Expand (a+b)(3+a/b): Multiplying Binomial Expressions Step-by-Step

Binomial Multiplication with Fractional Terms

(a+b)(3+ab)=? (a+b)(3+\frac{a}{b})=\text{?}

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:07 Let's simplify the expression.
00:10 Open the parentheses carefully. Multiply each term inside by each term outside.
00:33 Next, calculate the products.
00:37 Then, reduce the expressions where possible.
00:41 Identify and mark the important variables.
00:45 Use the commutative law to rearrange the variables nicely.
00:54 Now, collect like terms to simplify further.
01:01 And there you have it! That's the solution.

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

(a+b)(3+ab)=? (a+b)(3+\frac{a}{b})=\text{?}

2

Step-by-step solution

To solve this problem, we'll use the distributive property, which states that for any numbers a a , b b , and c c , a(b+c)=ab+ac a(b+c) = ab + ac .

Let's break it down step by step:

Step 1: Apply the distributive property
We will expand the expression (a+b)(3+ab)(a+b)(3+\frac{a}{b}) by distributing the terms in (a+b)(a+b) over (3+ab)(3+\frac{a}{b}).

Step 2: Expand the expression
(a+b)(3+ab)(a+b)(3+\frac{a}{b}) expands as follows:

  • First, distribute aa to each term in (3+ab)(3 + \frac{a}{b}):
    • a3=3aa \cdot 3 = 3a
    • aab=a2ba \cdot \frac{a}{b} = \frac{a^2}{b}
  • Next, distribute bb to each term in (3+ab)(3 + \frac{a}{b}):
    • b3=3bb \cdot 3 = 3b
    • bab=ab \cdot \frac{a}{b} = a (because bb cancels with the denominator)

Step 3: Combine and simplify the results
Putting it all together, we have:

3a+a2b+3b+a3a + \frac{a^2}{b} + 3b + a

Simplify the expression by combining like terms:

  • 3a+a=4a3a + a = 4a

Thus, the simplified result is:

4a+a2b+3b4a + \frac{a^2}{b} + 3b

Therefore, the solution to the problem is 4a+a2b+3b 4a + \frac{a^2}{b} + 3b .

3

Final Answer

4a+a2b+3b 4a+\frac{a^2}{b}+3b

Key Points to Remember

Essential concepts to master this topic
  • Distributive Property: Each term in first binomial multiplies each term in second
  • Technique: aab=a2b a \cdot \frac{a}{b} = \frac{a^2}{b} and bab=a b \cdot \frac{a}{b} = a
  • Check: Count four terms before combining: 3a+a2b+3b+a 3a + \frac{a^2}{b} + 3b + a

Common Mistakes

Avoid these frequent errors
  • Forgetting to distribute to all terms
    Don't just multiply a×3=3a a \times 3 = 3a and stop = missing three terms! This gives incomplete expansion. Always ensure each term from the first binomial multiplies each term from the second binomial for all four products.

Practice Quiz

Test your knowledge with interactive questions

\( 3x+4x+7+2=\text{?} \)

FAQ

Everything you need to know about this question

Why do I get four terms when multiplying two binomials?

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Each term in the first binomial must multiply each term in the second binomial. Since (a+b) (a+b) has 2 terms and (3+ab) (3+\frac{a}{b}) has 2 terms, you get 2 × 2 = 4 products total!

How do I handle the fraction ab \frac{a}{b} when multiplying?

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Treat it like any other term! When you multiply a×ab a \times \frac{a}{b} , you get a2b \frac{a^2}{b} . When you multiply b×ab b \times \frac{a}{b} , the b's cancel to give you just a a .

Why does bab=a b \cdot \frac{a}{b} = a ?

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Think of it as bab \frac{b \cdot a}{b} . The b in the numerator cancels with the b in the denominator, leaving just a a . This is the same as bab=bab=a \frac{ba}{b} = \frac{\cancel{b}a}{\cancel{b}} = a !

How do I combine like terms in the final answer?

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Look for terms with the same variables and exponents. In this problem, 3a+a=4a 3a + a = 4a because both are just 'a' terms. The terms a2b \frac{a^2}{b} and 3b 3b can't be combined because they have different variables.

Can I check my answer by substituting numbers?

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Yes! Try a=2,b=1 a = 2, b = 1 . The original gives (2+1)(3+21)=3×5=15 (2+1)(3+\frac{2}{1}) = 3 \times 5 = 15 . Your answer gives 4(2)+221+3(1)=8+4+3=15 4(2) + \frac{2^2}{1} + 3(1) = 8 + 4 + 3 = 15

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