Rich example

## Warm-up solution

We begin by expanding the brackets using the grid method.

1. $(x+5)(x+2)$
$x^2$ $+5x$
$+2x$ $+10$

Therefore $(x+5)(x+2)=x^2+7x+10$.

1. $(x-1)(2x+3)$
$2x^2$ $-2x$
$+3x$ $-3$

Therefore $(x-1)(2x+3)=2x^2+x-3$.

1. $(2x-3)(3x-2)$
$6x^2$ $-9x$
$-4x$ $+6$

Therefore $(2x-3)(3x-2)=6x^2-13x+6$.

1. $(px+r)(qx+s)$
$pqx^2$ $+rqx$
$+psx$ $+rs$

Therefore $(px+r)(qx+s)=pqx^2+rqx+psx+rs$.

#### There seem to be some common factors in some of the rows and columns

Did you notice that the heading for each row and column is a highest common factor of that row and column? For example, in question 3, $2x$ is a highest common factor (hcf) of $6x^2$ and $-4x$, $-3$ is an hcf of $-9x$ and $+6$, $3x$ is an hcf of $6x^2$ and $-9x$ and $-2$ is an hcf of $-4x$ and $+6$.

#### There seems to be some relationship between the two diagonals in each grid

Did you notice the products of the terms on the diagonals are equal to each other? For example, in question 1, one diagonal has $x^2$ and $+10$, the product of these terms is $10x^2$. The other diagonal has $+5x$ and $+2x$, the product of these terms is also $10x^2$.

The generalised form in question 4 can help explain this pattern:

• The $x^2$ and constant terms multiply to $pqx^2\times rs=pqrsx^2$.

• The other two cells (the $x$ terms) multiply to $psx\times qrx=pqrsx^2$.

So the products of the terms on the diagonals are the same.

This will help us when factorising quadratics.