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Chapter 5: Problem 13

Find each product. $$ 6 k^{2}\left(3 k^{2}+2 k+1\right) $$

Short Answer

Expert verified
The product is \18k^4 + 12k^3 + 6k^2\.

Step by step solution

01

- Distribute the first term

Multiply the first term, which is the entire monomial outside the parenthesis, by each term inside the parenthesis. This means each term in \((3k^2 + 2k + 1)\) gets multiplied by \(6k^2\).
02

- Multiply the first term inside the parenthesis

Multiply \(6k^2\) by \(3k^2\). This results in \((6k^2) * (3k^2) = 18k^4\).
03

- Multiply the second term inside the parenthesis

Multiply \(6k^2\) by \(2k\). This results in \((6k^2) * (2k) = 12k^3\).
04

- Multiply the third term inside the parenthesis

Multiply \(6k^2\) by \(1\). This results in \((6k^2) * (1) = 6k^2\).
05

- Write the final expression

Combine all the results from the multiplication: \(18k^4 + 12k^3 + 6k^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distribution in Algebra
To multiply a polynomial by a monomial, we use the distributive property. This involves multiplying each term inside the parenthesis by the term outside the parenthesis. For example, in the expression \( 6k^2(3k^2 + 2k + 1) \), we distribute \( 6k^2 \) to each of the terms inside the parentheses, one at a time.

Using distribution:
  • First, multiply \( 6k^2 \) by \( 3k^2 \).
  • Next, multiply \( 6k^2 \) by \( 2k \).
  • Finally, multiply \( 6k^2 \) by \( 1 \).

This systematic approach ensures that we do not miss any terms, and we correctly apply the distributive property to the entire polynomial.
Multiplying Monomials
When multiplying monomials, we multiply their coefficients and add the exponents of like bases. In our example, the original monomial is \( 6k^2 \).

Here's how we multiply each term:
  • For \( 6k^2 \times 3k^2 \), multiply the coefficients \( (6 \times 3 = 18) \), and then add the exponents of \( k \) (\( 2+2 = 4 \)). Thus, the result is \( 18k^4 \).
  • For \( 6k^2 \times 2k \), multiply the coefficients \( (6 \times 2 = 12) \), and then add the exponents of \( k \) (\( 2+1 = 3 \)). Thus, the result is \( 12k^3 \).
  • For \( 6k^2 \times 1 \), the coefficient is \( 6 \) (since \( 6 \times 1 = 6 \)) and the exponent of \( k \) remains \( 2 \). Thus, the result is \( 6k^2 \).

By following these steps we ensure each multiplication is done correctly.
Combining Like Terms
After multiplying, we combine all the terms to form the final expression. In this exercise, we did not need to combine any like terms because each resulting term already had a different exponent.

Our results were:
  • \( 18k^4 \)
  • \( 12k^3 \)
  • \( 6k^2 \)

Writing them together, we get the final expression: \( 18k^4 + 12k^3 + 6k^2 \).
It is always important to look out for like terms that can be combined to simplify the polynomial further.
In this example, there were no additional like terms, so the expression was already in its simplest form.

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Most popular questions from this chapter

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Simplify each expression so that no negative exponents appear in the final result. Assume that all variables represent nonzero real numbers. $$ 3^{5} \cdot 3^{-6} $$

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