Factorising by Grouping in Pairs
Factorising by Grouping in Pairs
The taking out of common factors is the starting point in all factorisation problems. We know that the factors of \(3x+3\) are \(\text{3}\) and \((x+1)\). Similarly, the factors of \(2{x}^{2}+2x\) are \(2x\) and \((x+1)\). Therefore, if we have an expression:
\[2{x}^{2} + 2x + 3x + 3\]
there is no common factor to all four terms, but we can factorise as follows:
\[(2{x}^{2} + 2x) + (3x + 3) = 2x(x + 1) + 3(x + 1)\]
We can see that there is another common factor \((x+1)\). Therefore, we can write:
\[(x + 1)(2x + 3)\]
We get this by taking out the \((x+1)\) and seeing what is left over. We have \(2x\) from the first group and \(\text{+3}\) from the second group. This is called factorising by grouping.
Example
Question
Find the factors of \(7x + 14y + bx + 2by\).
There are no factors common to all terms
Group terms with common factors together
\(\text{7}\) is a common factor of the first two terms and \(b\) is a common factor of the second two terms. We see that the ratio of the coefficients \(7:14\) is the same as \(b:2b\).
\begin{align*} 7x + 14y + bx + 2by & = (7x + 14y) + (bx + 2by)\\ & = 7(x + 2y) + b(x + 2y) \end{align*}Take out the common factor \((x+2y)\)
\[7(x + 2y) + b(x + 2y) = (x + 2y)(7 + b)\]OR
Group terms with common factors together
\(x\) is a common factor of the first and third terms and \(2y\) is a common factor of the second and fourth terms \((7:b = 14:2b)\).
Rearrange the equation with grouped terms together
\begin{align*} 7x + 14y + bx + 2by & = (7x + bx) + (14y + 2by) \\ & = x(7 + b) + 2y(7 + b) \end{align*}Take out the common factor \((7+b)\)
\[x(7 + b) + 2y(7 + b) = (7 + b)(x + 2y)\]Write the final answer
The factors of \(7x + 14y + bx + 2by\) are \((7 + b)\) and \((x + 2y)\).
This lesson is part of:
Algebraic Expressions Overview