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Expanding triple brackets: worked examples and method
Higher · Grade 6 · Non-calculator
This page sets out the method for expanding triple brackets in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Expanding triple brackets topic page.
Key facts
(x+a)(x+b)=x2+(a+b)x+ab
Multiplying a quadratic (3 terms) by a linear bracket (2 terms) gives 6 terms before simplifying.
A cubic expansion has the form ax3+bx2+cx+d.
The constant term is the product of the three constants; the x3 coefficient is the product of the three x coefficients — use these as checks.
(x+a)3=(x+a)(x+a)(x+a) — expand two brackets first, never cube each term separately.
The method, step by step
Method: Two then one
When you see three brackets multiplied together, pick the two easiest to expand, then multiply the quadratic by the third bracket in a grid or line by line.
Expand the first pair of brackets and simplify to a quadratic.
Write the quadratic next to the third bracket.
Multiply every term of the quadratic by the first term of the bracket.
Multiply every term of the quadratic by the second term of the bracket.
Collect like terms and order by powers of x.
Check the x3 coefficient and the constant term match the products of the original coefficients.
Method: check by substituting a number
When you have expanded three brackets and want a 10-second check before moving on.
Pick a small number for x, such as x=1 or x=2 (avoid 0, which hides terms).
Work out the original product of the three brackets at that value.
Work out your expanded expression at the same value.
If the two numbers match, the expansion is very likely right. If they differ, an x term has slipped.
How the lesson explains it
Three brackets
Multiply two brackets first, then multiply by the third. That product is a cubic.
Cube
Expanding (x+1)^3 gives x^3+3x^2+3x+1. The middle coefficients match the row 1, 3, 3, 1.
x^2 coefficient
In (x+a)(x+b)(x+c) the coefficient of x^2 is a+b+c.
Two brackets first
A pair such as (x+4)(x+5) expands to the quadratic you get before a third bracket joins.
Signs
A minus in one bracket changes the sign of every term that uses that number.
Check the ends
The x^3 coefficient stays 1, and the constant is the product of the three numbers.
Six products, then collect
Multiply (x^2+3x+2) by (x+3) one term at a time. The x in the last bracket gives x^3+3x^2+2x. The +3 gives 3x^2+9x+6. You should see six products, not three. Set them out in a line or a grid, then collect: x^3+(3x^2+3x^2)+(2x+9x)+6=x^3+6x^2+11x+6. The x^2 coefficient is 1+2+3=6, and the constant is 1 × 2 × 3=6. Those two shortcuts check the ends. They do not replace the x term, which is 11 here.
A minus sign belongs to the whole bracket. (x^2+3x+2)(x-3) is x(x^2+3x+2)-3(x^2+3x+2). The second block changes every sign: -3x^2-9x-6. Students who only change the first sign of that block lose the constant and the x term together. If a question says "show that" the cubic equals a given expression, write the six products before you collect. The products are the method marks. The collected cubic is the accuracy mark.
Method: 1. List the six products rather than trying to hold them in your head. 2. Watch the sign on every negative end. 3. Add the like powers only after the six products are written. 4. The constant term should equal the product of the three ends.
Cubes and a number in front of x
(x+2)^3 means (x+2)(x+2)(x+2). First, (x+2)^2=x^2+4x+4. Then multiply by (x+2): x^3+4x^2+4x+2x^2+8x+8=x^3+6x^2+12x+8. The pattern is x^3+3(2)x^2+3(2)^2 x+2^3. Halving the work by cubing each end gives x^3+8, which is missing 6x^2+12x. That mistake is so common that it is worth saying the row 1, 3, 3, 1 out loud before you start.
A bracket such as (2x-1) changes the x^3 coefficient. Expand (2x-1)(x+4)(x-3) by doing (x+4)(x-3)=x^2+x-12 first. Then (2x-1)(x^2+x-12)=2x^3+2x^2-24x-x^2-x+12=2x^3+x^2-25x+12. The x^3 coefficient is 2, not 1, because of the 2x. Check with x=0: the product of the constants is (-1)(4)(-3)=12, which matches the constant. Check with x=1: (1)(5)(-2)=-10, and 2+1-25+12=-10. Two checks catch a sign error that one check can miss.
Method: 1. Use the row 1, 3, 3, 1 when a bracket is cubed. 2. The middle terms carry 3a and 3a^2, not just a. 3. Do not stop at x^3+a^3. 4. Check with x=1 if the arithmetic looks light.
A check that earns the last mark
Substitute after you collect, not before. Pick x=1 because powers of 1 stay 1, so the cubic collapses to the sum of its coefficients. For x^3+6x^2+11x+6 that sum is 1+6+11+6=24. The brackets give (2)(3)(4)=24. If those disagree, do not fiddle the constant. Go back to the six products. A second check with x=2 is worth it on a show-that question: (3)(4)(5)=60, and 8+24+22+6=60.
Volume questions use the same check. Edges (x+2) cm, (x+3) cm and (x+4) cm give the volume (x+2)(x+3)(x+4) cubic centimetres. The first pair (x+2)(x+3)=x^2+5x+6, then times (x+4) is x^3+9x^2+26x+24. At x=1 the box is 3 by 4 by 5, volume 60, and 1+9+26+24=60. Write the unit on the volume if the question asks for a value, and leave the cubic in centimetres cubed if it asks for an expression. Descending powers, like terms collected, and one substitution shown: that is a complete answer.
Method: 1. Expand in stages, even when the question is a volume. 2. The highest power is x^3 for three linear brackets. 3. The constant term is the product of the three lengths with x=0. 4. A matching substitution is the check that the cubic is complete.
Signs in a triple bracket
Most slips in a triple expansion are sign slips, so it is worth knowing the pattern in advance. For (x+a)(x+b)(x+c) the answer is x^3+(a+b+c)x^2+(ab+ac+bc)x+abc. The number of negative signs in the brackets controls the constant term. With (x-1)(x-2)(x-3) there are three negatives, so the constant is -6, not +6. With two negatives in the brackets the constant is positive.
This gives you a built-in check. If a bracket pair has both signs negative, the middle of their product is negative and the end is positive: (x-2)(x-3)=x^2-5x+6. Multiplying by (x+4) then gives x^3-5x^2+6x+4x^2-20x+24=x^3-x^2-14x+24. The x^2 coefficient is -5+4=-1, which is the sum of the three numbers -2-3+4=-1. Spotting that match tells you the working is on track.
Method: 1. Count the minus signs in the three brackets to predict the sign of the constant. 2. Expand two brackets first and collect. 3. Multiply every term of that result by each term of the third bracket. 4. Compare the x^2 coefficient with the sum of the three numbers.
Tip: The constant term is always the product of the three numbers in the brackets, signs included.
Shortcuts: a difference of two squares or a repeated bracket
Look for a pair of brackets of the form (x-3)(x+3). These multiply to x^2-9 with no middle term, which makes the final step much shorter. For (x-3)(x+3)(x+5) write (x^2-9)(x+5)=x^3+5x^2-9x-45. Choosing which two brackets to multiply first can save you a whole line of working and removes the chance of an error in the middle terms.
A squared bracket, such as (x-2)^2(x+5), is a triple bracket too. Write it out in full as (x-2)(x-2)(x+5), or expand the square first as x^2-4x+4. Never write (x-2)^2=x^2-4. After you have the square, multiply by (x+5) term by term to get x^3+5x^2-4x^2-20x+4x+20=x^3+x^2-16x+20. Showing the square on its own line earns the first method mark.
Method: 1. Scan for a pair (x-a)(x+a) and multiply that pair first. 2. For a squared bracket, write it as two identical brackets and expand it on its own line. 3. Multiply the result by the remaining bracket, term by term. 4. Collect like terms and write the powers in descending order.
Show that, and the "form" questions
Triple brackets appear in exam papers as "show that (n+5)(2n-3)(3n+1) can be written in the form an^3+bn^2+cn+d, where a, b, c and d are integers". The question tells you the shape of the answer, so you only have to find four numbers. Expand carefully, collect, and write the final line as 6n^3+23n^2-38n-15. A final line with the letters matched to the form is what the mark is for.
In a "find a, b and c" version, you are given the expansion in a form like x^3+ax^2+bx+c and told to find the values. Use the sum pattern to check: a is the sum of the numbers, b is the sum of the pairwise products, and c is the product. If the question says "show that", set the answer out as a proof from the left-hand side, not by starting with the answer and working backwards.
Method: 1. Read the target form and identify what letters you must find. 2. Expand in stages and keep the working visible. 3. Collect terms and write the answer in the printed form. 4. Name the coefficients, for example "so a=6, b=23, c=-38, d=-15".
Tip: A single numerical check, such as n = 2 in both the original and the expansion, takes ten seconds and catches nearly every slip.
Triple brackets in volume and proof questions
A cuboid with sides (x+5), (x-2) and (2x+1) has volume equal to the product of the three. Multiply two sides first, then the third: (x+5)(x-2)=x^2+3x-10, and (x^2+3x-10)(2x+1)=2x^3+7x^2-17x-10. If the question gives a value, substitute it into your expanded form and also into the original product. With x=4 both give 162. Units for volume are cubic units, so include cm^3.
Proof questions use the same skill. The product of three consecutive integers n-1, n and n+1 is n(n-1)(n+1)=n(n^2-1)=n^3-n. Spotting the difference of two squares (n-1)(n+1) gives a short route. A "hence" part, such as working out 19× 20× 21 without a calculator, uses n=20: 20^3-20=8000-20=7980. Link the second part to the first, as the question asks.
Method: 1. Set up the product of the three expressions that make up the volume or the proof. 2. Expand using the shortcut or the staged method. 3. Substitute the given value into the final form, and check against the original product. 4. Give units for a volume and explain a proof in words.
Remember
Three brackets give a cubic.
The x^2 coefficient is the sum of the three added numbers.