Question

fully simplify.
\\(\frac{9x^3 - 63x^2 - 72x}{x + 1} cdot \frac{x^2 - 14x + 40}{x^2 - 18x + 80} cdot \frac{x^2 - 16}{x^2 - 16}\\)

Explanation:

Step1: Factor each polynomial

• For \(9x^{3}-63x^{2}-72x\), factor out \(9x\): \(9x(x^{2}-7x - 8)\), then factor the quadratic: \(9x(x - 8)(x+1)\)
• For \(x^{2}-14x + 40\), factor: \((x - 4)(x - 10)\)
• For \(x^{2}-18x + 80\), factor: \((x - 8)(x - 10)\)
• For \(x^{2}-16\), factor (difference of squares): \((x - 4)(x + 4)\) (applies to both numerator and denominator instances)

So the expression becomes:
\(\frac{9x(x - 8)(x + 1)}{x + 1}\cdot\frac{(x - 4)(x - 10)}{(x - 8)(x - 10)}\cdot\frac{(x - 4)(x + 4)}{(x - 4)(x + 4)}\)

Step2: Cancel common factors

• Cancel \(x + 1\) (numerator and denominator)
• Cancel \(x - 8\) (numerator and denominator)
• Cancel \(x - 10\) (numerator and denominator)
• Cancel \((x - 4)(x + 4)\) (numerator and denominator)

After canceling, we are left with \(9x\)

Answer:

\(9x\)