In the following exercises, factor each trinomial of the form $x^2+bx+c$.

$n^2+19n+48$.

We have $n^2+19n+48$.

We will first find any common factor. if present, in the equation. Then we will any two numbers whose product comes out to be the product of the first and last term of the equation and the sum comes out to be the middle term.

Then by splitting the middle term, we will factor out the common terms and will get the factors of the trinomial.

We can see that there is no common factor in the equation. So we will proceed finding the numbers whose product comes out to be $48$and sum results in $19$. The factors of$48$are-

Through this we can say that two numbers are$3,16$.

Two numbers which we have are $3,16$.

Splitting the middle terms gives us,

$$\begin{gathered} =n^2+19n+48 \\ =n^2+3n+16n+48 \\ =n(n+3)+16(n+3) \\ =(n+3)(n+16) \end{gathered}$$

This shows that factors are$(n+3)(n+16)$.

We will check the solution by multiplying. If we get the same given equation, our calculations are right.

So,

$$\begin{gathered} =(n+3)(n+16) \\ =n^2+16n+3n+48 \\ =n^2+19n+48 \end{gathered}$$