Home Stage 5 Factorising

FACTORISING PRACTICE

Common Factor | Difference of two Squares | Quadratic Trinomials | Grouping Method

[1] Common Factor

The method of common factors is doing the opposite of expanding brackets. When you expand brackets, you multiply everything inside the brackets by the factor outside the bracket:

\[ 3 (2x + 5 ) = 3 \times 2x + 3 \times 5 = 6x + 15\]

When you use common factors, you are taking out the biggest number or letter that is common in all the terms.

\[ 6x + 15 = 3 \times 2x + 3 \times 5 = 3(2x + 5) \]

Example 1: Factorise \( 12ab + 18a\)

Step 1: Find the highest common factor (HCF) of \( 12ab\) and \(18a\): \( 6a \)

Step 2: Write each term as a product using the HCF: \( 12ab + 18a = 6a \times 2b + 6a \times 3\)

Step 3: Write HCF outside the bracket: \( 6a \times 2b + 6a \times 3 = 6a( \hspace{4em}) \)

Step 4: Write the other factors from Step 2 inside the brackets: \( 6a \times 2b + 6a \times 3 = 6a(2b+3) \)

\[12ab + 18a = 6a(2b+3) \]

Example 2: Factorise \( 4ab - 20a^{2}bc\)

Step 1: Find the highest common factor (HCF) of \( 4ab\) and \(20a^{2}bc\): \( 4ab \)

Step 2: Write each term as a product using the HCF: \( 4ab - 20a^{2}bc = 4ab \times 1 - 4ab \times 5ac\)

Step 3: Write HCF outside the bracket: \( 4ab \times 1 - 4ab \times 5ac = 4ab( \hspace{4em}) \)

Step 4: Write the other factors from Step 2 inside the brackets: \( 4ab \times 1 - 4ab \times 5ac = 4ab(1-5c) \)

\[ 4ab - 20a^{2}bc = 4ab(1-5c) \]

The common factor may be a binomial term:

Example 3: Factorise \( a(x+y) + b(x+y)\)

Step 1: HCF: \( (x+y) \)

Step 2: \( a(x+y) + b(x+y) = (x+y) \times a + (x+y) \times b\)

Step 3: \( (x+y) \times a + (x+y) \times b\ = (x+y)( \hspace{4em}) \)

Step 4: \( (x+y) \times a + (x+y) \times b\ = (x+y)(a+b) \)

\[ a(x+y) + b(x+y) = (x+y)(a+b) \]

Practice

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[2] Difference of 2 Squares

The difference of two squares is a special type of factorisation. It happens when you have two squared terms being subtracted. The rule is:

\[ a^{2} - b^{2} = (a - b)(a +b) \]

Example 4: Factorise \( x^{2} - 16 \)

Step 1: Recognise that \( x^{2} \) and 16 are square terms, i.e., \( x^{2} = (x)^{2} \) and \( 16 = (4)^{2} \)

Step 2: Use the rule: \( x^{2} - 16 = (x)^{2} - (4)^{2} = (x - 4)(x + 4) \)

Example 5: Factorise \( 9x^{2} - 25y^{2} \)

Step 1: Recognise that \( 9x^{2} \) and \( 25y^{2} \) are square terms, i.e., \( 9x^{2} = (3x)^{2} \) and \( 25y^{2} = (5y)^{2} \)

Step 2: Use the rule: \( 9x^{2} - 25y^{2} = (3x)^{2} - (5y)^{2} = (3x + 5y)(3x - 5y) \)

Practice

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[3] Quadratic Trinomials (Monic)

Monic quadratic trinomials like \( x^{2} + bx + c \) can be factorised by finding factors of the constant, c, term which add (or subtract) to the (unsigned) coefficient of the middle term, b.

Example 6: Factorise \( {x}^{2} + 12x + 20 \)

Step 1: Find factors of the constant term (20): \( 1 \times 20, 2 \times 10, 4 \times 5 \)

Step 2: which add (c > 0) to 12 (unsigned coeffiecent b): \( 10 + 2 = 12 \)

Step 3: The larger factor has the same sign as the middle term b: \( (x + 10)(x \hspace{1em} \hspace{1em} ) \)

Step 4: Factors have same sign as c is positive: \( (x + 10)(x + 2) \)

Example 7: Factorise \( {x}^{2} - 5x - 24 \)

Step 1: Find factors of the constant term (24): \( 1 \times 24, 2 \times 12, 3 \times 8, 4 \times 6 \)

Step 2: which subtract (c < 0) to 5 (unsigned coeffiecent b): \( 8 - 3 = 5 \)

Step 3: The larger factor has the same sign as the middle term b: \( (x - 8)(x \hspace{1em} \hspace{1em} ) \)

Step 4: Factors have different signs as c is negative: \( (x - 8)(x + 3) \)

Practice

Non-monic quadratic trinomials like \( ax^{2} + bx + c \) can be factorised in a way similar to how monic quadatic trinomials are factorised.

Example 7: Factorise \( {x}^{2} + 12x + 20 \)

Step 1: Find factors of the constant term (20): \( 1 \times 20, 2 \times 10, 4 \times 5 \)

Step 2: which add (c > 0) to 12 (unsigned coeffiecent b): \( 10 + 2 = 12 \)

Step 3: The larger factor has the same sign as the middle term b: \( (x + 10)(x \hspace{1em} \hspace{1em} ) \)

Step 4: Factors have same sign as c is positive: \( (x + 10)(x + 2) \)

Example 8: Factorise \( {x}^{2} - 5x - 24 \)

Step 1: Find factors of the constant term (24): \( 1 \times 24, 2 \times 12, 3 \times 8, 4 \times 6 \)

Step 2: which subtract (c < 0) to 5 (unsigned coeffiecent b): \( 8 - 3 = 5 \)

Step 3: The larger factor has the same sign as the middle term b: \( (x - 8)(x \hspace{1em} \hspace{1em} ) \)

Step 4: Factors have different signs as c is negative: \( (x - 8)(x + 3) \)

Practice

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[4] Grouping Method

The Grouping Method is actually just two applications of the Common Factor Method.

Example 3: Simplify: \( \frac{x+3}{x^{2} - 4} \div \frac{x^{2} + 7x + 12}{3x - 6} \)

Step 2: Factorise the fractions: \( \frac{x+3}{x^{2} - 4} \times \frac{3x - 6}{x^{2} + 7x + 12} = \frac{(x + 3)}{(x-2)(x+2)} \times \frac{3(x-2)}{(x + 3)(x + 4)} \)

Step 3: Cancel common factors: \( = \frac{ \cancel{(x + 3)}}{\cancel{(x-2)}(x+2)} \times \frac{3\cancel{(x-2)}}{\cancel{(x + 3)}(x + 4)} = \frac{1}{(x+2)} \times \frac{3}{(x + 4)} \)

Step 4: Multiply numerators & multiply denominators: \( \frac{1 \times 3}{(x+2) \times (x+4)} = \frac{3}{(x+2)(x + 4)} \)

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"The only way to learn mathematics is to do mathematics" - Paul R Halmos