How can we use the multiplication grid method to help us factorise quadratics?
How can we factorise \(6x^2+7x+2\) using a multiplication grid?
The \(6x^2\) term is the product of the \(x\) terms, so goes in the top left cell.
The constant term \(+2\) is the product of constant terms, so goes in the bottom right cell.
So our grid starts out as
\(\ldots x\)
\(\ldots\)
\(\ldots x\)
\(6x^2\)
\(\ldots\)
\(+2\)
The missing two entries inside the grid must add up to \(+7x\).
From the warm-up we know the products of the terms on the two diagonals are equal.
The product of \(6x^2\) and \(+2\) gives \(+12x^2\).
Therefore we need to find a pair of \(x\) terms that has a product of \(+12x^2\) and a sum of \(+7x\).
Pairs of \(x\) terms with a product of \(+12x^2\) are \(+12x\) and \(+x\), \(-12x\) and \(-x\), \(+6x\) and \(+2x\), \(-6x\) and \(-2x\), \(+4x\) and \(+3x\), \(-4x\) and \(-3x\).
Only \(+4x\) and \(+3x\) sum to \(+7x\) so we can now fill in the grid completely.
\(\ldots x\)
\(\ldots\)
\(\ldots x\)
\(6x^2\)
\(+4x\)
\(\ldots\)
\(+3x\)
\(+2\)
Does it matter which way round the \(+4x\) and \(+3x\) go? Try both ways and see what happens.
We fill in the outside of the grid by looking for the highest common factor of the first column.
\(3x\)
\(\ldots\)
\(\dots x\)
\(6x^2\)
\(+4x\)
\(\ldots\)
\(+3x\)
\(+2\)
We then fill in the rest of the grid to make the multiplication work.
\(3x\)
\(+2\)
\(2x\)
\(6x^2\)
\(+4x\)
\(+1\)
\(+3x\)
\(+2\)
So what is the factorisation of \(6x^2+7x+2\)?
\[6x^2+7x+2 = (3x+2)(2x+1)\]
Factorising quadratics
Here are a few quadratic expressions you can use for practising this method.
\(4x^2+7x+3\)
\(3x^2+13x-10\)
\(6x^2-5x+1\)
\(4x^2-2x-12\)
\(2x^2-3xy-2y^2\)
Reflecting
We filled in the outside of the grid by finding the highest common factor of the first column. Did we need to use the first column or could we have used the first row? Could we have used another row or column? Can you explain your reasoning.
How does this method of factorising quadratics compare to other methods you know?
