4B - Factorising quadratics
4B - Content video: An overview of factorisation
This video provides an overview of factorisation.
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An overview of factorisation:
Factorising is the reverse process of expanding. When you are factorising you are removing the highest common factor (HCF). The highest common factor can be:
Factorising is the reverse process of expanding. When you are factorising you are removing the highest common factor (HCF). The highest common factor can be:
- A constant term, \(k\), such as the number \(5\) in \(5x+15=5(x+3)\)
- A linear term, \(ax+b\), such as \(x-1\) in \(x^2+x-2=(x-1)(x+2)\)
- A quadratic term (or higher), such as \(x^2\) in \(2x^3-x^2=x^2(2x-1)\)
- Factorisation by removing the highest common factor.
- Factorisation by grouping.
- Factorisation using the difference of perfect squares.
- Factorisation of quadratics by inspection.
- Factorisation by completing the square.
Deciding which method to use:
If there are two terms present:
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If there are three terms present:
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If there are four terms present:
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Methods of factorisation
Case 1: Factorising using common factors
It is always a good idea to inspect each term for a common factor such as a constant, \(k\), or a simple linear term, \(ax\). If all terms have the same common factor it can be removed and the expression factorised. |
■ Remember, you can always check if the factorised form is correct by expanding. |
4B - Example 1: Factorising by removing a common factor
Factorise the following expressions:
(a) \(12x+4\) (b) \(15x+5xy\) (c) \(6x+15xy+9xz\) |
4B - Example 1: Video solution
4B - Example 1: Practice
Question 1: Factorise each of the following expressions (a) \(6x+24\) (b) \(5x-3x^2\) (c) \(15xy^2+10xy\) 4B - Example 1: Solutions
Question 1: Factorise each of the following expressions (a) \(6(x+4)\) (b) \(x(5-3x)\) (c) \(5xy(3y+2)\) |
Case 2: Factorising using perfect squares
To factorise using perfect squares you must be able to recognise expressions where:
To factorise using perfect squares you must be able to recognise expressions where:
- The first and last terms appear to be squared (such as: \(x^2\), \(81\), \(16x^2\) etc.)
- The middle term is twice the value of the square root of the two outside terms.
4B - Example 2: Factorising using perfect squares
Factorise the following expression: \[x^2-6x+9\]
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4B - Example 2: Video solution
4B - Example 2: Practice
Question 1: Factorise \(x^2+14x+49\) Question 2: Factorise \(x^2-10x+25\) 4B - Example 2: Solutions
Question 1: \(x^2+14x+49=(x+7)^2\) Question 2: \(x^2-10x+25=(x-5)^2\) |
4B - Example 3: Factorising using perfect squares
Factorise the following expression: \[4x^2+12x+9\]
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4B - Example 3: Video solution
4B - Example 3: Practice
Question 1: Factorise \(2x^2+12x+18\) Question 2: Factorise \(4x^2+20x+25\) 4B - Example 3: Solutions
Question 1: Factorise \(2x^2+12x+18=2(x+3)^2\) Question 2: Factorise \(4x^2+20x+25=(2x+5)^2\) |
Case 3: Factorising using the difference of perfect squares (DOPS)
To factorise using the difference of perfect squares (DOPS) you must be able to recognise expressions where:
To factorise using the difference of perfect squares (DOPS) you must be able to recognise expressions where:
- The two terms appear to be squared (such as: \(x^2\), \(81\), \(16x^2\) etc.) with one subtracted from the other.
- No ‘middle term’ can be present!
4B - Example 4: Factorising using the difference of perfect squares (DOPS)
Factorise the following expression: \[x^2-25\]
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4B - Example 4: Video solution
4B - Example 4: Practice
Question 1: Factorise \(x^2-81\) Question 2: Factorise \(144-x^2\) 4B - Example 4: Solutions
Question 1: \(x^2-81=(x-9)(x+9)\) Question 2: \(144-x^2=(12-x)(12+x)\) |
4B - Example 5: Factorising using the difference of perfect squares (DOPS)
Factorise the following expression: \[x^2-225z^2\]
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4B - Example 5: Video solution
4B - Example 5: Practice
Question 1: Factorise \(36x^2-49\) Question 2: Factorise \(100-12x^2\) Question 3: Factorise \(6-3x^2\) 4B - Example 5: Solutions
Question 1: \(36x^2-49=(6x-7)(6x+7)\) Question 2: \(100-12x^2=4(25-4x^2)=4(5-2x)(5+2x)\) Question 3: Factorise \(6-3x^2=3(2-x^2)=3(\sqrt{2}+x)(\sqrt{2}-x)\) |
Case 4: Factorising quadratic trinomials by inspection
A quadratic trinomial has the general form: \(ax^2+bx+c\)
A quadratic trinomial has the general form: \(ax^2+bx+c\)
- If \(a=1\) it is considered a monic trinomial. To factorise we simply look for factors of \(c\) that add to give \(b\). We will call this process "factorisation by inspection - easy".
- If \(a\neq1\) it is considered a non-monic trinomial. To factorise we look for factors of \(a \times c\) that add to give \(b\). We will call this process "factorisation by inspection - hard".
4B - Example 6: Factorisation by inspection "easy"
Factorise the following expression: \[x^2-x-42\]
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4B - Example 6: Video solution
4B - Example 6: Practice
Question 1: Factorise \(x^2+8x+12\) Question 2: Expand \(x^2+5x-24\) 4B - Example 6: Solutions
Question 1: Factorise \(x^2+8x+12=(x+2)(x+6)\) Question 2: Expand \(x^2+5x-24=(x+8)(x-3)\) |
4B - Example 7: Factorisation by inspection "easy"
Factorise the following expression: \[x^2-13x+36\]
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4B - Example 7: Video solution
4B - Example 7: Practice
Question 1: Factorise \(x^2+6x-27\) Question 2: Factorise \(x^2-11x+28\) 4B - Example 7: Solutions
Question 1: \(x^2+6x-27=(x+9)(x-3)\) Question 2: \(x^2-11x+28=(x-7)(x-4)\) |
4B - Example 8: Factorisation by inspection "easy"
Factorise the following expression: \[2x^2+24x+70\]
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4B - Example 8: Video solution
4B - Example 8: Practice
Question 1: Factorise \(3x^2+27x+60\) Question 2: Factorise \(-5x^2-20x-15\) 4B - Example 8: Solutions
Question 1: \(3x^2+27x+60=3(x+4)(x+5)\) Question 2: \(-5x^2-20x-15=-5(x+1)(x+3)\) |
4B - Example 9: Factorisation by inspection "hard"
Factorise the following expression: \[2x^2+3x-20\]
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4B - Example 9: Video solution
4B - Example 9: Practice
Question 1: Factorise \(2x^2+x-6\) Question 2: Factorise \(-4x^2-7x+15\) 4B - Example 9: Solutions
Question 1: \(2x^2+x-6=(x+2)(2x-3)\) Question 2: \(-4x^2-7x+15=-(x+3)(4x-5)\) |
Case 5: Factorising by grouping
When four or more terms are present, we need to factorise by grouping terms together. Which terms to group is done by inspection or trial and error.
When four or more terms are present, we need to factorise by grouping terms together. Which terms to group is done by inspection or trial and error.
- You are aiming to remove a common term from each group and leave the same factor behind.
- Often when grouping we need to group so that other methods of factorisation can be used, such as: difference of perfect squares (DOPS), perfect squares and inspection.
4B - Example 10: Factorising by grouping
Factorise the following expression (by grouping 2 and 2):
\[x^2-7x-y^2+7y\] |
4B - Example 10: Video solution
4B - Example 10: Practice
Question 1: ABC Question 2: ABC 4B - Example 10: Solutions
Question 1: ABC Question 2: ABC |
4B - Example 11: Factorising by grouping
Factorise the following expression (by grouping 2 and 2):
\[64-8g-16h+2gh\] |
4B - Example 11: Video solution
4B - Example 11: Practice
Question 1: ABC Question 2: ABC 4B - Example 11: Solutions
Question 1: ABC Question 2: ABC |
4B - Example 12: Factorising by grouping
Factorise the following expression (by grouping 3 and 1):
\[4x^2+12xy+9y^2-16z^2\] |
4B - Example 12: Video solution
4B - Example 12: Practice
Question 1: ABC Question 2: ABC 4B - Example 12: Solutions
Question 1: ABC Question 2: ABC |
Case 5: Factorising by grouping
CAS calculators can readily factorise polynomials using the following commands:
CAS calculators can readily factorise polynomials using the following commands:
- Factor - use when the factors are rational.
- rFactor - used when the factors are irrational.
4B - Example 13: Factorising using a CAS calculator
Factorise each of the following expressions using CAS:
(a) \(-6x^2+25x+91\) (b) \(x^2-3x+1\) |
4B - Example 13: Video solution
4B - Example 13: Practice
Question 1: Factorise \(-15x^2-4x+4\) using a CAS calculator. Question 2: Factorise \(x^2-6x+7\) using a CAS calculator. 4B - Example 13: Solutions
Question 1: \(-15x^2-4x+4=-(5x-2)(3x+2)\) Question 2: \(x^2-6x+7=(x-3-\sqrt{2})(x-3+\sqrt{2})\) |