Factorising quadratics
In a quadratic expression, the highest power of \(x\) is \(x^2\).
A quadratic expressionAn expression where the maximum power of any variable is 2. For example, 2x2 is a quadratic expression as the power of x is 2. can sometimes be factorised into two brackets in the form of \((x + a)(x + b)\) where \(a\) and \(b\) can be any term, positive, negative or zero. \(a\) and \(b\) can be found by using a productTo multiply. The product of two numbers is the answer to the multiplication of the numbers. The product of 5 and 8 is 40. and sumTo add. The sum of 8 and 4 is 12 as 8 + 4 = 12. method.
Expanding the brackets \((x + 2)(x + 3)\) gives \(x^2 + 3x + 2x + 6\)
Which simplifies to: \(x^2 + 5x + 6\)
Factorising is the reverse process of expanding brackets, so factorising \(x^2 + 5x + 6\) gives \((x + 2)(x + 3)\).
Example 1
Factorise \(x^2 + 7x + 10\).
To factorise this expression, find two numbers that have a product of +10 and a sum of +7.
There are a couple of ways of making +10 by multiplying two numbers. These are \(1 \times 10\) and \(2 \times 5\). Only the combination of 2 and 5 will also give a sum of +7, so the two numbers are 2 and 5.
\(x^2 + 7x + 10 = (x + 2)(x + 5)\)
To check this answer is correct, expand out the brackets.
\((x + 2)(x + 5) = x \times x + x \times 5 + 2 \times x + 2 \times 5\)
\(= x^2 + 5x + 2x + 10 = x^2 + 7x + 10\)
This is the original expression, so \((x + 2)(x + 5)\) is the correct factorisation.
Example 2
Factorise \(y^2 - 12y + 11\).
As there is a negative sign in the quadratic, it is helpful to remember the sign rules, ie two same signs make a positive and two different signs make a negative.
To factorise this quadratic, find two numbers that have a product of +11 and a sum of -12.
11 is a primeA number that only has two factors - itself and one. number, so the only way of multiplying two numbers to make 11 is \(11 \times 1\).
\(11 + 1\) will add up to +12, not -12, so the numbers must be -11 and -1 as \(-11 \times -1 = +11\).
\(= (y - 11)(y - 1)\)
To check this answer is correct, expand out the brackets:
\((y - 11)(y - 1) = (y \times y) + (y \times -1) + (-11 \times y) + (-11 \times -1)\)
\(= y^2 - y - 11y + 11\) which simplifies to \(y^2 - 12y + 11\).
This is the original expression, so \((y - 11)(y - 1)\) is the correct factorisation.
The difference of two squares
The differenceThe answer to the subtraction of two numbers. For example, the difference between 8 and 2 is 6 as 8 - 2 = 6. of two numbers is found by subtracting. The difference of two squares means one squared term subtract another squared term. For example, \(x^2 - 9\) would be the difference of two squares as \(x^2\) is a squared term (\(x\) has been multiplied by itself) and 9 is a square number (\(3 \times 3\)).
The difference of two squares can be factorised into brackets using the method above for factorising quadratics.
Example
Factorise \(x^2 - 4\).
Quadratics can be factorised into the form \((x + a)(x + b)\).
\(x^2 - 4\) can be written as \(x^2 + 0x -4\).
To factorise this quadratic, find two numbers that have a product of -4 and a sum of 0.
The factor pairsPairs of factors multiplied together to get a certain product. that make -4 are either \(-1 \times 4\), \(1 \times -4\) or \(-2 \times 2\). The factor pair that has a product of -4 but a sum of 0 is \(-2 \times 2\) because \(-2 \times 2 = -4\) and \(-2 + 2 = 0\).
Therefore, the two numbers to go into the brackets are -2 and 2.
This gives \((x - 2)(x + 2)\).
To check this answer, expand the brackets to see if the answer gives the original expression. If it does, then this is a correct factorisation.
\((x - 2)(x + 2) = (x \times x) + (x \times 2) + (-2 \times x) + (-2 \times 2)\)
\(= x^2 + 2x - 2x - 4\). Collecting the like terms gives the original expression of \(x^2 - 4\).
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