Question

question
fully simplify the expression below and write your answer as a single fraction.
\\(\frac{x^2 + 3x - 10}{6(x + 5)^2} \cdot \frac{x^4 + 3x^3 - 10x^2}{x(x - 2)^2}\\)

Explanation:

Step1: Factor numerators and denominators

First, factor \(x^{2}+3x - 10\). We need two numbers that multiply to \(- 10\) and add to \(3\). Those numbers are \(5\) and \(-2\), so \(x^{2}+3x - 10=(x + 5)(x-2)\).

Next, factor \(x^{4}+3x^{3}-10x^{2}\). Factor out \(x^{2}\) first: \(x^{2}(x^{2}+3x - 10)\). Then factor \(x^{2}+3x - 10\) as before, so \(x^{4}+3x^{3}-10x^{2}=x^{2}(x + 5)(x - 2)\).

The original expression becomes \(\frac{(x + 5)(x - 2)}{6(x + 5)^{2}}\cdot\frac{x^{2}(x + 5)(x - 2)}{x(x - 2)^{2}}\).

Step2: Cancel common factors

Cancel out common factors in the numerators and denominators.
- Cancel one \((x + 5)\) from the first numerator and one from the second denominator: \(\frac{(x - 2)}{6(x + 5)}\cdot\frac{x^{2}(x + 5)(x - 2)}{x(x - 2)^{2}}\)
- Cancel one \((x - 2)\) from the first numerator and one from the second denominator: \(\frac{1}{6(x + 5)}\cdot\frac{x^{2}(x + 5)(x - 2)}{x(x - 2)}\)
- Cancel one \((x + 5)\) from the second numerator and one from the first denominator: \(\frac{1}{6}\cdot\frac{x^{2}(x - 2)}{x(x - 2)}\)
- Cancel one \(x\) from the second numerator and one from the second denominator, and cancel one \((x - 2)\) from the second numerator and one from the second denominator: \(\frac{1}{6}\cdot\frac{x}{1}\)

Step3: Multiply the remaining factors

Multiply the remaining factors: \(\frac{x}{6}\) (Wait, no, let's re - check. Wait, after canceling:

From \(\frac{(x + 5)(x - 2)}{6(x + 5)^{2}}\cdot\frac{x^{2}(x + 5)(x - 2)}{x(x - 2)^{2}}\)

Cancel \((x + 5)\) : numerator has \((x + 5)\) (first) and \((x + 5)\) (second), denominator has \((x + 5)^{2}\) (first) and \(x\) (second). So cancel one \((x + 5)\) from first numerator and one from first denominator, one \((x + 5)\) from second numerator and one from first denominator.

Cancel \((x - 2)\): first numerator has \((x - 2)\), second numerator has \((x - 2)\), second denominator has \((x - 2)^{2}\). So cancel one \((x - 2)\) from first numerator and one from second denominator, one \((x - 2)\) from second numerator and one from second denominator.

Cancel \(x\): second numerator has \(x^{2}\), second denominator has \(x\). So cancel one \(x\) from second numerator and one from second denominator.

After canceling, we have \(\frac{1}{6}\cdot\frac{x(x + 5)(x - 2)}{(x + 5)(x - 2)}\) Wait, no, let's do it step by step:

First fraction numerator: \((x + 5)(x - 2)\), denominator: \(6(x + 5)^{2}\)

Second fraction numerator: \(x^{2}(x + 5)(x - 2)\), denominator: \(x(x - 2)^{2}\)

Multiply numerators: \((x + 5)(x - 2)\cdot x^{2}(x + 5)(x - 2)=x^{2}(x + 5)^{2}(x - 2)^{2}\)

Multiply denominators: \(6(x + 5)^{2}\cdot x(x - 2)^{2}=6x(x + 5)^{2}(x - 2)^{2}\)

Now cancel common factors: \(x^{2}(x + 5)^{2}(x - 2)^{2}\div[6x(x + 5)^{2}(x - 2)^{2}]=\frac{x}{6}\)

Wait, I made a mistake earlier. Let's do it correctly.

Numerator: \((x + 5)(x - 2)\times x^{2}(x + 5)(x - 2)=x^{2}(x + 5)^{2}(x - 2)^{2}\)

Denominator: \(6(x + 5)^{2}\times x(x - 2)^{2}=6x(x + 5)^{2}(x - 2)^{2}\)

Cancel \(x^{2}\) with \(x\) (leaving \(x\)), cancel \((x + 5)^{2}\) and \((x - 2)^{2}\). So we get \(\frac{x}{6}\)

Wait, no, \(x^{2}\div x=x\), and \((x + 5)^{2}\div(x + 5)^{2}=1\), \((x - 2)^{2}\div(x - 2)^{2}=1\), and \(1\div6=\frac{1}{6}\). So \(x\times\frac{1}{6}=\frac{x}{6}\)

Wait, another way:

\(\frac{(x + 5)(x - 2)}{6(x + 5)^{2}}\cdot\frac{x^{2}(x + 5)(x - 2)}{x(x - 2)^{2}}\)

Group the factors:

\(=\frac{(x + 5)\times(x - 2)\times x^{2}\times(x + 5)\times(x - 2)}{6\times(x + 5)^{2}\times x\times(x - 2)^{2}}\)

Now, for the \((x + 5)\) terms: numerator has \((x + 5)\times(x + 5)\), denominator has \((x + 5)^{2}\), so they cancel out.

For the \((x - 2)\) terms: numerator has \((x - 2)\times(x - 2)\), denominator has \((x - 2)^{2}\), so they cancel out.

For the \(x\) terms: numerator has \(x^{2}\), denominator has \(x\), so \(x^{2}\div x=x\)

For the constant term: numerator has \(1\), denominator has \(6\)

So after canceling, we have \(\frac{x}{6}\)

Wait, no, wait:

Wait, the first numerator: \((x + 5)(x - 2)\), second numerator: \(x^{2}(x + 5)(x - 2)\)

First denominator: \(6(x + 5)^{2}\), second denominator: \(x(x - 2)^{2}\)

So when we multiply the fractions, it's \(\frac{(x + 5)(x - 2)\times x^{2}(x + 5)(x - 2)}{6(x + 5)^{2}\times x(x - 2)^{2}}\)

Now, count the exponents:

- For \((x + 5)\): numerator has \(1 + 1=2\), denominator has \(2\), so \((x + 5)^{2}\div(x + 5)^{2}=1\)
- For \((x - 2)\): numerator has \(1+1 = 2\), denominator has \(2\), so \((x - 2)^{2}\div(x - 2)^{2}=1\)
- For \(x\): numerator has \(2\), denominator has \(1\), so \(x^{2}\div x=x\)
- For the constant: numerator has \(1\), denominator has \(6\), so \(1\div6=\frac{1}{6}\)

So the result is \(\frac{x}{6}\)

Wait, no, I think I messed up. Let's do it again.

Original expression:

\(\frac{x^{2}+3x - 10}{6(x + 5)^{2}}\cdot\frac{x^{4}+3x^{3}-10x^{2}}{x(x - 2)^{2}}\)

Factor each part:

\(x^{2}+3x - 10=(x + 5)(x - 2)\)

\(x^{4}+3x^{3}-10x^{2}=x^{2}(x^{2}+3x - 10)=x^{2}(x + 5)(x - 2)\)

So substitute back:

\(\frac{(x + 5)(x - 2)}{6(x + 5)^{2}}\cdot\frac{x^{2}(x + 5)(x - 2)}{x(x - 2)^{2}}\)

Now, multiply the numerators: \((x + 5)(x - 2)\times x^{2}(x + 5)(x - 2)=x^{2}(x + 5)^{2}(x - 2)^{2}\)

Multiply the denominators: \(6(x + 5)^{2}\times x(x - 2)^{2}=6x(x + 5)^{2}(x - 2)^{2}\)

Now, divide numerator by denominator:

\(\frac{x^{2}(x + 5)^{2}(x - 2)^{2}}{6x(x + 5)^{2}(x - 2)^{2}}\)

Cancel out \((x + 5)^{2}\) and \((x - 2)^{2}\) from numerator and denominator:

\(\frac{x^{2}}{6x}\)

Simplify \(\frac{x^{2}}{6x}=\frac{x}{6}\) (since \(x eq0,x eq2,x eq - 5\) to avoid division by zero)

Answer:

\(\boxed{\dfrac{x}{6}}\)