0580 · 2.2Algebraic manipulation

Equivalent expressions and like terms

0:00The three expressions on screen look different, but they give exactly the same value for every value of x. Algebraic manipulation changes how an expression is written without changing what it means. One form makes addition easier, while another reveals factors or a square. By the end, you will understand how to move between these forms, and why each method works. An expression represents a value, rather than asking us to find an unknown.

0:25Our task here is to rewrite expressions, not solve equations. You will meet solving equations in the video on that subtopic. Today, collecting terms, expanding, factorising and completing the square all follow one principle: keep the expression equivalent throughout. A term is a part of an expression separated by addition or subtraction. Its coefficient is the numerical multiplier, and a constant term contains no variable. For example, −3x² has coefficient negative three.

0:55Remember that x² means x × x, while 2x means 2 × x. Like terms have exactly the same variables raised to exactly the same powers. Two a squared and 5a² are like terms. Three a times b and −9a × b form another matching pair. However, a squared and a × b are different kinds of term, so their coefficients cannot simply be combined. To simplify this expression, combine the coefficients of the a² terms, giving 7a².

1:27Collecting the a × b terms gives −6a × b. The constants combine to give three. Our simplest form is seven a squared minus six a times b plus three, because no remaining terms are alike. This works just like combining quantities of the same kind. Two lengths plus five lengths gives seven lengths, but lengths cannot be combined with areas in that way. Also, collecting terms does not change their powers.

1:522x² + 5x² gives 7x², not 7x⁴.

Expanding brackets systematically

1:58Expanding uses the distributive law: a multiplier applies to every term inside a bracket. The rectangle illustrates this with a height of three and a width of x + 2. Its total area can be calculated using the whole width or the two smaller areas. Both methods give 3x + 6. Consider 3x multiplied by the bracket 2x − 4y. Multiplying 3x by 2x gives 6x². Applying the multiplier to −4y gives −12x × y.

2:28The expanded expression is six x squared minus twelve x times y, with both original terms accounted for. Signs belong to their terms, so include them when multiplying. A positive factor multiplied by a negative factor gives a negative result. Two negative factors give a positive result. Missing a term or losing its sign changes the expression, so draw a separate multiplication connection to every term. For two brackets, every term in the first must multiply every term in the second.

2:56Here, those products are 3x², −12x × y, x × y, and −4y². Collecting the two middle terms gives −11x × y. The final expression is three x squared minus eleven x times y minus four y squared. With three brackets, expand two first and leave the third unchanged. Multiplying x − 2 by x + 3 gives x² + 3x − 2x − 6. Collecting like terms produces x² + x − 6.

3:28We then multiply this entire expression by the remaining bracket, 2x + 1. Multiplying the x² term gives 2x³ + x². The x term contributes 2x² + x, while negative six contributes −12x − 6. Collecting like terms gives 2x³ + 3x² − 11x − 6. Working in stages keeps all six products visible.

Common factors and special factorisation

3:52Factorising reverses expanding by writing an expression as a product of factors. A factor is something multiplied by another quantity to make the expression. The instruction factorise means factorise fully. That means checking whether any numerical factor, variable factor or bracket can still be taken further. In nine x squared plus fifteen x times y, both terms contain a factor of 3x. Dividing 9x² by 3x leaves 3x. The second term divided by that common factor leaves 5y.

4:23Therefore, the fully factorised expression is 3x multiplied by the bracket 3x + 5y. Taking out only x would be a correct first step, but it would not finish the factorisation. The coefficients remaining inside would still share a factor of three. Use the highest common numerical factor and every variable factor shared by all terms. Expanding your answer should reproduce the original expression exactly. Sometimes a common factor appears only after grouping terms.

4:51In this expression, the first pair becomes x multiplied by the bracket a plus b. The second pair becomes k × y multiplied by that same bracket. We can then extract a plus b, leaving the other bracket x plus k times y. For an exam-style version, factorise 6x × y, plus 9x, plus 4y, plus six. The first pair has the common factor 3x, leaving the bracket 2y + 3.

5:17The second pair has the common factor two, leaving that same bracket. Extracting the shared bracket gives 2y + 3, multiplied by 3x + 2. A difference of two squares has a special factorisation because its middle terms cancel. Expanding p − q, multiplied by p + q, gives two middle terms: plus p × q, and minus p × q. These cancel, leaving p² − q². For a squared x² − b²y², the squared quantities are a times x and b × y.

5:50For 25x² − 9y², recognise the squares of 5x and 3y. The factors are 5x − 3y and 5x + 3y. Their opposite signs make the middle terms cancel. A sum of squares does not factorise using this rule, because that cancellation would produce subtraction, not addition. A perfect square expression comes from multiplying an identical pair of brackets. Expanding a plus b, all squared, produces a squared, two copies of a × b, and b².

6:21This explains the middle term in a squared plus two a times b plus b squared. For x² + 10x + 25, the matching bracket is x + 5, squared. The middle term is essential when recognising a perfect square. X² + 25 alone is not the square of x + 5. With subtraction inside the bracket, the middle term becomes negative. Therefore, x² − 10x + 25 is the square of x − 5.

Factorising quadratic expressions

6:48For a quadratic beginning with x², expanding explains how to choose the two bracket constants. Their sum gives the coefficient of x, while their product gives the constant term. In x² + 5x + 6, we need a sum of five and a product of six. Two and three satisfy both conditions, giving the brackets x + 2 and x + 3. An exam-style quadratic with a different leading coefficient is 6x² + 11x + 3.

7:16Multiplying six by three gives eighteen. Nine and two have product eighteen and sum eleven, so split 11x into 9x + 2x. The expression becomes 6x² + 9x + 2x + 3. Factorising the first pair gives 3x multiplied by the bracket 2x + 3. The second pair is one multiplied by that same bracket. Extracting the shared bracket gives three x plus one multiplied by two x plus three.

7:43Splitting the middle term works because it creates matching bracket factors without changing the expression. Negative constants require the same reasoning, with careful attention to signs. For x² + x − 6, the required product is negative six and the sum is one. Three and negative two satisfy both conditions. The factorisation is therefore x plus three multiplied by x minus two. A cubic of the listed form has an x factor in every term.

8:10For 2x³ + 10x² + 12x, extract the common factor 2x first. This leaves x² + 5x + 6 inside the bracket. Factorising that quadratic gives two x multiplied by x plus two multiplied by x plus three. This example shows why factorise fully matters. Stopping after extracting 2x would leave a quadratic that can still be factorised. A useful order is common factors first, then any recognisable patterns inside.

8:39Some quadratics have no integer bracket factors, so do not force a pair that fails when expanded.

Completing the square

8:45Completing the square deliberately builds a perfect square, even when the original expression is not one. The diagram shows x² with two strips of area 3x added around it. A corner of area nine completes the larger square. Its side is x + 3, so its area is x + 3, all squared. To complete the square for x² + 6x + 2, half the x coefficient is three. Squaring three gives nine, which is the missing constant for the perfect square.

9:12Add nine and subtract nine, leaving the expression unchanged. The first three terms become x + 3, all squared, while −9 + 2 gives negative seven. Halving works because expanding x plus a constant, all squared, creates twice that constant as the coefficient of x. Adding the missing square alone would change the expression, so compensation is essential. A negative coefficient follows exactly the same reasoning. For x² − 8x + 3, use x − 4, all squared, minus thirteen.

9:44For the exam-style expression 2x² + 12x + 7, first extract two from the terms containing x. This gives two multiplied by the bracket x² + 6x, then plus seven. Inside, completing the square gives x + 3, all squared, minus nine. The outside factor multiplies both parts. This produces two times x plus three, all squared, then minus 18 + 7. Combining the constants gives two times x plus three, all squared, minus eleven.

10:15The important detail is that the correction becomes negative eighteen, not negative nine. For a general quadratic, extract the leading coefficient from the first two terms before completing the square. That multiplier must also apply to the compensating constant. An odd x coefficient produces fractions, which are perfectly valid. For x² + 3x + 1, half of three is 3/2. Its square is 9/4, so write x + 3/2, all squared, minus 9/4, plus one.

10:45Combining the constants gives x + 3/2, all squared, minus 5/4.

Exam commands and final checks

10:50In an examination, the command word tells you which form to produce. Simplify means give the answer in its simplest form, while expand requires multiplying out the brackets. Factorise means factorise fully, and complete the square requires a squared bracket with the appropriate multiplier and constant. Show the intermediate lines, especially middle-term splitting and the compensation when completing a square. These methods must work without a calculator on Paper two, and they remain relevant on calculator Paper four.

11:20Algebraic manipulation normally gives exact expressions, so retain fractions rather than rounding them into decimals. Numerical accuracy instructions matter only when a numerical answer is requested. Unless instructed otherwise, those answers are generally given to three significant figures, without rounding intermediate calculations. The main mistakes are combining unlike terms, missing products during expansion, and stopping factorisation too early. Squaring a bracket also creates a middle term, so square both terms and include their cross products.

11:50When completing the square, compensate for every added quantity, including any outside multiplier. Expanding a proposed answer is a reliable final check. All these methods change form while keeping value. Collecting like terms combines matching quantities, and expanding distributes multiplication across addition or subtraction. Factorising reverses that process by exposing shared factors and patterns. Completing the square builds an equivalent perfect square expression, giving you another useful way to see the same algebra.