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.