Q. \(x^2 + 2x\)

Q: Solve x^2+2x=8.

x^2+2x-8=0. Factoring gives (x+4)(x-2)=0 so x=2 or x=-4

Q: Find the vertex and minimum of y=x^2+2x.

Vertex at x=-1. Using y=(x+1)^2-1, the minimum value is -1

Answer

We factor the expression \(x^2 + 2x\) by taking out the common factor \(x\):

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

Detailed Explanation

We want to simplify the expression \(x^2+2x\). A common first step is to factor out the greatest common factor.

Step 1: Identify the greatest common factor (GCF).

Both terms \(x^2\) and \(2x\) contain a factor of \(x\). So the GCF is \(x\).

Step 2: Factor out the GCF.

Factor \(x\) out of both terms:

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

Step 3: Write the factored form as a single product.

Now combine what remains inside the parentheses:

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

Final Answer:

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

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Algebra FAQ

Factor \(x^2+2x\).

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

Find the roots of \(x^2+2x=0\).

\(x(x+2)=0\) so \(x=0\) or \(x=-2\)

Solve \(x^2+2x=8\).

\(x^2+2x-8=0\). Factoring gives \((x+4)(x-2)=0\) so \(x=2\) or \(x=-4\)

Complete the square for \(x^2+2x\).

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

Find the vertex and minimum of \(y=x^2+2x\).

Vertex at \(x=-1\). Using \(y=(x+1)^2-1\), the minimum value is \(-1\)

Expand \(x(x+2)\) to verify.

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

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