Hi Ashley,

You didn't give any instructions but I think you are to factor $n^2 + 3n - 270.$ If it is possible you should be able to find two positive integers $a$ and $b$ so that

\[n^2 + 3n - 270 = (x + a)(x - b).\]

Expanding $(x + a)(x - b)$ you get $x^2 +(a - b)x - ab$ and hence $ab = 270$ and $a - b = 3.$

To find $a$ and $b$ I would write 270 in terms of its prime factors that is $270 = 2 \times 3^3 \times 5.$ For example you might write $270 = (2 \times 5) \times (3^3) = 10 \times 27$ but the difference between 27 and 10 is 17 not 3. You might try $ 270 = (2 \times 3) \times (3^2 \times 5) = 6 \times 45$ which again doesn't work.

Can you see how to write $270 = a \times b$ where $a - b = 3?$

Penny