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Multiply. \[ (x+5 w+2)(4 x-5 w) \] Simplify. your answer. पI

Ask by Weston Herrera. in the United States
Jan 27,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The simplified expression is \(4x^2 + 15wx - 25w^2 + 8x - 10w\).

Solution

Sure, let's multiply and simplify the expression: \[ (x + 5w + 2)(4x - 5w) \] ### Step 1: Distribute Each Term Multiply each term in the first polynomial by each term in the second polynomial: \[ \begin{align*} x \cdot 4x &= 4x^2 \\ x \cdot (-5w) &= -5wx \\ 5w \cdot 4x &= 20wx \\ 5w \cdot (-5w) &= -25w^2 \\ 2 \cdot 4x &= 8x \\ 2 \cdot (-5w) &= -10w \\ \end{align*} \] ### Step 2: Combine Like Terms Now, add all the products together and combine like terms: \[ 4x^2 - 5wx + 20wx - 25w^2 + 8x - 10w \] Combine the \( wx \) terms: \[ -5wx + 20wx = 15wx \] So, the expression simplifies to: \[ 4x^2 + 15wx - 25w^2 + 8x - 10w \] ### Final Simplified Form \[ 4x^2 + 15wx - 25w^2 + 8x - 10w \]

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Beyond the Answer

To multiply the expressions \((x+5)(w+2)(4x-5w)\), we'll apply the distributive property, also known as the FOIL method for binomials. First, distribute \((x + 5)(w + 2)\): \[ = x(w + 2) + 5(w + 2) \\ = xw + 2x + 5w + 10 \] Now, we have \(xw + 2x + 5w + 10\). Next, we'll multiply this result by \((4x - 5w)\): \[ = (xw + 2x + 5w + 10)(4x - 5w) \] Distributing \(4x\) and \(-5w\) to each term: \[ = 4x(xw) + 4x(2x) + 4x(5w) + 4x(10) - 5w(xw) - 5w(2x) - 5w(5w) - 5w(10) \] This results in: \[ = 4x^2w + 8x^2 + 20x + 40x - 5w^2x - 10wx - 25w^2 - 50w \] Now, combine like terms: \[ = (4x^2w - 5w^2x) + (8x^2 - 10wx) + (20x - 25w^2 - 50w) \] Putting together common terms, our final expression simplifies to: \[ = 4x^2w - 5w^2x + 8x^2 - 10wx + 20x - 25w^2 - 50w \] This is a simplified version of the product \((x+5)(w+2)(4x-5w)\).

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7. Efectueaza sis scrie rezultatul sub formă de putere: \( \begin{array}{lll}\text { a) } \frac{18}{5} \cdot\left(\frac{18}{5}\right)^{2}= & \text { b) }\left(\frac{6}{5}\right)^{2} \cdot\left(\frac{6}{5}\right)^{3} \cdot \frac{6}{5}= & \text { c) }\left(\frac{19}{5}\right)^{5} \cdot\left(\frac{19}{5}\right)^{16}= \\ \begin{array}{lll}\text { d) } \frac{3}{2} \cdot\left(\frac{3}{2}\right)^{3} \cdot\left(\frac{3}{2}\right)^{0} \cdot\left(\frac{3}{2}\right)^{4}= & \text { e) }\left[\left(\frac{28}{5}\right)^{2}\right]^{3}= & \text { f) }\left[\left(\frac{5}{6}\right)^{6}\right]^{7}= \\ \text { g) }\left[\left(\frac{24}{5}\right)^{2} \cdot\left(\frac{24}{5}\right)^{3}\right]^{8}= & \text { h) }\left[\frac{5}{7} \cdot\left(\frac{5}{7}\right)^{0} \cdot\left(\frac{5}{7}\right)^{4}\right]^{5}= & \text { i) }\left(\frac{29}{10}\right)^{10}:\left(\frac{29}{10}\right)^{7}=\end{array} \\ \left.\left.\begin{array}{lll}\text { j) }\left(\frac{1}{3}\right)^{17}: \frac{1}{3}= & \left.\text { k) }\left(\frac{3}{7}\right)^{11} \cdot\left(\frac{9}{49}\right)^{3}:\left(\frac{3}{7}\right)^{15}=1\right)\end{array}\right]\left(1 \frac{1}{2}\right)^{2}\right]^{8}:\left(\frac{3}{2}\right)^{13}= \\ \text { m) }\left(\frac{9}{10}\right)^{7} \cdot\left(\frac{1}{5}\right)^{7}= & \text { n) }\left(\frac{5}{2}\right)^{10} \cdot\left(\frac{8}{5}\right)^{10}: 2^{10}= & \text { o) } 9^{3} \cdot\left(\frac{7}{10}\right)^{3}:\left(\frac{63}{10}\right)^{3}= \\ \text { p) }\left[\left(\frac{1}{5}\right)^{7}\right]^{2} \cdot 6^{14}:\left(\frac{6}{5}\right)^{14}= & \text { q) }\left(\frac{5}{2}\right)^{7}:\left(\frac{5}{2}\right)^{5}= & \end{array} \)
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7. Efectueaza sis scrie rezultatul sub formă de putere: \( \begin{array}{lll}\text { a) } \frac{18}{5} \cdot\left(\frac{18}{5}\right)^{2}= & \text { b) }\left(\frac{6}{5}\right)^{2} \cdot\left(\frac{6}{5}\right)^{3} \cdot \frac{6}{5}= & \text { c) }\left(\frac{19}{5}\right)^{5} \cdot\left(\frac{19}{5}\right)^{16}= \\ \begin{array}{lll}\text { d) } \frac{3}{2} \cdot\left(\frac{3}{2}\right)^{3} \cdot\left(\frac{3}{2}\right)^{0} \cdot\left(\frac{3}{2}\right)^{4}= & \text { e) }\left[\left(\frac{28}{5}\right)^{2}\right]^{3}= & \text { f) }\left[\left(\frac{5}{6}\right)^{6}\right]^{7}= \\ \text { g) }\left[\left(\frac{24}{5}\right)^{2} \cdot\left(\frac{24}{5}\right)^{3}\right]^{8}= & \text { h) }\left[\frac{5}{7} \cdot\left(\frac{5}{7}\right)^{0} \cdot\left(\frac{5}{7}\right)^{4}\right]^{5}= & \text { i) }\left(\frac{29}{10}\right)^{10}:\left(\frac{29}{10}\right)^{7}=\end{array} \\ \left.\left.\begin{array}{lll}\text { j) }\left(\frac{1}{3}\right)^{17}: \frac{1}{3}= & \left.\text { k) }\left(\frac{3}{7}\right)^{11} \cdot\left(\frac{9}{49}\right)^{3}:\left(\frac{3}{7}\right)^{15}=1\right)\end{array}\right]\left(1 \frac{1}{2}\right)^{2}\right]^{8}:\left(\frac{3}{2}\right)^{13}= \\ \text { m) }\left(\frac{9}{10}\right)^{7} \cdot\left(\frac{1}{5}\right)^{7}= & \text { n) }\left(\frac{5}{2}\right)^{10} \cdot\left(\frac{8}{5}\right)^{10}: 2^{10}= & \text { o) } 9^{3} \cdot\left(\frac{7}{10}\right)^{3}:\left(\frac{63}{10}\right)^{3}= \\ \text { p) }\left[\left(\frac{1}{5}\right)^{7}\right]^{2} \cdot 6^{14}:\left(\frac{6}{5}\right)^{14}= & \text { q) }\left(\frac{5}{2}\right)^{7}:\left(\frac{5}{2}\right)^{5}= & \end{array} \)
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