To factor the given polynomials using the difference of squares, we can use the formula \( a^2 - b^2 = (a - b)(a + b) \).
Let's factor each binomial step by step:
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Binomial: \( 9x^2 - 25 \)
Here, \( a^2 = 9x^2 \) and \( b^2 = 25 \).
Thus, \( a = 3x \) and \( b = 5 \).
Factored form:
\[ 9x^2 - 25 = (3x - 5)(3x + 5) \] -
Binomial: \( 9 - 25x^2 \)
Here, \( a^2 = 9 \) and \( b^2 = 25x^2 \).
Thus, \( a = 3 \) and \( b = 5x \).
Factored form:
\[ 9 - 25x^2 = (3 - 5x)(3 + 5x) \] -
Binomial: \( 4x^2 - 36 \)
Here, \( a^2 = 4x^2 \) and \( b^2 = 36 \).
Thus, \( a = 2x \) and \( b = 6 \).
Factored form:
\[ 4x^2 - 36 = (2x - 6)(2x + 6) \] We can further factor it using \( 2 \) as a common factor:
\[ = 2(2x - 6)(2x + 6) = 2(2(x - 3))(2(x + 3)) = 4(x - 3)(x + 3) \] -
Binomial: \( 81 - 49x^2 \)
Here, \( a^2 = 81 \) and \( b^2 = 49x^2 \).
Thus, \( a = 9 \) and \( b = 7x \).
Factored form:
\[ 81 - 49x^2 = (9 - 7x)(9 + 7x) \] -
Binomial: \( 81x^2 - 49 \)
Here, \( a^2 = 81x^2 \) and \( b^2 = 49 \).
Thus, \( a = 9x \) and \( b = 7 \).
Factored form:
\[ 81x^2 - 49 = (9x - 7)(9x + 7) \] -
Binomial: \( 4x^2 + 36 \)
This is a sum of squares, not a difference, and cannot be factored using the difference of squares formula. So, it is already in its simplest form.
In summary, the factored forms for the provided binomials are:
- \( 9x^2 - 25 = (3x - 5)(3x + 5) \)
- \( 9 - 25x^2 = (3 - 5x)(3 + 5x) \)
- \( 4x^2 - 36 = 4(x - 3)(x + 3) \)
- \( 81 - 49x^2 = (9 - 7x)(9 + 7x) \)
- \( 81x^2 - 49 = (9x - 7)(9x + 7) \)
- \( 4x^2 + 36 \) cannot be factored using difference of squares.