Q. \(x^2 + 6x + 9\)
Answer
Rewrite the quadratic as a perfect square:
\[
x^2+6x+9=(x+3)^2
\]
So the expression factors as
\[
(x+3)(x+3)
\]
Final result:
\[
x^2+6x+9=(x+3)^2
\]
Detailed Explanation
We want to simplify the expression \(x^2 + 6x + 9\) step-by-step.
Step 1: Recognize the expression as a perfect square trinomial.
A perfect square trinomial has the form:
\[
a^2 + 2ab + b^2
\]
For \(x^2 + 6x + 9\), compare terms:
- \(x^2\) matches \(a^2\), so \(a = x\).
- \(9\) matches \(b^2\), so \(b = 3\) (since \(3^2 = 9\)).
- Then \(2ab = 2(x)(3) = 6x\), which matches the middle term.
Step 2: Rewrite as a perfect square.
Since it matches the pattern \(a^2 + 2ab + b^2 = (a+b)^2\), we have:
\[
x^2 + 6x + 9 = (x+3)^2
\]
Final Answer:
\[
x^2 + 6x + 9 = (x+3)^2
\]
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Algebra FAQ
What is the factorization of \(x^2+6x+9\)?
\(x^2+6x+9=(x+3)^2\)
Complete the square for \(x^2+6x+9\).
\(x^2+6x+9=(x+3)^2\) because \(x^2+6x+9=x^2+6x+9=(x+3)^2\)
Solve \(x^2+6x+9=0\).
\((x+3)^2=0\) so \(x=-3\) (double root)
What are the roots and their multiplicity?
The only root is \(x=-3\) with multiplicity \(2\) since the polynomial is a perfect square
Find the vertex of \(y=x^2+6x+9\).
Vertex is at \(x=-\frac{b}{2a}=-\frac{6}{2}=-3\), and \(y=(-3+3)^2=0\) so \((-3,0)\)
Determine whether the quadratic opens up or down.
Since \(a=1>0\), it opens upward, and the minimum value is \(0\) at \(x=-3\)
Solve (x^2+6x+9) here.
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