Question

Factor completely.
\[81x^2+180x+100=\]

Answer 1

To factor the expression \(81x^2 + 180x + 100\), we will start by identifying if it can be expressed in the form of a perfect square trinomial.

The expression is of the form \(ax^2 + bx + c\), where:
- \(a = 81\)
- \(b = 180\)
- \(c = 100\)

Next, we can check if this trinomial is a perfect square. A perfect square trinomial follows the formula:
\[
(a + b)^2 = a^2 + 2ab + b^2
\]

We need to check if there exists some \(A\) and \(B\) such that:
1. \(A^2 = 81\)
2. \(B^2 = 100\)
3. \(2AB = 180\)

From \(A^2 = 81\), we find:
\[
A = 9 \quad (\text{since } 9^2 = 81)
\]

From \(B^2 = 100\), we find:
\[
B = 10 \quad (\text{since } 10^2 = 100)
\]

Now we check:
\[
2AB = 2 \cdot 9 \cdot 10 = 180
\]

Since all conditions are satisfied, we can express our original trinomial as a perfect square:
\[
81x^2 + 180x + 100 = (9x + 10)^2
\]

Thus, the completely factored form is:
\[
\boxed{(9x + 10)^2}
\]