Q. \((x+3)^2\)
Answer
We expand \((x+3)(x+3)\) as a square:
\[
(x+3)^2 = x^2 + 2\cdot x \cdot 3 + 3^2
\]
So,
\[
(x+3)^2 = x^2 + 6x + 9
\]
Detailed Explanation
We are asked to simplify the expression \((x+3)(x+3)\). This means we multiply the two binomials together.
Step 1: Recognize the expression
The expression \((x+3)(x+3)\) is the same factor written twice. So it is a “square of a binomial.” We can treat it like \((x+3)^2\).
Step 2: Use the distributive property (FOIL)
Multiply each term in the first parentheses by each term in the second parentheses:
\((x+3)(x+3)\) has these products:
- \(x \cdot x = x^2\)
- \(x \cdot 3 = 3x\)
- \(3 \cdot x = 3x\)
- \(3 \cdot 3 = 9\)
Step 3: Write all the terms together
Add the results from all four products:
\[
\begin{aligned}
(x+3)(x+3) &= x^2 + 3x + 3x + 9
\end{aligned}
\]
Step 4: Combine like terms
The like terms are \(3x\) and \(3x\). Combine them:
\[
\begin{aligned}
x^2 + 3x + 3x + 9 &= x^2 + 6x + 9
\end{aligned}
\]
Final Answer
\[
(x+3)(x+3) = x^2 + 6x + 9
\]
Graph
Algebra FAQ
What is \( (x+3)(x+3) \) simplified?
Expand \( (x+3)^2 \).
Which formula matches \( (a+b)^2 \) for this problem?
Why can \( (x+3)(x+3) \) be written as a square?
What is the coefficient of \(x\) in \( x^2+6x+9 \)?
What is the constant term after expanding?
How do you verify your expansion quickly?
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