Let's factor each expression step by step.
1. Factor \(9b^2 - 16d^2\)
This expression is a difference of squares, which can be factored using the formula \(a^2 - b^2 = (a - b)(a + b)\).
Here, we can identify:
- \(a^2 = 9b^2\) → \(a = 3b\)
- \(b^2 = 16d^2\) → \(b = 4d\)
Applying the difference of squares formula: \[ 9b^2 - 16d^2 = (3b - 4d)(3b + 4d) \]
Final Answer:
\[ 9b^2 - 16d^2 = (3b - 4d)(3b + 4d) \]
2. Factor \(x^2 - 4x - 32\)
To factor this quadratic, we need to find two numbers that multiply to \(-32\) (the constant term) and add up to \(-4\) (the coefficient of \(x\)).
The numbers \(-8\) and \(4\) work because:
- \(-8 \times 4 = -32\)
- \(-8 + 4 = -4\)
Thus, we can factor the quadratic as: \[ x^2 - 4x - 32 = (x - 8)(x + 4) \]
Final Answer:
\[ x^2 - 4x - 32 = (x - 8)(x + 4) \]
3. Factor \(2b^2 - 0b - 15\)
This expression simplifies to \(2b^2 - 15\).
To factor this quadratic, let's look for two numbers that multiply to \(-15 \times 2 = -30\) (the product of the leading coefficient and the constant) and add to \(0\) (the coefficient of \(b\)).
The numbers \(5\) and \(-6\) work because:
- \(5 \times (-6) = -30\)
- \(5 - 6 = -1\) (this would be incorrect for our middle term, so let’s revise this approach)
Instead, since this has only two terms which are a difference of squares as well, we will use: \[ 2b^2 - 15 = 2(b^2 - \frac{15}{2}) \]
However, because the expression isn't a clean reduction using integers, we note this can also be reinterpreted as: \[ 2b^2 - 15 = \frac{1}{2}(4b^2 - 30) = (b - \sqrt{15})(b + \sqrt{15}) \]
If we want integers, we leave it as: The expression can remain: \[ 2b^2 - 15 = (2b - 5\sqrt{3})(2b + 5\sqrt{3}) \]
Final Answer:
\[ 2b^2 - 15 \text{ (not factorable into integers beyond quadratic)} \]
In conclusion, factoring it precisely pure, as polynomial roots will give indicating complex for pure integers. But please clarify if that is desired.
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