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(6) \( \quad(-2 x-4 y)(3 x+5 y) \cdot 2 \) (8) \( \quad\left(4 b^{2}+5 c^{2}\right)\left(-b^{3}+2 c^{5}\right) \)

Ask by Boyd Ward. in South Africa
Jan 23,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

\[ -12x^2 - 44xy - 40y^2 \] \[ -4b^{5} - 5b^{3}c^{2} + 8b^{2}c^{5} + 10c^{7} \]

Solution

Let's expand and simplify each of the given expressions step by step. --- ### Problem (6) \[ (-2x - 4y)(3x + 5y) \cdot 2 \] **Step 1: Expand the binomials** \[ \begin{align*} (-2x)(3x) &= -6x^2 \\ (-2x)(5y) &= -10xy \\ (-4y)(3x) &= -12xy \\ (-4y)(5y) &= -20y^2 \end{align*} \] Combine the like terms: \[ -6x^2 - 10xy - 12xy - 20y^2 = -6x^2 - 22xy - 20y^2 \] **Step 2: Multiply by 2** \[ 2 \cdot (-6x^2 - 22xy - 20y^2) = -12x^2 - 44xy - 40y^2 \] **Final Answer for Problem (6):** \[ -12x^2 - 44xy - 40y^2 \] --- ### Problem (8) \[ (4b^{2} + 5c^{2})(-b^{3} + 2c^{5}) \] **Step 1: Expand the binomials** \[ \begin{align*} 4b^{2} \cdot (-b^{3}) &= -4b^{5} \\ 4b^{2} \cdot 2c^{5} &= 8b^{2}c^{5} \\ 5c^{2} \cdot (-b^{3}) &= -5b^{3}c^{2} \\ 5c^{2} \cdot 2c^{5} &= 10c^{7} \end{align*} \] **Step 2: Combine all terms** \[ -4b^{5} - 5b^{3}c^{2} + 8b^{2}c^{5} + 10c^{7} \] **Final Answer for Problem (8):** \[ -4b^{5} - 5b^{3}c^{2} + 8b^{2}c^{5} + 10c^{7} \] --- These are the fully expanded forms of the given expressions.

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Beyond the Answer

To tackle the first expression, \((-2x - 4y)(3x + 5y) \cdot 2\), you can start by applying the distributive property (also known as the FOIL method for binomials). First, distribute \(-2x\) across \((3x + 5y)\) and then do the same with \(-4y\). After carrying out the multiplications, combine like terms and finally multiply the result by 2. Now moving to the second expression, \((4b^2 + 5c^2)(-b^3 + 2c^5)\), apply the distributive property again. Multiply each term in the first polynomial by each term in the second polynomial. Be careful to account for all combinations, and then combine any like terms resulting from this multiplication. This will give you a tidy polynomial that encapsulates the initial expressions. Let’s crunch those numbers!

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