Question

Factor trinomials using trials and error 81a^2 + 153a - 18

Answer 1

We need to facto the trinomial:

$81a^2+153a-18$

Notice that the first term can be factored as:

$(9a)(9a)$

Thus, one possible way of factoring the trinomial is:

$(9a+?)(9a+??)$

When we multiply the terms represented by "?" and "??", we need to obtain -18. Thus, we need to factor -18 and try each pair of possible factors.

We have:

$\text{factors of -18: }\pm1,\pm2,\pm3,\pm6,\pm18$

Notice that:

$\begin{gathered} (9a+?)(9a+??)=81a^2+9a(?)+9a(??)+(?)(??) \\ \\ (9a+?)(9a+??)=81a^2+9a(?+??)+(?)(??) \end{gathered}$

Then, the pair of factors of -18 must satisfy:

$\begin{gathered} (?)(??)=-18 \\ \\ 9a(?+??)=153a \\ 9(?+??)=153 \\ (?+??)=\frac{153}{9} \\ (?+??)=17 \end{gathered}$

One possible pair of factors of -18, whose sum is 17, is: -1 and 18.

Using those numbers, we obtain:

$\begin{gathered} (9a+?)(9a+??) \\ \\ (9a-1)(9a+18) \\ \\ 81a^2+9\cdot18a-9a-18 \\ \\ 81a^2+162a-9a-18 \\ \\ 81a^2+153a-18 \end{gathered}$

Then, we see that:

$(9a-1)(9a+18)$

is a way of factoring the given trinomial. Notice, though, that terms of the second factor can both be divided by 9. Then, we can write the expression as:

$9a+18=9(a+2)$

Then, the trinomial can be factored as:

$9(9a-1)(a+2)$