To factor the perfect square trinomials, we use the formula:
\[ a^2 + 2ab + b^2 = (a + b)^2 \]
Now, let's factor the given trinomials:
- \( x^2 + 6x + 9 \)
Here, \( a = x \) and \( b = 3 \) (because \( 3^2 = 9 \) and \( 2ab = 6x \) implies \( 2 \cdot x \cdot 3 = 6x \)). Thus, we have: \[ x^2 + 6x + 9 = (x + 3)^2 \]
- \( x^2 + 10x + 25 \)
Here, \( a = x \) and \( b = 5 \) (because \( 5^2 = 25 \) and \( 2ab = 10x \)). Thus, we have: \[ x^2 + 10x + 25 = (x + 5)^2 \]
- \( 9x^2 + 12x + 4 \)
Here, \( a = 3x \) and \( b = 2 \) (because \( 2^2 = 4 \) and \( 2ab = 12x \) implies \( 2 \cdot 3x \cdot 2 = 12x \)). Thus, we have: \[ 9x^2 + 12x + 4 = (3x + 2)^2 \]
- \( 9x^2 - 30x + 25 \)
Here, \( a = 3x \) and \( b = 5 \) (because \( 5^2 = 25 \) and \( 2ab = -30x \)). Thus, we have: \[ 9x^2 - 30x + 25 = (3x - 5)^2 \]
So the factorizations are:
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\( (x + 3)^2 \)
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\( (x + 5)^2 \)
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\( (3x + 2)^2 \)
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\( (3x - 5)^2 \)