Q. \( \dfrac{x^3 + 7x^2 + 10x}{x^2 + 2x} \).

Problem

Simplify the rational expression
\( \displaystyle \frac{x^3+7x^2+10x}{x^2+2x} \).

Step 1 — Factor out the greatest common factors

Look for a common factor in each polynomial.

Numerator: each term \(x^3\), \(7x^2\), \(10x\) has a factor \(x\). Factor it out:
\[ x^3+7x^2+10x = x\bigl(x^2+7x+10\bigr). \]

Denominator: each term \(x^2\), \(2x\) has a factor \(x\). Factor it out:
\[ x^2+2x = x\bigl(x+2\bigr). \]

Step 2 — Factor the quadratic \(x^2+7x+10\)

We need two numbers whose product is \(10\) (the constant term) and whose sum is \(7\) (the coefficient of \(x\)). The numbers \(5\) and \(2\) work because \(5\cdot 2 = 10\) and \(5+2 = 7\).

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

Step 3 — Rewrite the whole expression using the factorizations

Substitute the factored forms found above:
\[ \frac{x^3+7x^2+10x}{x^2+2x} = \frac{x\bigl(x^2+7x+10\bigr)}{x\bigl(x+2\bigr)} = \frac{x(x+5)(x+2)}{x(x+2)}. \]

Step 4 — Cancel common (nonzero) factors

Cancel the common factors \(x\) and \(x+2\) from numerator and denominator. Cancellation is valid only when those factors are not zero, so we must note the excluded values first.

After cancellation we get
\[ \frac{x(x+5)(x+2)}{x(x+2)} = x+5, \]
provided the cancelled factors are nonzero.

Step 5 — State the domain restriction and the final simplified form

The original expression is undefined when the denominator equals zero. From the factored denominator \(x(x+2)\), the values to exclude are \(x=0\) and \(x=-2\).

Therefore the simplified form is
\[ \boxed{x+5 \quad \text{for } x \neq 0,\ -2.} \]