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

$w^2+10x+21$.

Consider the given trinomial in a trinomial of a single variable.

Then by simply using the ac method where two numbers are found whose product is the product of the first and last term of the equation and whose sum is the middle term, we can find the factors.

Considering the equation as $w^2+10w+21$.

Now we will proceed finding the numbers whose product comes out to be $21$and sum results in $10$.So the factors of $21$are-

Through this we can say that the numbers are$3,7$.

Two numbers we have are $3,7$.

Splitting the middle term will give us

$$\begin{gathered} =w^2+10w+21 \\ =w^2+3w+7w+21 \\ =w(w+3)+7(w+3) \\ =(w+3)(w+7) \end{gathered}$$

This shows that the factors are$(w+3)(w+7)$.

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

So,

$$\begin{gathered} =(w+3)(w+7) \\ =w^2+7w+3w+21 \\ =w^2+10w+21 \end{gathered}$$