Q. Convert the expression \(x^2-1\) into factored form.
Answer
\(x^2 – 1\) is a difference of squares, since \(x^2 – 1^2\). Using the identity \(a^2 – b^2 = (a-b)(a+b)\) with \(a=x\) and \(b=1\):
\[
x^2 – 1 = (x-1)(x+1)
\]
Detailed Explanation
We want to factor the expression \(x^2 – 1\). This is a difference of squares.
Step 1: Recognize the pattern of a difference of squares.
The formula for a difference of squares is:
\[
a^2 – b^2 = (a-b)(a+b)
\]
Step 2: Match \(x^2 – 1\) to the pattern.
Here, \(x^2\) can be written as \((x)^2\), and \(1\) can be written as \((1)^2\). So we identify:
\[
a = x,\quad b = 1
\]
Step 3: Substitute into the difference of squares formula.
Using \(a^2 – b^2 = (a-b)(a+b)\), we get:
\[
x^2 – 1 = (x-1)(x+1)
\]
Final factored form:
\[
x^2 – 1 = (x-1)(x+1)
\]
See full solution
Algebra FAQ
What does \(x^2-1\) factor into?
\(x^2-1=(x-1)(x+1)\).
Is there a difference of squares pattern for \(x^2-1\)?
Yes. Use \(a^2-b^2=(a-b)(a+b)\) with \(a=x\), \(b=1\). So \((x-1)(x+1)\).
How do you verify the factorization \( (x-1)(x+1)\)?
Expand: \((x-1)(x+1)=x^2+x-x-1=x^2-1\).
What are the solutions to \(x^2-1=0\) using the factors?
\((x-1)(x+1)=0\) gives \(x-1=0\) or \(x+1=0\). So \(x=1\) or \(x=-1\).
Can you factor \(x^2-1\) over integers?
Yes. Since it matches \(a^2-b^2\), it factors over integers as \((x-1)(x+1)\).
What is the factored form of \(1-x^2\)?
\(1-x^2=-(x^2-1)=-(x-1)(x+1)=(x+1)(1-x)\).
Solve the factored form clearly.
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