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The expression \(16v^{2} - 9\) can be factored as \((4v - 3)(4v + 3)\).
Solución
We start with the expression
\[
16v^2 - 9.
\]
Notice that both terms are perfect squares:
\[
16v^2 = (4v)^2 \quad \text{and} \quad 9 = 3^2.
\]
This expression is a difference of squares, which can be factored using the formula
\[
a^2 - b^2 = (a - b)(a + b).
\]
Here, we have \(a = 4v\) and \(b = 3\). Thus, the factorization is
\[
(4v-3)(4v+3).
\]
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Simplifique esta solución The Deep Dive
To factor the expression \( 16v^{2}-9 \), we can recognize that it is a difference of squares. This expression can be rewritten as \( (4v)^{2} - 3^{2} \). According to the difference of squares formula, which states that \( a^{2} - b^{2} = (a - b)(a + b) \), we can apply it here: \[ 16v^{2} - 9 = (4v - 3)(4v + 3) \] So, the factored form of the expression is \( (4v - 3)(4v + 3) \).
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