We have looked at factorising quadratic where the coeffect of the \( x^2 \) term is 1. If we have \( 5x^2 \), we will need to follow a slightly different method to factorise.
Ensure the quadratic is written as \( ax^2 + bx + c \).
Recognise the values of \( a \) (the coefficient of \( x^2 \)), \( b \) (the coefficient of \( x \)), and \( c \) (the constant term).
Multiply \( a \) and \( c \). Find two numbers that multiply to this \( ac \) value and add up to \( b \). These two numbers will help split the middle term.
Replace \( bx \) with two terms using the numbers found in the previous step. This splits the quadratic into four terms. For example, if the quadratic is \( 6x^2 + 7x + 2 \), find two numbers that multiply to \( 6 \times 2 = 12 \) and add to \( 7 \), which are \( 3 \) and \( 4 \). Rewrite \( 6x^2 + 7x + 2 \) as \( 6x^2 + 3x + 4x + 2 \).
Group the terms into two pairs: \( (6x^2 + 3x) + (4x + 2) \). Factor out the common factor from each pair: \( 3x(2x + 1) + 2(2x + 1) \).
If both groups have a common factor (a binomial), factor it out. For the example, factor out \( (2x + 1) \): \( (2x + 1)(3x + 2) \).
Expand the factors to confirm they multiply to the original quadratic.
Factorise \( 6x^2 + 7x + 2 \).
\( 6x^2 + 7x + 2 \).
The values are \( a = 6 \), \( b = 7 \), and \( c = 2 \).
The numbers \( 3 \) and \( 4 \) work.
Rewrite \( 6x^2 + 7x + 2 \) as \( 6x^2 + 3x + 4x + 2 \).
\( (6x^2 + 3x) + (4x + 2) \).
Factor out \( 3x(2x + 1) + 2(2x + 1) \).
\( (2x + 1)(3x + 2) \).
\( (2x + 1)(3x + 2) \).
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