Q. \(\,x^2 – x\,\)

Q: Can x^2-x be factored by grouping?

Not needed. First factor out x: x^2-x=x(x-1). Grouping would still reduce to the same result.

Q: What are the zeros of x^2-x?

Solve x^2-x=0\Rightarrow x(x-1)=0. So x=0 or x=1.

Q: What is the vertex of y=x^2-x and its minimum value?

For y=ax^2+bx+c with a=1,b=-1: x=-\frac{b}{2a}=\frac{1}{2}. y\left(\frac{1}{2}\right)=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}.

Q: What is x^2-x rewritten in completed-square form?

x^2-x=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}.

Q: What is the derivative of x^2-x?

Use power rule: \frac{d}{dx}(x^2-x)=2x-1.

Answer

We factor the quadratic by finding two numbers that multiply to \( -1 \) and add to \( -1 \). Those numbers are \( -1 \) and \( 1 \).

\[
x^2 – x = x(x-1)
\]

Detailed Explanation

We want to simplify the expression \(x^2 – x\).

Step 1: Factor out the greatest common factor.
Both terms \(x^2\) and \(-x\) share a common factor of \(x\).

Write each term using the shared factor \(x\):

\[x^2 – x = x\cdot x – x\cdot 1\]

Step 2: Factor using distributive property in reverse.
The expression \(x\cdot x – x\cdot 1\) can be rewritten as \(x\) times \((x – 1)\).

\[x\cdot x – x\cdot 1 = x(x – 1)\]

Final Answer:

\[x^2 – x = x(x – 1)\]

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Algebra FAQ

What are the factors of \(x^2-x\)?

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

Can \(x^2-x\) be factored by grouping?

Not needed. First factor out \(x\): \(x^2-x=x(x-1)\). Grouping would still reduce to the same result.

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

Solve \(x^2-x=0\Rightarrow x(x-1)=0\). So \(x=0\) or \(x=1\).

What is the vertex of \(y=x^2-x\) and its minimum value?

For \(y=ax^2+bx+c\) with \(a=1,b=-1\): \(x=-\frac{b}{2a}=\frac{1}{2}\). \(y\left(\frac{1}{2}\right)=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}\).

What is \(x^2-x\) rewritten in completed-square form?

\(x^2-x=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\).

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

Use power rule: \(\frac{d}{dx}(x^2-x)=2x-1\).

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