Q. \[ x^2+7x+10 \]

We are going to work with the expression \(x^2+7x+10\). A common first step is to determine whether it can be factored into two binomials.

Step 1: Identify the coefficients.

Write the expression in the standard quadratic form:

\[
x^2+7x+10
\]

So we have \(a=1\), \(b=7\), and \(c=10\).

Step 2: Find two numbers that multiply to the constant term.

When factoring a quadratic of the form \(x^2+bx+c\) with leading coefficient \(1\), we look for two numbers \(m\) and \(n\) such that:

• \(m+n=b\)
• \(mn=c\)

Here, we need:

• \(m+n=7\)
• \(mn=10\)

Step 3: Choose the correct pair.

List factor pairs of \(10\):

• \(1 \cdot 10\) gives sum \(1+10=11\)
• \(2 \cdot 5\) gives sum \(2+5=7\)

The pair \(2\) and \(5\) works because \(2+5=7\) and \(2\cdot 5=10\).

Step 4: Factor the quadratic.

Now split the middle term \(7x\) into \(2x\) and \(5x\):

\[
x^2+7x+10=x^2+2x+5x+10
\]

Group terms:

\[
x^2+2x+5x+10=(x^2+2x)+(5x+10)
\]

Factor each group:

\[
(x^2+2x)=x(x+2)
\]
\[
(5x+10)=5(x+2)
\]

Combine the results:

\[
(x^2+2x)+(5x+10)=x(x+2)+5(x+2)
\]

Factor out the common factor \((x+2)\):

\[
x(x+2)+5(x+2)=(x+2)(x+5)
\]

Final Answer:

The expression \(x^2+7x+10\) factors as:

\[
x^2+7x+10=(x+2)(x+5)
\]