Solve these problems:

Question

3(8x+4) -7(6x^2+2x-1)

(-5/2x^3y+2/3x^2y^3-1/6x^2y) divided by (-2/1x^2y)

1/3x^2+3 and X^2-6x+15

-3x^7+6x^4-18x^2+3 divided by (-1/3)

What fits into 5x^2-45?

Answer 1

Let's solve each problem one by one.

1. Problem: Simplify \(3(8x+4) - 7(6x^2+2x-1)\).

   Solution: \[ 3(8x + 4) = 24x + 12 \] \[ 7(6x^2 + 2x - 1) = 42x^2 + 14x - 7 \] So, we can rewrite the expression: \[ 3(8x + 4) - 7(6x^2 + 2x - 1) = (24x + 12) - (42x^2 + 14x - 7) \] Combine like terms: \[ = -42x^2 + 24x - 14x + 12 + 7 \] \[ = -42x^2 + 10x + 19 \] Final Answer: \(-42x^2 + 10x + 19\)

2. Problem: \(\frac{-\frac{5}{2}x^3y + \frac{2}{3}x^2y^3 - \frac{1}{6}x^2y}{-\frac{2}{1}x^2y}\).

   Solution: First, to simplify, we flip the division into multiplication: \[ = \left(-\frac{5}{2}x^3y + \frac{2}{3}x^2y^3 - \frac{1}{6}x^2y\right) \cdot \left(-\frac{1}{2}x^{-2}y^{-1}\right) \] This means distributing \(-\frac{1}{2}x^{-2}y^{-1}\):

   • For each term:
     • The first term: \(-\frac{5}{2}x^3y \cdot -\frac{1}{2}x^{-2}y^{-1} = \frac{5}{4}x^{3-2}y^{1-1} = \frac{5}{4}x\)
     • The second term: \(\frac{2}{3}x^2y^3 \cdot -\frac{1}{2}x^{-2}y^{-1} = -\frac{1}{3}y^{3-1} = -\frac{1}{3}y^{2}\)
     • The third term: \(-\frac{1}{6}x^2y \cdot -\frac{1}{2}x^{-2}y^{-1} = \frac{1}{12}y^{1-1} = \frac{1}{12}\)

   Combine: \[ \frac{5}{4}x - \frac{1}{3}y^2 + \frac{1}{12} \]

   Since this is an expression, we present it as: Final Answer: \(\frac{5}{4}x - \frac{1}{3}y^2 + \frac{1}{12}\)

3. Problem: Solve \( \frac{1}{3}x^2 + 3 \) and \( x^2 - 6x + 15 \).

   This looks like a two-part comparison without an operation mentioned. If you want to set the two equations equal, it would be: \[ \frac{1}{3}x^2 + 3 = x^2 - 6x + 15 \] setting them equal and rearranging to set to zero might be the intent.

   Rearranging gives us: \[ 0 = x^2 - \frac{1}{3}x^2 - 6x + 15 - 3 \] \[ 0 = \left(1 - \frac{1}{3}\right)x^2 - 6x + 12 \] \[ 0 = \frac{2}{3}x^2 - 6x + 12 \] Multiply through by 3 to clear the fraction: \[ 0 = 2x^2 - 18x + 36 \] Divide by 2: \[ 0 = x^2 - 9x + 18 \] Final Answer: The quadratic \(x^2 - 9x + 18\) can be solved using the quadratic formula or factoring directly.

4. Problem: What fits into \(5x^2 - 45\)?

   Solution: We can factor \(5x^2 - 45\): \[ 5(x^2 - 9) \] Recognizing \(x^2 - 9\) as a difference of squares, we can factor further: \[ 5(x - 3)(x + 3) \]

   Final Answer: \(5(x - 3)(x + 3)\) is the factored form of \(5x^2 - 45\).

Please let me know if you need further clarification on any of these or additional assistance!