Question

rational expressions
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simplify: \\(\frac{3x^2 + 11x + 10}{3x^2 + 20x + 25}\\)
\\(\frac{2}{5}\\)
\\(\frac{11x + 10}{20x + 25}\\)
\\(\frac{3x + 2}{3x + 5}\\)
\\(\frac{x + 2}{x + 5}\\)
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Explanation:

Step1: Factor numerator and denominator

Factor \(3x^2 + 11x + 10\): We need two numbers that multiply to \(3\times10 = 30\) and add to \(11\). The numbers are \(5\) and \(6\). So, \(3x^2 + 5x + 6x + 10 = x(3x + 5) + 2(3x + 5) = (3x + 5)(x + 2)\) (Wait, no, correction: Wait, \(3x^2 + 11x + 10\), let's do it properly. \(a = 3\), \(b = 11\), \(c = 10\). \(ac = 30\), factors of \(30\) that add to \(11\) are \(5\) and \(6\). So split middle term: \(3x^2 + 5x + 6x + 10 = x(3x + 5) + 2(3x + 5) = (3x + 5)(x + 2)\)? Wait, no, \(3x^2 + 11x + 10\): Let's check \((3x + 2)(x + 5)=3x^2 + 15x + 2x + 10 = 3x^2 + 17x + 10\), no. Wait, \((3x + 5)(x + 2)=3x^2 + 6x + 5x + 10 = 3x^2 + 11x + 10\). Yes, that's correct.

Now factor \(3x^2 + 20x + 25\): \(a = 3\), \(b = 20\), \(c = 25\). \(ac = 75\), factors of \(75\) that add to \(20\) are \(15\) and \(5\). Split middle term: \(3x^2 + 15x + 5x + 25 = 3x(x + 5) + 5(x + 5) = (3x + 5)(x + 5)\).

So the rational expression becomes \(\frac{(3x + 5)(x + 2)}{(3x + 5)(x + 5)}\)

Step2: Cancel common factors

We can cancel the common factor \((3x + 5)\) (assuming \(3x + 5 eq0\)) from the numerator and the denominator. So we get \(\frac{x + 2}{x + 5}\)? Wait, no, wait: Wait, numerator was \((3x + 2)(x + 5)\)? Wait, I made a mistake earlier. Let's re - factor \(3x^2 + 11x + 10\) correctly.

Let's use the quadratic formula for \(3x^2+11x + 10 = 0\). \(x=\frac{-11\pm\sqrt{121 - 120}}{6}=\frac{-11\pm1}{6}\). So \(x=\frac{-11 + 1}{6}=\frac{-10}{6}=-\frac{5}{3}\) and \(x=\frac{-11 - 1}{6}=-2\). So the factors are \((3x + 5)(x + 2)\)? Wait, when \(x=-2\), \(3(-2)^2+11(-2)+10 = 12-22 + 10 = 0\), and \(3x + 5=3(-2)+5=-1 eq0\), so \((x + 2)\) is a factor. When \(x =-\frac{5}{3}\), \(3(-\frac{5}{3})^2+11(-\frac{5}{3})+10=\frac{25}{3}-\frac{55}{3}+10=\frac{25 - 55 + 30}{3}=0\), so \((3x + 5)\) is a factor. So numerator is \((3x + 5)(x + 2)\).

Denominator \(3x^2+20x + 25\): Quadratic formula \(x=\frac{-20\pm\sqrt{400 - 300}}{6}=\frac{-20\pm\sqrt{100}}{6}=\frac{-20\pm10}{6}\). So \(x=\frac{-20 + 10}{6}=-\frac{5}{3}\) and \(x=\frac{-20 - 10}{6}=-5\). So factors are \((3x + 5)(x + 5)\).

So the expression is \(\frac{(3x + 5)(x + 2)}{(3x + 5)(x + 5)}\). Cancel \((3x + 5)\) ( \(3x+5 eq0\) ), we get \(\frac{x + 2}{x + 5}\)? Wait, no, wait: Wait, \((3x + 2)(x + 5)=3x^2+15x + 2x + 10 = 3x^2+17x + 10\), which is not the numerator. Wait, I think I messed up the factoring of the numerator. Let's do numerator again: \(3x^2+11x + 10\). Let's try \((3x + 2)(x + 5)\): \(3x^2+15x+2x + 10=3x^2+17x + 10\), no. \((3x + 10)(x + 1)=3x^2+13x + 10\), no. Wait, \((3x + 5)(x + 2)=3x^2+6x+5x + 10=3x^2+11x + 10\), yes! That's correct. Denominator: \(3x^2+20x + 25=(3x + 5)(x + 5)\), because \(3x\times x=3x^2\), \(3x\times5 = 15x\), \(5\times x=5x\), \(5\times5 = 25\), \(15x + 5x=20x\). So numerator is \((3x + 5)(x + 2)\), denominator is \((3x + 5)(x + 5)\). Cancel \((3x + 5)\), we get \(\frac{x + 2}{x + 5}\)? Wait, no, wait: \((3x + 5)(x + 2)\) divided by \((3x + 5)(x + 5)\) is \(\frac{x + 2}{x + 5}\)? Wait, no, \((3x + 5)\) cancels, so it's \(\frac{x + 2}{x + 5}\)? But let's check the options. Wait, one of the options is \(\frac{3x + 2}{3x + 5}\). Wait, I must have factored the numerator wrong. Let's try again.

Wait, numerator: \(3x^2+11x + 10\). Let's use the AC method correctly. \(a = 3\), \(b = 11\), \(c = 10\). \(ac=30\). We need two numbers that multiply to \(30\) and add to \(11\). The numbers are \(5\) and \(6\). So split the middle term: \(3x^2+5x+6x + 10\). Now group: \((3x^2+5x)+(6x + 10)=x(3x + 5)+2(3x + 5)=(3x + 5)(x + 2)\). Denominator: \(3x^2+20x + 25\), \(ac = 75\), numbers that multiply to \(75\) and add to \(20\) are \(15\) and \(5\). Split middle term: \(3x^2+15x+5x + 25=3x(x + 5)+5(x + 5)=(3x + 5)(x + 5)\). So when we cancel \((3x + 5)\), we get \(\frac{x + 2}{x + 5}\)? But that's one of the options (the last option: \(\frac{x + 2}{x + 5}\)). Wait, but let's check with \(x = 0\). Original expression: \(\frac{0 + 0 + 10}{0 + 0 + 25}=\frac{10}{25}=\frac{2}{5}\). \(\frac{x + 2}{x + 5}\) when \(x = 0\) is \(\frac{2}{5}\), which matches. \(\frac{3x + 2}{3x + 5}\) when \(x = 0\) is \(\frac{2}{5}\), also matches. Wait, that's a problem. Wait, let's check with \(x = 1\). Original expression: \(\frac{3 + 11 + 10}{3 + 20 + 25}=\frac{24}{48}=\frac{1}{2}\). \(\frac{x + 2}{x + 5}\) when \(x = 1\) is \(\frac{3}{6}=\frac{1}{2}\). \(\frac{3x + 2}{3x + 5}\) when \(x = 1\) is \(\frac{5}{8}\), which is not \(\frac{1}{2}\). Wait, so my factoring was wrong. Wait, let's re - factor the numerator. Let's try \((3x + 2)(x + 5)\): \(3x^2+15x+2x + 10=3x^2+17x + 10\), no. \((3x + 10)(x + 1)=3x^2+13x + 10\), no. Wait, maybe I made a mistake in the quadratic formula. For numerator \(3x^2+11x + 10\), discriminant \(D = 121-120 = 1\), so roots are \(\frac{-11\pm1}{6}\), so \(x=\frac{-10}{6}=-\frac{5}{3}\) and \(x=-2\). So factors are \((x + 2)(3x + 5)\) (since when \(x=-2\), \(3x+5=-1\), so \((x + 2)\) is a factor, and when \(x =-\frac{5}{3}\), \(x + 2=\frac{1}{3}\), so \((3x + 5)\) is a factor). Denominator: \(3x^2+20x + 25\), discriminant \(D = 400 - 300 = 100\), roots are \(\frac{-20\pm10}{6}\), so \(x=-\frac{5}{3}\) and \(x=-5\), so factors are \((3x + 5)(x + 5)\). So the expression is \(\frac{(3x + 5)(x + 2)}{(3x + 5)(x + 5)}=\frac{x + 2}{x + 5}\), which when \(x = 1\) gives \(\frac{3}{6}=\frac{1}{2}\), and original expression when \(x = 1\) is \(\frac{3 + 11 + 10}{3 + 20 + 25}=\frac{24}{48}=\frac{1}{2}\), which matches. \(\frac{3x + 2}{3x + 5}\) when \(x = 1\) is \(\frac{5}{8} eq\frac{1}{2}\), so my initial factoring was correct. Wait, but let's check the options again. The options are:

1. \(\frac{2}{5}\)
2. \(\frac{11x + 10}{20x + 25}\)
3. \(\frac{3x + 2}{3x + 5}\)
4. \(\frac{x + 2}{x + 5}\)

When \(x = 0\), original expression is \(\frac{10}{25}=\frac{2}{5}\), option 1 is \(\frac{2}{5}\), option 3 is \(\frac{2}{5}\), option 4 is \(\frac{2}{5}\). When \(x = 2\), original expression: \(\frac{12 + 22 + 10}{12 + 40 + 25}=\frac{44}{77}=\frac{4}{7}\). Option 4: \(\frac{4}{7}\), option 3: \(\frac{8}{11}\approx0.727\), which is not \(\frac{4}{7}\approx0.571\). So option 4 is correct.

Wait, but let's do the factoring again carefully. Numerator: \(3x^2+11x + 10\). Let's use the method of factoring by grouping correctly.

\(3x^2+11x + 10\)

Multiply \(a\) and \(c\): \(3\times10 = 30\)

Find two numbers that multiply to \(30\) and add to \(11\): \(5\) and \(6\)

Rewrite the middle term: \(3x^2+5x+6x + 10\)

Group the first two and last two terms: \((3x^2+5x)+(6x + 10)\)

Factor out the GCF from each group: \(x(3x + 5)+2(3x + 5)\)

Factor out the common binomial factor: \((3x + 5)(x + 2)\)

Denominator: \(3x^2+20x + 25\)

Multiply \(a\) and \(c\): \(3\times25 = 75\)

Find two numbers that multiply to \(75\) and add to \(20\): \(15\) and \(5\)

Rewrite the middle term: \(3x^2+15x+5x + 25\)

Group the first two and last two terms: \((3x^2+15x)+(5x + 25)\)

Factor out the GCF from each group: \(3x(x + 5)+5(x + 5)\)

Factor out the common binomial factor: \((3x + 5)(x + 5)\)

So the rational expression is \(\frac{(3x + 5)(x + 2)}{(3x + 5)(x + 5)}\)

Cancel the common factor \((3x + 5)\) (assuming \(3x+5 eq0\)): \(\frac{x + 2}{x + 5}\)

Answer:

\(\frac{x + 2}{x + 5}\) (the last option, which is D if we consider the options as A. \(\frac{2}{5}\), B. \(\frac{11x + 10}{20x + 25}\), C. \(\frac{3x + 2}{3x + 5}\), D. \(\frac{x + 2}{x + 5}\))