Q. \[ (x+1)^2 \]

Q: What is the derivative of (x+1)^2?

\frac{d}{dx}(x+1)^2=2(x+1)=2x+2.

Q: How do I solve (x+1)^2=0?

(x+1)^2=0 \Rightarrow x+1=0 \Rightarrow x=-1 (double root).

Q: What are the zeros of (x+1)^2?

The zero is x=-1 with multiplicity 2.

Answer

Expand using \((a+b)^2=a^2+2ab+b^2\) with \(a=x\) and \(b=1\).

\[
(x+1)^2=x^2+2x+1
\]

Detailed Explanation

We want to expand the expression \( (x+1)^2 \). The exponent \(2\) means we multiply the quantity by itself.

Step 1: Rewrite the square as a product.

\[
(x+1)^2 = (x+1)(x+1)
\]

Step 2: Distribute (multiply) the first factor across the second factor.

Distribute \(x\) to both terms in \( (x+1) \), and then distribute \(1\) to both terms in \( (x+1) \):

\[
(x+1)(x+1) = x(x+1) + 1(x+1)
\]

Step 3: Expand each smaller product.

\[
x(x+1) = x\cdot x + x\cdot 1 = x^2 + x
\]

\[
1(x+1) = 1\cdot x + 1\cdot 1 = x + 1
\]

Step 4: Combine the results.

\[
(x+1)(x+1) = (x^2 + x) + (x + 1)
\]

Step 5: Combine like terms.

The like terms are the \(x\) terms: \(x + x = 2x\). So we get:

\[
(x+1)^2 = x^2 + 2x + 1
\]

Final Answer:

\[
(x+1)^2 = x^2 + 2x + 1
\]

See full solution

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Algebra FAQ

How do I expand \((x+1)^2\)?

\((x+1)^2=x^2+2x+1\).

What is \((x+1)(x+1)\)?

\((x+1)(x+1)=(x+1)^2=x^2+2x+1\).

How do I rewrite \((x+1)^2\) in standard quadratic form?

Standard form is \(x^2+2x+1\).

What is the derivative of \((x+1)^2\)?

\(\frac{d}{dx}(x+1)^2=2(x+1)=2x+2\).

How do I solve \((x+1)^2=0\)?

\((x+1)^2=0 \Rightarrow x+1=0 \Rightarrow x=-1\) (double root).

What are the zeros of \((x+1)^2\)?

The zero is \(x=-1\) with multiplicity \(2\).

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