Question

practice: multiply two rational expressions
\\(\frac{x^2 + 5x + 6}{x + 4} \cdot \frac{x^2 + 4x}{x + 3}\\)
which of the following is equivalent to the expression above?
choose 1 answer:
a \\(\frac{x}{x + 2}\\)
b \\(\frac{x + 2}{x}\\)
c \\(x + 2\\)
d \\(x^2 + 2x\\)

Explanation:

Step1: Factor the quadratics

Factor \(x^{2}+5x + 6\) as \((x + 2)(x+3)\) and \(x^{2}+4x\) as \(x(x + 4)\). So the expression becomes \(\frac{(x + 2)(x + 3)}{x + 4}\cdot\frac{x(x + 4)}{x + 3}\).

Step2: Cancel common factors

Cancel out \((x + 3)\) and \((x + 4)\) from numerator and denominator. We get \((x + 2)\cdot x=x^{2}+2x\)? Wait, no, wait: Wait, \(\frac{(x + 2)(x + 3)}{x + 4}\times\frac{x(x + 4)}{x + 3}=\ (x + 2)\times x=x^{2}+2x\)? Wait, no, let's re - check. Wait, \(\frac{(x + 2)(x + 3)}{x + 4}\times\frac{x(x + 4)}{x + 3}\), the \((x + 3)\) in the numerator of the first fraction and denominator of the second fraction cancel, the \((x + 4)\) in the denominator of the first fraction and numerator of the second fraction cancel. So we have \((x + 2)\times x=x^{2}+2x\)? Wait, but let's check the options. Option D is \(x^{2}+2x\), but wait, let's do it again. Wait, \(\frac{(x + 2)(x + 3)}{x + 4}\times\frac{x(x + 4)}{x + 3}\). After canceling \((x + 3)\) and \((x + 4)\), we have \((x + 2)\times x=x(x + 2)=x^{2}+2x\). Wait, but let's check the options. Wait, maybe I made a mistake. Wait, no, let's re - factor:

Wait, \(x^{2}+5x + 6=(x + 2)(x + 3)\), \(x^{2}+4x=x(x + 4)\). So the product is \(\frac{(x + 2)(x + 3)}{x + 4}\times\frac{x(x + 4)}{x + 3}\). Cancel \((x + 3)\) and \((x + 4)\), we get \(x(x + 2)=x^{2}+2x\). So the equivalent expression is \(x^{2}+2x\), which is option D? Wait, but let's check the options again. Option D is \(x^{2}+2x\). Wait, but let's check with a value. Let's take \(x = 1\). Original expression: \(\frac{1 + 5+6}{1 + 4}\times\frac{1 + 4}{1+3}=\frac{12}{5}\times\frac{5}{4}=3\). Now check option D: \(1^{2}+2\times1=3\). Option C: \(1 + 2 = 3\). Wait, that's a problem. Wait, wait, I must have made a mistake. Wait, \(\frac{(x + 2)(x + 3)}{x + 4}\times\frac{x(x + 4)}{x + 3}\). Wait, no, \(\frac{(x + 2)(x + 3)}{x + 4}\times\frac{x(x + 4)}{x + 3}\). The \((x + 3)\) cancels, \((x + 4)\) cancels, so we have \((x + 2)\times x\)? No, wait, \(\frac{(x + 2)(x + 3)}{x + 4}\times\frac{x(x + 4)}{x + 3}=\ (x + 2)\times x\)? No, wait, \(\frac{(x + 2)(x + 3)}{x + 4}\times\frac{x(x + 4)}{x + 3}=(x + 2)\times x\)? No, wait, the numerator of the first fraction is \((x + 2)(x + 3)\), numerator of the second is \(x(x + 4)\), denominator of the first is \(x + 4\), denominator of the second is \(x + 3\). So multiplying numerators: \((x + 2)(x + 3)\times x(x + 4)\), denominators: \((x + 4)\times(x + 3)\). Then cancel \((x + 3)\) and \((x + 4)\) from numerator and denominator, we get \((x + 2)\times x\)? No, wait, \((x + 2)(x + 3)x(x + 4)\div[(x + 4)(x + 3)]=(x + 2)x=x^{2}+2x\). But when \(x = 1\), original expression: \(\frac{1 + 5+6}{1 + 4}\times\frac{1 + 4}{1+3}=\frac{12}{5}\times\frac{5}{4}=3\), \(x^{2}+2x=1 + 2=3\), and option C: \(x + 2=3\) when \(x = 1\), option D: \(x^{2}+2x=3\) when \(x = 1\). Wait, let's take \(x=2\). Original expression: \(\frac{4 + 10+6}{2 + 4}\times\frac{4 + 8}{2+3}=\frac{20}{6}\times\frac{12}{5}=\frac{20\times12}{6\times5}=\frac{240}{30}=8\). Option C: \(x + 2=4
eq8\). Option D: \(x^{2}+2x=4 + 4 = 8\). Ah, so when \(x = 2\), option D gives 8, option C gives 4. So my initial mistake was in the substitution. So the correct simplification is \(x^{2}+2x\). Wait, but let's re - do the algebra:

\(\frac{x^{2}+5x + 6}{x + 4}\cdot\frac{x^{2}+4x}{x + 3}=\frac{(x + 2)(x + 3)}{x + 4}\cdot\frac{x(x + 4)}{x + 3}\)

Cancel \((x + 3)\) and \((x + 4)\):

\(=(x + 2)\cdot x=x^{2}+2x\)

Answer:

D. \(x^{2}+2x\)