Question

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

$u^2+101u+100$

Short answer

The factors are$(u+1)(u+100)$.

Step-by-step solution

Step 1. Writing the expression

Write the factors as two binomials with first terms $u$.

localid="1644544034773" $(u)(u)$.

Step 2. Finding the factors

We need to find two numbers whose product is $100$and sum is $101$.

Since $101$is a prime number, we know that $1$and $100$are the only numbers whose product is $100$and sum is $101$.

Hence, the factorization of the trinomial is$(u+1)(u+100)$.

Step 3. Check

Multiply the factorization to get the trinomial again.

$$\begin{gathered} (u+1)(u+100)=u^2+100u+u+100 \\ (u+1)(u+100)=u^2+101u+100 \end{gathered}$$

Thus, the obtained factorization is correct.