadratic equations later in the course.
A positive squared term is always at least zero; this makes the minimum of \((x-h)^2+k\) equal to \(k\). A negative squared term has a maximum instead.
Rules and important ideas
- \(x^2+bx+c=(x+\frac b2)^2+c-(\frac b2)^2\)
Worked example 26 · Medium
Worked method
Complete the square for \(x^2+8x+5\).
Half of 8 is 4. \(x^2+8x+5=(x+4)^2-16+5=\boxed{(x+4)^2-11}\).
Worked example 27 · Exam-style
Worked method
Express \(2x^2-12x+13\) as \(a(x-h)^2+k\), and state its minimum value.
\(2(x^2-6x)+13=2[(x-3)^2-9]+13=\boxed{2(x-3)^2-5}\). Minimum is \(\boxed{-5}\), at \(x=3\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
Chapter 20 practice
These are newly written practice questions, not past-paper reproductions. Questions progress from confidence-building to exam-style. Answers are hidden until you open each question’s Show Answer section.
Simplifying algebraic expressions
1Confidence practice[2 marks]
Simplify \(6x+4x\).
Show Answer
Step-by-step solution: \(\boxed{10x}\).
2Developing practice[3 marks]
Simplify \(5a+7b-2a-3b\).
Show Answer
Step-by-step solution: \(5a-2a+7b-3b=\boxed{3a+4b}\).
3Developing practice[3 marks]
Simplify \(2x\times5x^2\).
Show Answer
Step-by-step solution: \(2\times5\times x^{1+2}=\boxed{10x^3}\).
4Exam-style practice[4 marks]
Simplify \(6m^2n-2mn^2+3mn^2-5m^2n\).
Show Answer
Step-by-step solution: \((6-5)m^2n+(-2+3)mn^2=\boxed{m^2n+mn^2}\).
Simplifying more complex algebraic expressions
5Confidence practice[2 marks]
Simplify \(4ab+3ba\).
Show Answer
Step-by-step solution: \(ab=ba\), hence \(\boxed{7ab}\).
6Developing practice[3 marks]
Simplify \(8p^2-5p+3p^2+2p\).
Show Answer
Step-by-step solution: \(\boxed{11p^2-3p}\).
7Developing practice[3 marks]
Simplify \(5x^2y+2xy^2-3x^2y\).
Show Answer
Step-by-step solution: \(\boxed{2x^2y+2xy^2}\).
8Exam-style practice[4 marks]
Simplify \(-2ab^2+5a^2b+7ab^2-3a^2b\).
Show Answer
Step-by-step solution: \((-2+7)ab^2+(5-3)a^2b=\boxed{5ab^2+2a^2b}\).
Expanding a single bracket
9Confidence practice[2 marks]
Expand \(3(a+4)\).
Show Answer
Step-by-step solution: \(\boxed{3a+12}\).
10Developing practice[3 marks]
Expand \(-2(5x-3)\).
Show Answer
Step-by-step solution: \(\boxed{-10x+6}\).
11Developing practice[3 marks]
Expand \(4y(2y-5)\).
Show Answer
Step-by-step solution: \(\boxed{8y^2-20y}\).
12Exam-style practice[4 marks]
Expand and simplify \(-2a(3a^2-4a-6)\).
Show Answer
Step-by-step solution: \(\boxed{-6a^3+8a^2+12a}\).
Factorising algebraic expressions
13Confidence practice[2 marks]
Factorise \(5x+20\).
Show Answer
Step-by-step solution: \(\boxed{5(x+4)}\).
14Developing practice[3 marks]
Factorise \(12p^2-18p\).
Show Answer
Step-by-step solution: \(\boxed{6p(2p-3)}\).
15Developing practice[3 marks]
Factorise \(9ab+6ac\).
Show Answer
Step-by-step solution: \(\boxed{3a(3b+2c)}\).
16Exam-style practice[4 marks]
Fully factorise \(18x^3y-12x^2y^2\).
Show Answer
Step-by-step solution: Common factor \(6x^2y\): \(\boxed{6x^2y(3x-2y)}\).
Factorising expressions of the form \(ax+bx+kay+kby\)
17Confidence practice[2 marks]
Factorise \(ax+bx+ay+by\).
Show Answer
Step-by-step solution: \((a+b)(x+y)\).
18Developing practice[3 marks]
Factorise \(2ax+2bx+ay+by\).
Show Answer
Step-by-step solution: \(2x(a+b)+y(a+b)=\boxed{(a+b)(2x+y)}\).
19Developing practice[3 marks]
Factorise \(px-py+3x-3y\).
Show Answer
Step-by-step solution: \(p(x-y)+3(x-y)=\boxed{(p+3)(x-y)}\).
20Exam-style practice[4 marks]
Factorise \(6ax-9bx+4ay-6by\).
Show Answer
Step-by-step solution: \(3x(2a-3b)+2y(2a-3b)=\boxed{(2a-3b)(3x+2y)}\).
Expanding a pair of brackets
21Confidence practice[2 marks]
Expand \((x+1)(x+2)\).
Show Answer
Step-by-step solution: \(x^2+2x+x+2=\boxed{x^2+3x+2}\).
22Developing practice[3 marks]
Expand \((x-4)(x+3)\).
Show Answer
Step-by-step solution: \(x^2+3x-4x-12=\boxed{x^2-x-12}\).
23Developing practice[3 marks]
Expand \((3x+2)(x-1)\).
Show Answer
Step-by-step solution: \(3x^2-3x+2x-2=\boxed{3x^2-x-2}\).
24Exam-style practice[4 marks]
Expand and simplify \((2a-3b)(4a+b)\).
Show Answer
Step-by-step solution: \(8a^2+2ab-12ab-3b^2=\boxed{8a^2-10ab-3b^2}\).
Expanding more complex brackets
25Confidence practice[2 marks]
Expand \((2x+1)(x+3)\).
Show Answer
Step-by-step solution: \(2x^2+6x+x+3=\boxed{2x^2+7x+3}\).
26Developing practice[3 marks]
Expand \((2a-b)(a+3b)\).
Show Answer
Step-by-step solution: \(2a^2+6ab-ab-3b^2=\boxed{2a^2+5ab-3b^2}\).
27Developing practice[3 marks]
Expand \((x+2)^2\) by multiplying two brackets.
Show Answer
Step-by-step solution: \((x+2)(x+2)=x^2+2x+2x+4=\boxed{x^2+4x+4}\).
28Exam-style practice[4 marks]
Expand \((x-1)(x+1)(x+4)\).
Show Answer
Step-by-step solution: First \((x-1)(x+1)=x^2-1\). Then \((x^2-1)(x+4)=\boxed{x^3+4x^2-x-4}\).
Factorising expressions of the form \(a^2x^2-b^2y^2\)
29Confidence practice[2 marks]
Factorise \(x^2-16\).
Show Answer
Step-by-step solution: \(\boxed{(x-4)(x+4)}\).
30Developing practice[3 marks]
Factorise \(9a^2-25\).
Show Answer
Step-by-step solution: \(\boxed{(3a-5)(3a+5)}\).
31Developing practice[3 marks]
Factorise \(4x^2-49y^2\).
Show Answer
Step-by-step solution: \(\boxed{(2x-7y)(2x+7y)}\).
32Exam-style practice[4 marks]
Fully factorise \(2x^2-98\).
Show Answer
Step-by-step solution: \(2(x^2-49)=\boxed{2(x-7)(x+7)}\).
Factorising expressions of the form \(x^2+bx+c\)
33Confidence practice[2 marks]
Factorise \(x^2+5x+6\).
Show Answer
Step-by-step solution: 2 and 3 sum to 5 and multiply to 6. \(\boxed{(x+2)(x+3)}\).
34Developing practice[3 marks]
Factorise \(x^2-9x+20\).
Show Answer
Step-by-step solution: -4 and -5: \(\boxed{(x-4)(x-5)}\).
35Developing practice[3 marks]
Factorise \(x^2+2x-15\).
Show Answer
Step-by-step solution: 5 and -3: \(\boxed{(x+5)(x-3)}\).
36Exam-style practice[4 marks]
Factorise \(y^2-11y-26\).
Show Answer
Step-by-step solution: -13 and 2 give sum -11, product -26. \(\boxed{(y-13)(y+2)}\).
Factorising expressions of the form \(a^2+2ab+b^2\)
37Confidence practice[2 marks]
Factorise \(x^2+6x+9\).
Show Answer
Step-by-step solution: \(\boxed{(x+3)^2}\).
38Developing practice[3 marks]
Factorise \(y^2-14y+49\).
Show Answer
Step-by-step solution: \(\boxed{(y-7)^2}\).
39Developing practice[3 marks]
Factorise \(4x^2+12x+9\).
Show Answer
Step-by-step solution: \(\boxed{(2x+3)^2}\).
40Exam-style practice[4 marks]
Factorise \(25p^2-30pq+9q^2\).
Show Answer
Step-by-step solution: \(\boxed{(5p-3q)^2}\).
Factorising quadratic expressions of the form \(ax^2+bx+c\)
41Confidence practice[2 marks]
Factorise \(2x^2+5x+2\).
Show Answer
Step-by-step solution: \(\boxed{(2x+1)(x+2)}\).
42Developing practice[3 marks]
Factorise \(3x^2+10x+3\).
Show Answer
Step-by-step solution: \(\boxed{(3x+1)(x+3)}\).
43Developing practice[3 marks]
Factorise \(2x^2-x-6\).
Show Answer
Step-by-step solution: \(\boxed{(2x+3)(x-2)}\).
44Exam-style practice[4 marks]
Fully factorise \(4x^2-14x+6\).
Show Answer
Step-by-step solution: Take out 2: \(2(2x^2-7x+3)\). Then \(\boxed{2(2x-1)(x-3)}\).
Factorising expressions of the form \(ax^3+bx^2+cx\)
45Confidence practice[2 marks]
Factorise \(x^3+3x^2\).
Show Answer
Step-by-step solution: Common factor \(x^2\): \(\boxed{x^2(x+3)}\).
46Developing practice[3 marks]
Factorise \(x^3-9x\).
Show Answer
Step-by-step solution: \(x(x^2-9)=\boxed{x(x-3)(x+3)}\).
47Developing practice[3 marks]
Factorise \(2x^3+10x^2+12x\).
Show Answer
Step-by-step solution: \(2x(x^2+5x+6)=\boxed{2x(x+2)(x+3)}\).
48Exam-style practice[4 marks]
Fully factorise \(3x^3-12x^2-15x\).
Show Answer
Step-by-step solution: \(3x(x^2-4x-5)=\boxed{3x(x-5)(x+1)}\).
Completing the square
49Confidence practice[2 marks]
Complete the square for \(x^2+4x+1\).
Show Answer
Step-by-step solution: \((x+2)^2-4+1=\boxed{(x+2)^2-3}\).
50Developing practice[3 marks]
Complete the square for \(x^2-6x+7\).
Show Answer
Step-by-step solution: \((x-3)^2-9+7=\boxed{(x-3)^2-2}\).
51Developing practice[3 marks]
Write \(x^2+10x+30\) in completed-square form.
Show Answer
Step-by-step solution: \((x+5)^2-25+30=\boxed{(x+5)^2+5}\).
52Exam-style practice[4 marks]
Complete the square for \(3x^2+12x+7\) and state the minimum value.
Show Answer
Step-by-step solution: \(3(x^2+4x)+7=3[(x+2)^2-4]+7=\boxed{3(x+2)^2-5}\). Minimum \(\boxed{-5}\).
High-grade revision checklist
- I can explain each rule in the exact textbook topic order.
- I can select and apply the correct method, including signs, powers and brackets.
- I can show clear working in harder multi-step questions.
- I can check by reversing the operation, estimating, or substituting a value, as appropriate.