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

\[
(x+3)(x+3) = x^2 + 6x + 9
\]

See full solution

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Algebra FAQ

What is \( (x+3)(x+3) \) simplified?

\( (x+3)^2 \)

Expand \( (x+3)^2 \).

\( (x+3)(x+3)=x^2+6x+9 \)

Which formula matches \( (a+b)^2 \) for this problem?

Use \( (a+b)^2=a^2+2ab+b^2 \) with \(a=x\), \(b=3\)

Why can \( (x+3)(x+3) \) be written as a square?

Same factor twice: \( (x+3)(x+3)=(x+3)^2 \)

What is the coefficient of \(x\) in \( x^2+6x+9 \)?

\(6\)

What is the constant term after expanding?

\(9\)

How do you verify your expansion quickly?

Multiply distributively: \(x(x+3)+3(x+3)=x^2+3x+3x+9=x^2+6x+9\)
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