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Factorising into Single Brackets

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How to factorise a two-term expression into a single bracket by taking out the highest common factor, including letters and powers, how to check your answer by expanding, and how factorising shows that an expression is always a multiple of a number.

What factorising means

To factorise an expression means to write it with a bracket, as a multiplication. It is the reverse of expanding. Expanding 4(x+3) gives 4x+12, so factorising 4x+12 gives 4(x+3)

A common factor is a number or letter that divides exactly into every term. In 4x+12, the 4 divides into 4x and into 12. The common factor goes outside the bracket, and what is left of each term goes inside.

You can picture 4x+12 as the area of a rectangle 4 high. The part with area 4x is x wide, and the part with area 12 is 3 wide. So the whole width is x+3, and the area is 4(x+3)

In past papers, the expression to factorise into a single bracket has always had two terms. You are usually asked to factorise something like 5x+10 or x^2+4x, for 1 mark. A question that says factorise fully, where a number and a letter can both be taken out, such as 8x^2+12x, is usually worth 2 marks. Factorising gives an expression, not a value. There is no equals sign, so there is nothing to solve.

Worked example 1: Taking out a number

Factorise 5y+35

  1. Look for a number that divides into both terms. 5 goes into 5y and into 35, so 5 goes outside the bracket.
  2. Divide the first term by 5: 5y\div5=y

  3. Your turn. Now divide the second term by 5. Work out 35\div5

    35\div5=7

  4. Check by expanding: 5\times y=5y and 5\times7=35
  5. Answer: 5(y+7)

Factorise fully: use the highest common factor

An expression can have more than one common factor. In 12x-8, both 2 and 4 divide into each term. Factorise fully means take out the highest common factor (HCF), so that nothing more can be taken out of the bracket.

12x-8=2(6x-4) is correct, but it is not fully factorised: 6x and 4 still have a common factor of 2. Taking out 4 instead gives 4(3x-2), which is fully factorised.

Each term keeps its sign. 12x\div4=3x and -8\div4=-2, so the bracket is 3x-2

Check your answer in two ways. Expand it: 4(3x-2)=12x-8, the expression you started with. Then look inside the bracket: 3x and -2 have no common factor left.

Worked example 2: Finding the highest common factor

Factorise fully 20m-30

  1. The numbers are 20 and 30. Both 2 and 5 are common factors, but there may be a bigger one.

  2. Your turn. What is the highest common factor of 20 and 30?

    The HCF of 20 and 30 is 10

  3. Divide each term by 10: 20m\div10=2m and -30\div10=-3
  4. Check: 10(2m-3)=20m-30, and 2m and -3 have no common factor. Taking out only 5 would give 5(4m-6), which is not fully factorised.
  5. Answer: 10(2m-3)

Taking out a letter

A letter can be a common factor too. k^2 means k\times k, and 6k means 6\times k. Both terms contain k, so k^2+6k=k(k+6)

Dividing by a letter takes one copy of it away: k^2\div k=k and 6k\div k=6

When the powers are different, take out the lowest power. Every term in m^3+5m^2 has at least m^2, so m^3+5m^2=m^2(m+5)

Different letters work in the same way. h is in both terms of hx+hy, so hx+hy=h(x+y)

Worked example 3: Taking out a letter

Factorise n^2-9n

  1. n^2=n\times n and 9n=9\times n, so n is a common factor.
  2. Divide the first term by n: n^2\div n=n

  3. Your turn. Now divide the second term by n. What is -9n\div n?

    -9n\div n=-9

  4. Check: n\times n=n^2 and n\times(-9)=-9n
  5. Answer: n(n-9)

A number and a letter together

Often the highest common factor is a number and a letter together. Find it in parts:

  • The number part is the HCF of the numbers.
  • The letter part is each letter that is in every term, with the lowest power you see.
  • Multiply the parts together. That is the HCF, and it goes outside the bracket.

Then divide each term by the HCF to find what goes inside. For 6x^2+10x, the HCF of 6 and 10 is 2, and x is in both terms, so the HCF is 2x. That gives 6x^2+10x=2x(3x+5)

Taking out only the number, or only the letter, is not fully factorised. 2(3x^2+5x) and x(6x+10) are both correct, but each bracket still has a common factor.

Worked example 4: A number and a letter outside

Factorise fully 14x^2+21x

  1. The HCF of 14 and 21 is 7. The letter x is in both terms, and its lowest power is x. So the HCF is 7x
  2. Divide the first term by 7x: 14x^2\div7x=2x

  3. Your turn. Divide 21x by 7x as well. Then write 14x^2+21x fully factorised.

    Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

    21x\div7x=3, so 14x^2+21x=7x(2x+3)

  4. Check: 7x\times2x=14x^2 and 7x\times3=21x. Inside the bracket, 2x and 3 have no common factor.
  5. Answer: 7x(2x+3)

Worked example 5: Two letters and powers

Factorise fully 9x^3y+6x^2y^2

  1. Numbers: the HCF of 9 and 6 is 3.
  2. Letters: the lowest power of x is x^2, and the lowest power of y is y. So the HCF is 3x^2y
  3. Divide the first term: 9x^3y\div3x^2y=3x

  4. Your turn. Now divide the second term. Work out 6x^2y^2\div3x^2y

    Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

    6x^2y^2\div3x^2y=2y

  5. Check: 3x^2y\times3x=9x^3y and 3x^2y\times2y=6x^2y^2
  6. Answer: 3x^2y(3x+2y)

When a term is the whole factor

When a term is exactly the common factor, dividing it leaves 1. Write the 1 in the bracket: 5x^2+5x=5x(x+1). Without the 1, the bracket would multiply back to 5x^2 only.

The 1 can come first. In 3-12y, the HCF is 3, and 3\div3=1, so 3-12y=3(1-4y)

If there are like terms, collect them first. 7x^2+4x-x^2 is 6x^2+4x, which factorises fully to 2x(3x+2)

Using factorising to show a multiple

Factorising can show that an expression is always a multiple of a number. When a and b are whole numbers, 8a+12b=4(2a+3b). The bracket 2a+3b is a whole number, so 8a+12b is 4 times a whole number. That makes it a multiple of 4, whatever a and b are.

Trying one value, such as a=1 and b=2, gives 8+24=32, which is a multiple of 4. That is a good check, but it is not a reason: it shows only one case. The factorised form is the reason, because it works for every value.

Worked example 6: Showing an expression is a multiple

a and b are whole numbers. Sam says, "14a+21b is always a multiple of 7." Is Sam correct? Give a reason for your answer.

  1. To show a multiple of 7, write 14a+21b as 7 times something. Factorise with 7 outside the bracket.
  2. 7 goes into 14 and into 21. Divide the first term by 7: 14a\div7=2a

  3. Your turn. Divide 21b by 7 as well. Then write 14a+21b factorised, with 7 outside the bracket.

    Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

    21b\div7=3b, so 14a+21b=7(2a+3b)

  4. When a and b are whole numbers, 2a+3b is a whole number. So 14a+21b is 7 times a whole number.
  5. Answer: Yes, Sam is correct, because 14a+21b=7(2a+3b), which is 7 times a whole number.

Common mistakes

  • Adding unlike terms together. 4x+6 is not 10x; factorising it gives 2(2x+3)
  • Not factorising fully. Look inside the bracket: 3x(2x+4) still has a common factor of 2, so the full answer is 6x(x+2)
  • Using two brackets because there is a squared term. A common factor needs only one bracket: x^2+7x=x(x+7)
  • Leaving a decimal in the bracket. Take out a factor that divides exactly into every term: 8x-20 is 4(2x-5), not 8(x-2.5)
  • Leaving the letter in the bracket as well. Divide every term by the factor you take out: y^2-3y is y(y-3), not y(y^2-3y)
  • Not checking by expanding. Multiply your answer out: it must give the expression you started with.

Practice questions

More on this topic: Factorising into Single Brackets worksheet with full solutions