Question

fully simplify the expression below and write your answer as a single fraction.\\(\frac{10x^2 - 40}{5x^2 - 25x - 70} \cdot \frac{x^3 - 7x^2}{x - 2}\\)

Explanation:

Step1: Factor each expression

Factor \(10x^2 - 40\): \(10(x^2 - 4)=10(x - 2)(x + 2)\)
Factor \(5x^2 - 25x - 70\): \(5(x^2 - 5x - 14)=5(x - 7)(x + 2)\)
Factor \(x^3 - 7x^2\): \(x^2(x - 7)\)

Step2: Substitute factored forms

The expression becomes:
\(\frac{10(x - 2)(x + 2)}{5(x - 7)(x + 2)} \cdot \frac{x^2(x - 7)}{x - 2}\)

Step3: Cancel common factors

Cancel \(5\) (from \(10\) and \(5\)): \(10\div5 = 2\)
Cancel \((x + 2)\), \((x - 2)\), and \((x - 7)\):

After canceling, we have:
\(2 \cdot x^2 = 2x^2\) (Wait, let's re - check the cancellation:

\(\frac{10(x - 2)(x + 2)}{5(x - 7)(x + 2)}\cdot\frac{x^2(x - 7)}{x - 2}=\frac{10}{5}\cdot\frac{(x - 2)}{(x - 2)}\cdot\frac{(x + 2)}{(x + 2)}\cdot\frac{(x - 7)}{(x - 7)}\cdot x^2\)

\(\frac{10}{5}=2\), \(\frac{(x - 2)}{(x - 2)} = 1\), \(\frac{(x + 2)}{(x + 2)}=1\), \(\frac{(x - 7)}{(x - 7)} = 1\)

So the product is \(2\times1\times1\times1\times x^2=2x^2\)? Wait, no, wait:

Wait, the original expression is \(\frac{10x^2 - 40}{5x^2 - 25x - 70}\cdot\frac{x^3 - 7x^2}{x - 2}\)

After factoring:

\(\frac{10(x - 2)(x + 2)}{5(x - 7)(x + 2)}\cdot\frac{x^2(x - 7)}{x - 2}\)

Now, multiply the numerators: \(10(x - 2)(x + 2)\times x^2(x - 7)\)
Multiply the denominators: \(5(x - 7)(x + 2)\times(x - 2)\)

Now, cancel:

- \(10\) and \(5\): \(10\div5 = 2\)
- \((x - 2)\) in numerator and denominator: cancels
- \((x + 2)\) in numerator and denominator: cancels
- \((x - 7)\) in numerator and denominator: cancels

So we are left with \(2\times x^2=2x^2\)? Wait, no, let's do the multiplication step by step:

\(\frac{10(x - 2)(x + 2)\times x^2(x - 7)}{5(x - 7)(x + 2)\times(x - 2)}=\frac{10x^2(x - 2)(x + 2)(x - 7)}{5(x - 2)(x + 2)(x - 7)}\)

Divide \(10\) by \(5\) to get \(2\), and cancel \((x - 2)\), \((x + 2)\), \((x - 7)\) from numerator and denominator. So we have \(2x^2\). Wait, but let's check with a different approach. Let's take \(x = 3\) (a value that doesn't make any denominator zero: \(x eq2\), \(x eq7\), \(x eq - 2\))

Original expression:

First fraction: \(\frac{10(9)-40}{5(9)-25(3)-70}=\frac{90 - 40}{45-75 - 70}=\frac{50}{-100}=-\frac{1}{2}\)

Second fraction: \(\frac{27-63}{3 - 2}=\frac{-36}{1}=-36\)

Product: \(-\frac{1}{2}\times(-36) = 18\)

Now, \(2x^2\) when \(x = 3\) is \(2\times9 = 18\). So that works. Wait, but let's check another value, \(x = 4\)

Original expression:

First fraction: \(\frac{10(16)-40}{5(16)-25(4)-70}=\frac{160 - 40}{80 - 100 - 70}=\frac{120}{-90}=-\frac{4}{3}\)

Second fraction: \(\frac{64-112}{4 - 2}=\frac{-48}{2}=-24\)

Product: \(-\frac{4}{3}\times(-24)=32\)

\(2x^2\) when \(x = 4\) is \(2\times16 = 32\). So the simplification is correct.

Wait, but let's re - examine the factoring:

\(10x^2-40 = 10(x^2 - 4)=10(x - 2)(x + 2)\) (correct, since \(x^2-4=(x - 2)(x + 2)\))

\(5x^2-25x - 70=5(x^2 - 5x - 14)\). Now, factor \(x^2 - 5x - 14\): we need two numbers that multiply to \(-14\) and add to \(-5\). Those numbers are \(-7\) and \(2\). So \(x^2-5x - 14=(x - 7)(x + 2)\) (correct).

\(x^3 - 7x^2=x^2(x - 7)\) (correct).

So the factoring is correct. Then the cancellation is correct. So the simplified form is \(2x^2\). Wait, but let's write it as a single fraction. Wait, \(2x^2=\frac{2x^2}{1}\), but since the problem says "write your answer as a single fraction", and \(2x^2\) is equivalent to \(\frac{2x^2}{1}\), but actually, in the multiplication, after cancellation, we have \(2x^2\), which is a monomial, but as a fraction, it's \(\frac{2x^2}{1}\). But let's check the algebra again.

Wait, \(\frac{10(x - 2)(x + 2)}{5(x - 7)(x + 2)}\cdot\frac{x^2(x - 7)}{x - 2}\)

Multiply the numerators: \(10(x - 2)(x + 2)\times x^2(x - 7)=10x^2(x - 2)(x + 2)(x - 7)\)

Multiply the denominators: \(5(x - 7)(x + 2)\times(x - 2)=5(x - 2)(x + 2)(x - 7)\)

Now, divide numerator and denominator by \(5(x - 2)(x + 2)(x - 7)\):

\(\frac{10x^2(x - 2)(x + 2)(x - 7)}{5(x - 2)(x + 2)(x - 7)}=\frac{10}{5}x^2=2x^2\)

Yes, that's correct.

Answer:

\(\boxed{2x^2}\)