Chapter 3 of 8

All 8 chapters
  1. 1Expanding two brackets, readunderstood
  2. 2Factorising quadratics, readunderstood
  3. 3Using factorising, readunderstood
  4. 4Completing the square, readunderstood
  5. 5The quadratic formula, readunderstood
  6. 6Graphs of quadratics, readunderstood
  7. 7Quadratic sequences, readunderstood
  8. 8A line and a curve, readunderstood

M4.9.3 Chapter 3

Using factorising

I factorised x² − 5x + 6 = 0 into (x − 2)(x − 3) = 0. What does that say about x?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Solve x² − 5x + 6 = 0 by factorising
  • Explain why ab = 0 means a = 0 or b = 0
  • Simplify (x² − 9)/(x + 3) by factorising

Factorising gives a way to solve equations. Take x² − 5x + 6 = 0. The left side factorises to (x − 2)(x − 3). So the equation says that two numbers multiply to give 0.

Try some products. If neither number is 0, the product is not 0. The product is 0 only when at least one of the numbers is 0.

A table with the heads a, b and a × b. Rows: a is 3, b is 4, a × b is 12; a is 0, b is 9, a × b is 0; a is 6, b is 0, a × b is 0; a is −2, b is 5, a × b is −10.aba × b3412090600−25−10
The product is 0 in the two rows where a number is 0, and in no other row.

Here is why. Say a × b = 0 and a is not 0. Divide both sides by a. That gives b = 0. So if a is not 0, then b must be. At least one of a and b is 0.

Draw it first

Solve x² − 5x + 6 = 0

  1. A column of equations with arrows between them. The first is x² − 5x + 6 = 0. Step 2 adds an arrow with the note factorise, and (x − 2)(x − 3) = 0. Step 3 adds an arrow with the note one bracket must be 0, and x − 2 = 0 or x − 3 = 0. Step 4 adds an arrow with the note solve each, and x = 2 or x = 3.x² − 5x + 6 = 0

    1 Start with the equation. It already has 0 on the right.

  2. A column of equations with arrows between them. The first is x² − 5x + 6 = 0. Step 2 adds an arrow with the note factorise, and (x − 2)(x − 3) = 0. Step 3 adds an arrow with the note one bracket must be 0, and x − 2 = 0 or x − 3 = 0. Step 4 adds an arrow with the note solve each, and x = 2 or x = 3.x² − 5x + 6 = 0x² − 5x + 6 = 0(x − 2)(x − 3) = 0factorise

    2 Factorise the left side: (x − 2)(x − 3) = 0.

  3. A column of equations with arrows between them. The first is x² − 5x + 6 = 0. Step 2 adds an arrow with the note factorise, and (x − 2)(x − 3) = 0. Step 3 adds an arrow with the note one bracket must be 0, and x − 2 = 0 or x − 3 = 0. Step 4 adds an arrow with the note solve each, and x = 2 or x = 3.x² − 5x + 6 = 0x² − 5x + 6 = 0(x − 2)(x − 3) = 0factorisex² − 5x + 6 = 0(x − 2)(x − 3) = 0x − 2 = 0 or x − 3 = 0factoriseone bracket must be 0

    3 A product is 0 only if a factor is 0. So x − 2 = 0 or x − 3 = 0.

  4. A column of equations with arrows between them. The first is x² − 5x + 6 = 0. Step 2 adds an arrow with the note factorise, and (x − 2)(x − 3) = 0. Step 3 adds an arrow with the note one bracket must be 0, and x − 2 = 0 or x − 3 = 0. Step 4 adds an arrow with the note solve each, and x = 2 or x = 3.x² − 5x + 6 = 0x² − 5x + 6 = 0(x − 2)(x − 3) = 0factorisex² − 5x + 6 = 0(x − 2)(x − 3) = 0x − 2 = 0 or x − 3 = 0factoriseone bracket must be 0x² − 5x + 6 = 0(x − 2)(x − 3) = 0x − 2 = 0 or x − 3 = 0x = 2 or x = 3factoriseone bracket must be 0solve each

    4 Solve each one. x = 2 or x = 3. These are the roots of the equation.

Worked example

Solve x² − 5x + 6 = 0.

  1. Factorise the left side: (x − 2)(x − 3) = 0.
  2. A product is 0 only when one factor is 0. So x − 2 = 0 or x − 3 = 0.
  3. Solve each one: x = 2 or x = 3.
  4. Check x = 3 in the equation: 3 × 3 − 5 × 3 + 6 = 9 − 15 + 6 = 0.

Answer: x = 2 or x = 3

You can see the roots on a graph. The curve y = x² − 5x + 6 meets the x axis where y = 0. That is exactly where the equation is true.

A graph with the x axis x and the y axis y. A curve named y = x² − 5x + 6 dips below the x axis between the points A and B, where it crosses the x axis.ABy = x² − 5x + 612345−3−2−112345670xy
The curve meets the x axis at A, where x = 2, and at B, where x = 3. These are the roots.

The method needs 0 on one side. If it is not there, make it. Take everything to the left, so that the right side is 0, and then factorise.

Worked example

Solve x² + 12 = 7x.

  1. Take 7x from both sides, so that one side is 0: x² − 7x + 12 = 0.
  2. Two numbers with product 12 and sum −7 are −3 and −4. So (x − 3)(x − 4) = 0.
  3. Put each factor equal to 0 and solve: x − 3 = 0 gives x = 3, and x − 4 = 0 gives x = 4.
  4. Check x = 4 in the first equation: 4 × 4 + 12 = 28, and 7 × 4 = 28.

Answer: x = 3 or x = 4

See it three ways

With things

Try the whole numbers 1 to 4 in (x − 1)(x − 4). Write down what each gives.

As a picture

A table with the heads x and (x − 1)(x − 4). Rows: x is 1, the product is 0; x is 2, the product is −2; x is 3, the product is −2; x is 4, the product is 0.x(x − 1)(x − 4)102−23−240

In words

The product is 0 for x = 1 and for x = 4. It is 0 only when x − 1 = 0 or x − 4 = 0.

In symbols

(x − 1)(x − 4) = 0 gives x − 1 = 0 or x − 4 = 0, so x = 1 or x = 4.

A difference of two squares gives two answers with opposite signs. Take x² − 16 = 0. It factorises as (x + 4)(x − 4) = 0.

Worked example

Solve x² − 16 = 0.

  1. Write 16 as 4 × 4. The left side is a difference of two squares: (x + 4)(x − 4) = 0.
  2. Put each factor equal to 0: x + 4 = 0 or x − 4 = 0.
  3. So x = −4 or x = 4. Both work, because (−4) × (−4) = 16 and 4 × 4 = 16.

Answer: x = −4 or x = 4

Factorising also simplifies fractions made of expressions. Take (x² − 9)/(x + 3). The top is a difference of two squares, so it factorises. Then a bracket that is on the top and the bottom cancels.

A fraction. The top is (x + 3)(x − 3) and the bottom is (x + 3). The bracket (x + 3) is crossed out on the top and on the bottom. After the fraction is = x − 3.(x + 3)(x − 3)(x + 3)= x − 3
The bracket (x + 3) is a factor of the whole top and of the whole bottom, so it cancels.

Worked example

Simplify (x² − 9)/(x + 3).

  1. Factorise the top: x² − 9 = (x + 3)(x − 3).
  2. Write the fraction with the top factorised: (x + 3)(x − 3) over (x + 3).
  3. Cancel the bracket (x + 3) on the top and on the bottom.
  4. What is left is x − 3. This holds when x is not −3, because the bottom cannot be 0.
  5. Check with x = 5: (25 − 9)/(5 + 3) = 16/8 = 2, and 5 − 3 = 2.

Answer: x − 3

You can only cancel a whole bracket or a whole number that multiplies all of the top and all of the bottom. A term that is added cannot be cancelled.

Try it

Find where a quadratic is 0

You need: paper and a pencil.

  1. Work out x² − 5x + 6 for x = 1, 2, 3 and 4.
  2. Write each result in a table.
  3. Look for the values of x that give 0.

What you will see: The results are 2, 0, 0 and 2. The zeros are at x = 2 and x = 3, the roots you found by factorising.

If you cannot try it, this is what happens.
A table with the heads x and x² − 5x + 6. Rows: x is 1 and the value is 2; x is 2 and the value is 0; x is 3 and the value is 0; x is 4 and the value is 2.xx² − 5x + 612203042

A wrong idea many people have

To solve x² = 5x, divide both sides by x. So x = 5.

Why it feels right: Dividing both sides by the same thing keeps an equation true, and it makes the equation shorter.

The test that breaks it: Put x = 0. Then x² = 0 and 5x = 0, so x = 0 is also an answer. Dividing by x lost it.

What is true: Do not divide both sides by x: x may be 0, and that answer is lost. Move everything to one side: x² − 5x = 0. Take out x: x(x − 5) = 0. So x = 0 or x = 5.

A wrong idea many people have

(x − 1)(x − 2) = 6 gives x − 1 = 6 or x − 2 = 6.

Why it feels right: It copies the method for 0. Each bracket seems to be allowed to take the 6.

The test that breaks it: Take x = 7 from x − 1 = 6. Then (7 − 1)(7 − 2) = 6 × 5 = 30, not 6.

What is true: Splitting into brackets works only when the product is 0. Expand and make one side 0 first: x² − 3x − 4 = 0. Then (x − 4)(x + 1) = 0.

A wrong idea many people have

(x + 3)/3 = x.

Why it feels right: There is a 3 on the top and a 3 on the bottom, so they seem to cancel.

The test that breaks it: Put x = 3. Then (3 + 3)/3 = 2, but the answer x gives 3.

What is true: The 3 on the top is added to x. It does not multiply the whole top, so it cannot cancel. Only a factor of the whole top can cancel.

Key words

roots
the solutions of an equation like x² − 5x + 6 = 0: the values of x that make it true.
cancel
to divide the top and the bottom of a fraction by the same factor.

Back to your guess

The answer is A: x = 2 or x = 3.

Two numbers multiply to 0 only if one of them is 0. So x − 2 = 0 or x − 3 = 0, which gives x = 2 or x = 3. Both answers work, but not at the same time.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. Solve x² − 9x + 20 = 0.

    Tap your answer.

    1. Yes, that is right. Two numbers with product 20 and sum −9 are −4 and −5. So (x − 4)(x − 5) = 0. Then x − 4 = 0 or x − 5 = 0, which gives x = 4 or x = 5.

    2. Not this one. The brackets are (x − 4)(x − 5). Putting x − 4 = 0 gives x = 4, not −4. Go to M4.9.3. Try another.

    3. Not this one. These are the numbers in the equation. The solutions come from the brackets, and they are 4 and 5. Go to M4.9.2. Try another.

    4. Not this one. Both numbers in the brackets are negative, (x − 4)(x − 5), so both solutions are positive. Go to M4.9.3. Try another.

    See the answer

    x = 4 or x = 5

    Two numbers with product 20 and sum −9 are −4 and −5. So (x − 4)(x − 5) = 0. Then x − 4 = 0 or x − 5 = 0, which gives x = 4 or x = 5.

    • If you chose B: The brackets are (x − 4)(x − 5). Putting x − 4 = 0 gives x = 4, not −4. Go to M4.9.3.
    • If you chose C: These are the numbers in the equation. The solutions come from the brackets, and they are 4 and 5. Go to M4.9.2.
    • If you chose D: Both numbers in the brackets are negative, (x − 4)(x − 5), so both solutions are positive. Go to M4.9.3.

  2. Solve (x − 5)(x + 2) = 0.

    Tap your answer.

    1. Not this one. The signs are the wrong way round. x − 5 = 0 gives x = 5, and x + 2 = 0 gives x = −2. Go to M4.9.3. Try another.

    2. Not this one. The second bracket is x + 2. Putting x + 2 = 0 gives x = −2, not 2. Go to M4.9.3. Try another.

    3. Yes, that is right. One factor must be 0. x − 5 = 0 gives x = 5. x + 2 = 0 gives x = −2.

    4. Not this one. The number −10 is the last number when the brackets are expanded. It is not a solution. Put each bracket equal to 0. Go to M4.9.3. Try another.

    See the answer

    x = 5 or x = −2

    One factor must be 0. x − 5 = 0 gives x = 5. x + 2 = 0 gives x = −2.

    • If you chose A: The signs are the wrong way round. x − 5 = 0 gives x = 5, and x + 2 = 0 gives x = −2. Go to M4.9.3.
    • If you chose B: The second bracket is x + 2. Putting x + 2 = 0 gives x = −2, not 2. Go to M4.9.3.
    • If you chose D: The number −10 is the last number when the brackets are expanded. It is not a solution. Put each bracket equal to 0. Go to M4.9.3.

  3. Why can you split (x − 2)(x − 3) = 0 into x − 2 = 0 or x − 3 = 0?

    Tap your answer.

    1. Not this one. It cannot. (x − 2)(x − 3) = 6 cannot be split like that. Splitting works because the product is 0. Go to M4.9.3. Try another.

    2. Not this one. They cannot both be 0 at once, because x cannot be 2 and 3 together. Only one of them has to be 0. Go to M4.9.3. Try another.

    3. Not this one. The brackets are multiplied, not added. A product of numbers can be 0 without the numbers adding to 0. Go to M4.9.3. Try another.

    4. Yes, that is right. If a × b = 0 and a is not 0, divide both sides by a to get b = 0. So one of the two must be 0.

    See the answer

    If a product is zero, at least one of its factors is zero.

    If a × b = 0 and a is not 0, divide both sides by a to get b = 0. So one of the two must be 0.

    • If you chose A: It cannot. (x − 2)(x − 3) = 6 cannot be split like that. Splitting works because the product is 0. Go to M4.9.3.
    • If you chose B: They cannot both be 0 at once, because x cannot be 2 and 3 together. Only one of them has to be 0. Go to M4.9.3.
    • If you chose C: The brackets are multiplied, not added. A product of numbers can be 0 without the numbers adding to 0. Go to M4.9.3.

  4. Solve x² = 6x.

    Tap your answer.

    1. Not this one. Dividing by x loses the answer x = 0. Move 6x to the left: x² − 6x = 0, so x(x − 6) = 0. Go to M4.9.3. Try another.

    2. Yes, that is right. Move 6x to the left: x² − 6x = 0. Take out x: x(x − 6) = 0. So x = 0 or x − 6 = 0, which gives x = 6.

    3. Not this one. x = 0 is one answer, but x(x − 6) = 0 also gives x = 6. Put each factor equal to 0. Go to M4.9.3. Try another.

    4. Not this one. Try x = −6: the left side is 36 and the right side is −36. They are not equal, so −6 is not an answer. Go to M4.9.3. Try another.

    See the answer

    x = 0 or x = 6

    Move 6x to the left: x² − 6x = 0. Take out x: x(x − 6) = 0. So x = 0 or x − 6 = 0, which gives x = 6.

    • If you chose A: Dividing by x loses the answer x = 0. Move 6x to the left: x² − 6x = 0, so x(x − 6) = 0. Go to M4.9.3.
    • If you chose C: x = 0 is one answer, but x(x − 6) = 0 also gives x = 6. Put each factor equal to 0. Go to M4.9.3.
    • If you chose D: Try x = −6: the left side is 36 and the right side is −36. They are not equal, so −6 is not an answer. Go to M4.9.3.

  5. Simplify (x² − 25)/(x + 5).

    Tap your answer.

    1. Not this one. After cancelling the bracket (x + 5), the bracket (x − 5) is left, not (x + 5). Go to M4.9.3. Try another.

    2. Not this one. You cannot cancel part of a sum. Factorise the top first, then cancel a whole bracket. Go to M4.9.3. Try another.

    3. Yes, that is right. The top is a difference of two squares: x² − 25 = (x + 5)(x − 5). The bracket (x + 5) cancels, and x − 5 is left.

    4. Not this one. It can. The top x² − 25 is a difference of two squares, (x + 5)(x − 5), and then (x + 5) cancels. Go to M4.9.2. Try another.

    See the answer

    x − 5

    The top is a difference of two squares: x² − 25 = (x + 5)(x − 5). The bracket (x + 5) cancels, and x − 5 is left.

    • If you chose A: After cancelling the bracket (x + 5), the bracket (x − 5) is left, not (x + 5). Go to M4.9.3.
    • If you chose B: You cannot cancel part of a sum. Factorise the top first, then cancel a whole bracket. Go to M4.9.3.
    • If you chose D: It can. The top x² − 25 is a difference of two squares, (x + 5)(x − 5), and then (x + 5) cancels. Go to M4.9.2.

  6. Solve x² − 49 = 0.

    Tap your answer.

    1. Yes, that is right. x² − 49 = (x + 7)(x − 7) = 0. So x + 7 = 0 or x − 7 = 0, which gives x = −7 or x = 7.

    2. Not this one. The number −7 also works, because (−7) × (−7) = 49. The brackets (x + 7)(x − 7) give both answers. Go to M4.9.3. Try another.

    3. Not this one. Try x = 49: x² is 2 401, not 49. The solutions are the numbers whose square is 49. Go to M4.9.3. Try another.

    See the answer

    x = 7 or x = −7

    x² − 49 = (x + 7)(x − 7) = 0. So x + 7 = 0 or x − 7 = 0, which gives x = −7 or x = 7.

    • If you chose B: The number −7 also works, because (−7) × (−7) = 49. The brackets (x + 7)(x − 7) give both answers. Go to M4.9.3.
    • If you chose C: Try x = 49: x² is 2 401, not 49. The solutions are the numbers whose square is 49. Go to M4.9.3.

  7. Ravi simplifies (x + 6)/6 and gets x. What is wrong?

    Tap your answer.

    1. Not this one. Try x = 6: (6 + 6)/6 = 2, but Ravi’s answer gives 6. The two do not agree. Go to M4.9.3. Try another.

    2. Yes, that is right. To cancel, the number must multiply the whole top. Here the top is x + 6, and the 6 is added. A test with x = 6 shows the answer x is wrong.

    3. Not this one. Try x = 6: (6 + 6)/6 = 2, but x + 1 gives 7. Only a factor of the whole top can cancel. Go to M4.9.3. Try another.

    See the answer

    A term that is added to x cannot be cancelled with the number below.

    To cancel, the number must multiply the whole top. Here the top is x + 6, and the 6 is added. A test with x = 6 shows the answer x is wrong.

    • If you chose A: Try x = 6: (6 + 6)/6 = 2, but Ravi’s answer gives 6. The two do not agree. Go to M4.9.3.
    • If you chose C: Try x = 6: (6 + 6)/6 = 2, but x + 1 gives 7. Only a factor of the whole top can cancel. Go to M4.9.3.

  8. From an earlier chapter

    Factorise x² + 6x + 8.

    Tap your answer.

    1. Not this one. These multiply to 8, but they add to 9, not 6. Go to M4.9.2. Try another.

    2. Not this one. These add to 8, but they multiply to 15. You need product 8 and sum 6. Go to M4.9.2. Try another.

    3. Not this one. These multiply to 8, but they add to −6. The x number is +6, so both numbers are positive. Go to M4.9.2. Try another.

    4. Yes, that is right. Find two numbers with product 8 and sum 6. The pair 2 and 4 works: 2 × 4 = 8 and 2 + 4 = 6.

    See the answer

    (x + 2)(x + 4)

    Find two numbers with product 8 and sum 6. The pair 2 and 4 works: 2 × 4 = 8 and 2 + 4 = 6.

    • If you chose A: These multiply to 8, but they add to 9, not 6. Go to M4.9.2.
    • If you chose B: These add to 8, but they multiply to 15. You need product 8 and sum 6. Go to M4.9.2.
    • If you chose C: These multiply to 8, but they add to −6. The x number is +6, so both numbers are positive. Go to M4.9.2.

  9. Solve x² − 11x + 24 = 0. Give the smaller solution first.

    See the answer

    x = 3 or x = 8

    Two numbers with product 24 and sum −11 are −3 and −8. So (x − 3)(x − 8) = 0, and x = 3 or x = 8.

  10. Solve x² = 8x. Give the solution that is not 0.

    See the answer

    x = 8

    Move 8x to the left: x² − 8x = 0. Then x(x − 8) = 0, so x = 0 or x = 8. The one that is not 0 is 8.

Keep it

Copy this into your exercise book.

Make one side 0. Factorise. A product is 0 only if a factor is 0, so put each bracket equal to 0. Do not divide both sides by x: x may be 0. Cancel only whole brackets.

Draw: The equation x² − 5x + 6 = 0, the brackets (x − 2)(x − 3) = 0, and the two answers.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

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