Problem 68 - Beginning and Intermediate Algebra

Question

Find the product \((3 a+2)(a-4)(a-2)\) a) by first multiplying \((3 a+2)(a-4)\) and then multiplying that result by \((a-2)\) b) by first multiplying \((a-4)(a-2)\) and then multiplying that result by \((3 a+2)\) c) What do you notice about the results?

Step-by-Step Solution

Answer

a) When we multiply \((3a+2)(a-4)\) first, we get the simplified product \((3a^2-10a-8)(a-2)\). After multiplying this with \((a-2)\), the final product is \(3a^3 - 16a^2 + 12a + 16\). b) When we multiply \((a-4)(a-2)\) first, we get the simplified product \((a^2-6a+8)(3a+2)\). After multiplying this with \((3a+2)\), the final product is \(3a^3 - 16a^2 + 12a + 16\). c) The final products are the same in both cases: \(3a^3 - 16a^2 + 12a + 16 = 3a^3 - 16a^2 + 12a + 16\). This demonstrates the associative property of multiplication, as the factors can be grouped and multiplied in any order without affecting the final product.

Step 1: Multiply (3a+2)(a-4)

Distribute the first factor \((3a+2)\) into the second factor, \((a-4)\). \[ (3a+2)(a-4) = 3a(a-4) + 2(a-4) \]

Step 2: Simplify

Distribute the terms inside the parenthesis and combine like terms. \[ 3a(a-4) + 2(a-4) = 3a^2 - 12a + 2a - 8 \] \[ = 3a^2 - 10a - 8 \] Now, combine this result with the remaining factor, \((a-2)\).

Step 3: Multiply (3a^2-10a-8)(a-2)

Distribute the factor \((3a^2-10a-8)\) into the factor \((a-2)\). \[ (3a^2-10a-8)(a-2) = 3a^2(a-2) - 10a(a-2) - 8(a-2) \]

Step 4: Simplify

Distribute the terms inside the parenthesis and combine like terms. \[ 3a^3 - 6a^2 - 10a^2 + 20a - 8a + 16 = 3a^3 - 16a^2 + 12a + 16 \] Thus, the product becomes \(3a^3 - 16a^2 + 12a + 16\) when we multiply \((3a+2)(a-4)\) first and then the result with \((a-2)\). #b) Multiplying (a-4)(a-2) first#

Step 1: Multiply (a-4)(a-2)

Distribute the first factor \((a-4)\) into the second factor, \((a-2)\). \[ (a-4)(a-2) = a(a-2) - 4(a-2) \]

Step 2: Simplify

Distribute the terms inside the parenthesis and combine like terms. \[ a(a-2) - 4(a-2) = a^2 - 2a - 4a + 8 \] \[ = a^2 - 6a + 8 \] Now, combine this result with the remaining factor, \((3a+2)\).

Step 3: Multiply (3a+2)(a^2-6a+8)

Distribute the factor \((3a+2)\) into the factor, \((a^2-6a+8)\). \[ (3a+2)(a^2-6a+8) = 3a(a^2-6a+8) + 2(a^2-6a+8) \]

Step 4: Simplify

Distribute the terms inside the parenthesis and combine like terms. \[ 3a^3 - 18a^2 + 24a + 2a^2 - 12a + 16 = 3a^3 - 16a^2 + 12a + 16 \] Thus, the product becomes \(3a^3 - 16a^2 + 12a + 16\) when we multiply \((a-4)(a-2)\) first and then the result with \((3a+2)\). #c) Comparing the Results# Upon comparing the two results, we notice that the final product is the same for both methods: \[ 3a^3 - 16a^2 + 12a + 16 = 3a^3 - 16a^2 + 12a + 16 \] This shows that regardless of the order in which we combine the factors, the distributive property allows us to arrive at the same result. This also demonstrates the associative property of multiplication, as the factors can be grouped and multiplied in any order without affecting the final product.

Key Concepts

Distributive Property

The distributive property is a fundamental principle in algebra that allows us to rewrite and simplify expressions. When we apply the distributive property, we multiply a single term by each term within a parenthesis. For example, if you have an expression like

• \[ (x + y)(z) \]

The distributive property tells us to multiply \(z\) with both \(x\) and \(y\), giving:

• \[ xz + yz \]

This is exactly what we did in the exercise above when multiplying

• \((3a+2)(a-4)\)

by distributing each term of \(3a+2\) into \(a-4\). It helps to break down complicated expressions into manageable parts, allowing for easy simplification. This powerful property is used extensively in polynomial multiplication, making it easier to handle more complex algebraic expressions. When practicing problems involving the distributive property, remember to carefully distribute each term and combine like terms.

Associative Property

The associative property is another key principle of algebra, especially useful when dealing with multiplication and addition of numbers or expressions. It states that the way in which factors are grouped does not affect the final product. For instance, in numbers,

• \((a \times b) \times c = a \times (b \times c)\)

This property is crucial when multiplying more than two factors. In our problem, we used the associative property by rearranging the order of multiplication:

• \((3a+2)(a-4)(a-2)\)

To approach the problem, you can choose different ways to group and multiply the factors, thus breaking down polynomial multiplication into simpler tasks. By recognizing the associative property, you can group these factors in any order:

• First multiply \( (3a+2) \) by \( (a-4) \)
• Multiply the result by \( (a-2) \)

Or swap and multiply

• \((a-4)(a-2)\)
• Then multiply by \( (3a+2) \)

Either way, you end up with the same result. The associative property ensures that regardless of how the terms are grouped, the result remains consistent.

Polynomial Multiplication

Polynomial multiplication may seem daunting at first, but with tools like the distributive and associative properties, it becomes far more manageable. Polynomials are expressions that include variables raised to whole number powers and constants. To begin polynomial multiplication:

• Break down each polynomial into its constituent terms.
• Apply the distributive property to ensure each term in one polynomial is multiplied by each term in the other.

This step-by-step approach was demonstrated in our example, simplifying expressions like

• \((a-4)(a-2)\)
• and multiplying them with third term \((3a+2)\)

After distributing the terms:

• Combine like terms to condense the expression into its simplest form.

In doing so, you may encounter a few common tricks, like recognizing special expressions such as

• perfect squares or difference of squares, which have simplified multiplication shortcuts.

With practice, recognizing where and how these mathematical principles are applied makes polynomial multiplication much less complicated and will significantly enhance your algebra skills.