BIG IDEAS MATH \#3 i Find the quiotient. Write your answer in factored form. \( \frac{16 x^{2}-8 x+1}{x^{3}-7 x^{2}+12 x} \div \frac{20 x^{2}-5 x}{15 x^{3}}=\square, x \neq \square, x \neq \square 0 \) Previous

To solve the problem, we need to find the quotient of the two rational expressions given. The expression is: \[ \frac{16 x^{2}-8 x+1}{x^{3}-7 x^{2}+12 x} \div \frac{20 x^{2}-5 x}{15 x^{3}} \] ### Step 1: Rewrite the Division as Multiplication We can rewrite the division of fractions as multiplication by the reciprocal: \[ \frac{16 x^{2}-8 x+1}{x^{3}-7 x^{2}+12 x} \times \frac{15 x^{3}}{20 x^{2}-5 x} \] ### Step 2: Factor Each Expression Now, we will factor each part of the expression. 1. **Numerator of the first fraction**: \(16 x^{2}-8 x+1\) This is a quadratic expression. We can factor it as follows: \[ 16 x^{2}-8 x+1 = (4x-1)^{2} \] 2. **Denominator of the first fraction**: \(x^{3}-7 x^{2}+12 x\) We can factor out \(x\): \[ x(x^{2}-7x+12) \] Now, we can factor the quadratic \(x^{2}-7x+12\): \[ x^{2}-7x+12 = (x-3)(x-4) \] So, the complete factorization is: \[ x(x-3)(x-4) \] 3. **Numerator of the second fraction**: \(20 x^{2}-5 x\) We can factor out \(5x\): \[ 5x(4x-1) \] 4. **Denominator of the second fraction**: \(15 x^{3}\) This is already factored as \(15x^{3}\). ### Step 3: Substitute the Factored Forms Now we substitute the factored forms back into the expression: \[ \frac{(4x-1)^{2}}{x(x-3)(x-4)} \times \frac{15x^{3}}{5x(4x-1)} \] ### Step 4: Simplify the Expression Now we can simplify the expression: 1. Cancel \(5x\) in the numerator and denominator: \[ \frac{(4x-1)^{2}}{(x-3)(x-4)} \times \frac{3x^{2}}{(4x-1)} \] 2. Cancel one \(4x-1\): \[ \frac{(4x-1)}{(x-3)(x-4)} \times 3x^{2} \] 3. Combine the fractions: \[ \frac{3x^{2}(4x-1)}{(x-3)(x-4)} \] ### Final Answer The final answer in factored form is: \[ \frac{3x^{2}(4x-1)}{(x-3)(x-4)} \] ### Restrictions The restrictions based on the original denominators are: 1. \(x \neq 0\) 2. \(x \neq 3\) 3. \(x \neq 4\) Thus, the final answer is: \[ \frac{3x^{2}(4x-1)}{(x-3)(x-4)}, \quad x \neq 0, x \neq 3, x \neq 4 \]