Question

Find all the integers,b, that make up the trinomial x2 + bx + 10 factorable.

Answer 1

By the fundamental theorem of algebra, we can write

$x^2+bx+10=(x-r_1)(x-r_2)$

Expanding the right hand side, we get

$x^2+bx+10=x^2-(r_1+r_2)x+r_1r_2\implies\begin{cases}r_1+r_2=-b\\r_1r_2=10\end{cases}$

We want $b$ to be an integer, which means $r_1,r_2$ must also be integers. This means $r_1,r_2$ must be factors of 10. There are several possibilities:

$r_1=\pm10,r_2=\pm1\implies b=-(10+1)=-11\text{ or }b=-(-10-1)=11$

$r_1=\pm5,r_2=\pm2\implies b=-(5+2)=-7\text{ or }b=-(-5-2)=7$

So there are 4 possible values for $b$: -11, -7, 7, and 11.