Q. \(x^{2}-4x+4\)
Answer
\(x^2-4x+4\) is a perfect square:
\[x^2-4x+4=(x-2)^2.\]
So the factorization is \((x-2)(x-2)\), and the vertex/double root is at \(x=2\).
Detailed Explanation
We are given the expression \(x^2 – 4x + 4\). The goal is typically to rewrite it in a simpler factored or completed-square form.
Step 1: Recognize a perfect square pattern
Notice that the first term is \(x^2\), and the last term is \(4\), which is \(2^2\). Also, the middle term is \(-4x\), which matches \(-2 \cdot x \cdot 2\). This suggests the expression is a perfect square of the form:
\[
x^2 – 4x + 4 = (x – 2)^2
\]
Step 2: Verify by expanding
Expand \((x – 2)^2\) to confirm it matches the original expression.
\[
(x – 2)^2 = (x – 2)(x – 2)
\]
Distribute each term:
\[
(x – 2)(x – 2) = x(x – 2) – 2(x – 2)
\]
Now distribute again:
\[
x(x – 2) = x^2 – 2x
\]
\[
-2(x – 2) = -2x + 4
\]
Add the results:
\[
x^2 – 2x – 2x + 4 = x^2 – 4x + 4
\]
Conclusion
Since the expansion matches the original expression, we conclude:
\[
x^2 – 4x + 4 = (x – 2)^2
\]
Graph
Algebra FAQ
Factor \(x^2-4x+4\)?
Complete the square for \(x^2-4x+4\)?
Find the roots of \(x^2-4x+4=0\)?
What is the vertex and minimum value of \(y=x^2-4x+4\)?
Expand \((x-2)^2\) to match \(x^2-4x+4\)?
Determine whether \(x^2-4x+4\) is always nonnegative?
Try three tools for instant help.
Math, Geometry, Trigonometry, etc.