Problem 18 - Algebra 2

Question

Multiply and simplify. Assume that all variables are positive. $$ \sqrt{8 y^{5}} \cdot \sqrt{40 y^{2}} $$

Step-by-Step Solution

Answer

The product and final simplified form is \(8y^{4}\sqrt{5y}\)

Step 1: Simplify the square roots

First, the square roots need to be simplified. \nThe prime factors of 8 are 2, 2, and 2 and \(y^{5}\) can be written as \(y^{2} \cdot y^{2} \cdot y\). Hence, \(\sqrt{8 y^{5}}\) can be simplified to \(2y^{2}\sqrt{2y}\). \n Similarly, for \( \sqrt{40 y^{2}} \), the prime factors for 40 are 2, 2, 2, and 5. Thus, \(\sqrt{40 y^{2}}\) simplifies to \(2y\sqrt{10}\).

Step 2: Multiply the simplified roots

Next, multiplying \(2y^{2}\sqrt{2y}\) and \(2y\sqrt{10}\) gives \(4y^{3}\sqrt{20y}\). This can be simplified further as \(4y^{3}\sqrt{4 \cdot 5 \cdot y}\).

Step 3: Simplify the product

Finally, simplify \(4y^{3}\sqrt{4 \cdot 5 \cdot y}\) to derive the final answer as \(8y^{4}\sqrt{5y}\). The square root of \(4y\) is \(2y\) and \(4y^{3}\) multiplied by \(2y\) is \(8y^{4}\).

Key Concepts

Simplifying Square Roots

To simplify square roots, we must break down the numbers into their prime factors and look for perfect squares within those. The prime factorization method helps in doing this. For instance, with the number 8, we have the prime factors 2, 2, and 2. Meanwhile, the expression \(y^5\) can be rewritten as \(y^2 \cdot y^2 \cdot y\). By grouping these factors into pairs, we can easily identify which portions can be taken out of the square root.

When simplifying \(\sqrt{8y^5}\), we notice that \(8 = 2 \cdot 2 \cdot 2\), which allows us to take out \(2\) from the square root. For \(y^5\), we can simplify it by taking out \(y^2\), leaving \(2y^2\sqrt{2y}\). This process helps in reducing the complexity of our expressions, making further operations like multiplication much easier.

Multiplying Radicals

Multiplying radicals involves handling the expressions both inside and outside the square root separately. Once each radical is simplified, the process can become straightforward. For example, to multiply \(2y^2\sqrt{2y}\) by \(2y\sqrt{10}\), take the following into account:

• Multiply the constants and the variable components outside the square roots: \(2y^2 \cdot 2y = 4y^3\).
• Multiply the terms inside the square roots: \(\sqrt{2y} \cdot \sqrt{10} = \sqrt{20y}\).

After multiplying, the result is \(4y^3\sqrt{20y}\). It's essential to multiply the terms inside the square roots together to form one expression under the square root, but be mindful of attempting simplifications whenever possible.

Prime Factorization

Prime factorization is a fundamental skill necessary for simplifying complex expressions like square roots. It involves breaking down a number into its prime components, which are numbers only divisible by 1 and themselves. Returning to our earlier example, 8 breaks down into 2, 2, and 2. This technique is useful when simplifying square roots, as it helps identify perfect squares.

Similarly, for the number 40, its prime factors are 2, 2, 2, and 5, pointing out that \(40 = 2^3 \cdot 5\). Identifying these factors allows us to simplify expressions such as \(\sqrt{40y^2}\) to \(2y\sqrt{10}\). Working with prime factors is a crucial step when dealing with radicals because it simplifies the process of recognizing which terms can be extracted from under the square root.