Expand (1/3a + b)(9b + 12): Binomial Multiplication with Fractions

Q: Why do I get 3ab instead of 3ab when multiplying the fractions?

When you multiply $ \frac{1}{3}a \times 9b $, think of it as $ \frac{1 \times 9 \times a \times b}{3} = \frac{9ab}{3} = 3ab $. The 9 and 3 cancel out perfectly!

Q: How do I know which terms to multiply together?

Use FOIL: First terms ($ \frac{1}{3}a \times 9b $), Outer terms ($ \frac{1}{3}a \times 12 $), Inner terms ($ b \times 9b $), Last terms ($ b \times 12 $).

Q: Do I need to combine like terms at the end?

In this problem, all terms are different ($ 9b^2, 3ab, 4a, 12b $), so no combining is needed. Just arrange them in descending order of powers.

Q: What if I get the order of terms wrong in my final answer?

The mathematical value is the same regardless of order, but it's good practice to write terms in descending order: highest degree terms first, like $ 9b^2 + 3ab + 4a + 12b $.

Q: How can I check if my expansion is correct?

Try factoring your answer back to the original form, or substitute simple values like $ a = 3, b = 1 $ into both the original expression and your answer to see if they match.

Distributive Property with Fractional Coefficients

$(\frac{1}{3}a+b)(9b+12)=$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Solve

00:03 Open parentheses properly, multiply each factor by each factor

00:25 Calculate the products

00:49 Calculate the quotients

01:04 Arrange the expression

01:13 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

$(\frac{1}{3}a+b)(9b+12)=$

Step-by-step solution

The given expression is $(\frac{1}{3}a + b)(9b + 12)$ . We will expand this expression using the distributive property:

Step 1: Multiply the first terms of each binomial:

$\frac{1}{3}a \times 9b = 3ab$

Step 2: Multiply the outer terms:

$\frac{1}{3}a \times 12 = 4a$

Step 3: Multiply the inner terms:

$b \times 9b = 9b^2$

Step 4: Multiply the last terms:

$b \times 12 = 12b$

Combine all the resulting terms together:

$9b^2 + 3ab + 4a + 12b$

Therefore, the solution to the problem is $9b^2 + 3ab + 4a + 12b$ .

Final Answer

$9b^2+3ab+4a+12b$

Key Points to Remember

Essential concepts to master this topic

FOIL Method: First, Outer, Inner, Last terms multiply systematically
Technique: $\frac{1}{3}a \times 9b = 3ab$ by canceling common factors
Check: Expand backwards: factor out common terms to verify original form ✓

Common Mistakes

Avoid these frequent errors

Forgetting to multiply all four term combinations
Don't just multiply $\frac{1}{3}a \times 9b = 3ab$ and stop there = missing three terms! This gives incomplete expansions like $3ab + 12b$ . Always multiply each term in the first binomial by each term in the second binomial using FOIL.

Practice Quiz

Test your knowledge with interactive questions

$(3+20)\times(12+4)=$

FAQ

Everything you need to know about this question

Why do I get 3ab instead of 3ab when multiplying the fractions?

When you multiply $\frac{1}{3}a \times 9b$ , think of it as $\frac{1 \times 9 \times a \times b}{3} = \frac{9ab}{3} = 3ab$ . The 9 and 3 cancel out perfectly!

How do I know which terms to multiply together?

Use FOIL: First terms ( $\frac{1}{3}a \times 9b$ ), Outer terms ( $\frac{1}{3}a \times 12$ ), Inner terms ( $b \times 9b$ ), Last terms ( $b \times 12$ ).

Do I need to combine like terms at the end?

In this problem, all terms are different ( $9b^2, 3ab, 4a, 12b$ ), so no combining is needed. Just arrange them in descending order of powers.

What if I get the order of terms wrong in my final answer?

The mathematical value is the same regardless of order, but it's good practice to write terms in descending order: highest degree terms first, like $9b^2 + 3ab + 4a + 12b$ .

How can I check if my expansion is correct?

Try factoring your answer back to the original form, or substitute simple values like $a = 3, b = 1$ into both the original expression and your answer to see if they match.

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