Question

Factor the given polynomials using difference of squares.

Binomial: 9x^2 - 25
Binomial: 9 - 25x^2
Binomial: 4x^2 -36
Binomial: 81-49x^2
Binomial: 81x^2 -49
Binomial: 4x^2 + 36

Answer 1

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:

1. **Binomial: \( 9x^2 - 25 \)**
Here, \( a^2 = 9x^2 \) and \( b^2 = 25 \).
Thus, \( a = 3x \) and \( b = 5 \).
Factored form:
\[
9x^2 - 25 = (3x - 5)(3x + 5)
\]

2. **Binomial: \( 9 - 25x^2 \)**
Here, \( a^2 = 9 \) and \( b^2 = 25x^2 \).
Thus, \( a = 3 \) and \( b = 5x \).
Factored form:
\[
9 - 25x^2 = (3 - 5x)(3 + 5x)
\]

3. **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:
\[
= 2(2x - 6)(2x + 6) = 2(2(x - 3))(2(x + 3)) = 4(x - 3)(x + 3)
\]

4. **Binomial: \( 81 - 49x^2 \)**
Here, \( a^2 = 81 \) and \( b^2 = 49x^2 \).
Thus, \( a = 9 \) and \( b = 7x \).
Factored form:
\[
81 - 49x^2 = (9 - 7x)(9 + 7x)
\]

5. **Binomial: \( 81x^2 - 49 \)**
Here, \( a^2 = 81x^2 \) and \( b^2 = 49 \).
Thus, \( a = 9x \) and \( b = 7 \).
Factored form:
\[
81x^2 - 49 = (9x - 7)(9x + 7)
\]

6. **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:

1. \( 9x^2 - 25 = (3x - 5)(3x + 5) \)
2. \( 9 - 25x^2 = (3 - 5x)(3 + 5x) \)
3. \( 4x^2 - 36 = 4(x - 3)(x + 3) \)
4. \( 81 - 49x^2 = (9 - 7x)(9 + 7x) \)
5. \( 81x^2 - 49 = (9x - 7)(9x + 7) \)
6. \( 4x^2 + 36 \) cannot be factored using difference of squares.