iii)

Method 1: Splitting the Middle Term

We need a pair of numbers that multiply to $a c$ which in this case is $10 \times 9 = 90$ , and add to $b$ , which in this case is -23

-18 and -5 satisfy these conditions

Rewrite the middle term using -18 and -5

$10 r^{2} - 23 r + 9 = 0$
$10 r^{2} - 5 r - 18 r + 9 = 0$

Factorise the first two terms, using $5 r$ as the highest common factor, and the second two terms using -9 as the factor

$5 r (2 r - 1) - 9 (2 r - 1) = 0$

Note that these now have the same factor of $(2 r - 1)$ so this can be used as a common factor

$(2 r - 1) (5 r - 9) = 0$

[2]

Once you get to this spot it may be worth expanding the brackets to check your answer

When two numbers are multiplied and the result is zero, at least one of them must be zero, so we can say that

$2 r - 1 = 0 or 5 r - 9 = 0$

Solving both of these equations by; adding 1 and dividing by 2, or adding 9 and dividing by 5 respectively;

$r = \frac{1}{2} or r = \frac{9}{5}$ [1]

iii)

Method 2: Factorising with a grid

We need a pair of numbers that multiply to $a c$ which in this case is $10 \times 9 = 90$ , and add to $b$ , which in this case is -23

-18 and -5 satisfy these conditions

Write the quadratic equation in a grid (as if you had used a grid to expand the brackets), splitting the middle term as $- 18 r$ and $- 5 r$

Write a heading for the first column, using $5 r$ as the highest common factor

You can then use this to find the headings for the rows, e.g. "What does $5 r$ need to be multiplied by to give $10 r^{2}$ ?"

We can then fill in the remaining column heading using the same idea; e.g. "What does $2 r$ need to be multiplied by to give $- 18 r$ ?"

We can now read-off the factors from the column and row headings