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What is the sum of (w6+23w2) \left(w^6+\dfrac{2}{3}w^2\right) and (12w6+13w2+14) \left(\dfrac{1}{2}w^6+\dfrac{1}{3}w^2+\dfrac{1}{4}\right) ?

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Evaluate definite integrals using the chain ruleright chevron icon

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Q. What is the sum of (w6+23w2) \left(w^6+\dfrac{2}{3}w^2\right) and (12w6+13w2+14) \left(\dfrac{1}{2}w^6+\dfrac{1}{3}w^2+\dfrac{1}{4}\right) ?
  1. Add coefficients of w^66: Add the coefficients of the w6w^6 terms.\newlineWe have w6w^6 from the first polynomial and 12w6\frac{1}{2}w^6 from the second polynomial. Adding these together gives us:\newlinew6+12w6=22w6+12w6=32w6 w^6 + \frac{1}{2}w^6 = \frac{2}{2}w^6 + \frac{1}{2}w^6 = \frac{3}{2}w^6
  2. Add coefficients of w^22: Add the coefficients of the w2w^2 terms.\newlineWe have 23w2\frac{2}{3}w^2 from the first polynomial and 13w2\frac{1}{3}w^2 from the second polynomial. Adding these together gives us:\newline23w2+13w2=33w2=w2 \frac{2}{3}w^2 + \frac{1}{3}w^2 = \frac{3}{3}w^2 = w^2
  3. Add constant terms: Add the constant terms.\newlineThere is no constant term in the first polynomial, but there is a 14\frac{1}{4} in the second polynomial. Since there is nothing to add to 14\frac{1}{4}, it remains as it is.\newline0+14=14 0 + \frac{1}{4} = \frac{1}{4}
  4. Combine results for final sum: Combine the results from Steps 11, 22, and 33 to get the final sum.\newlineThe sum of the two polynomials is:\newline32w6+w2+14 \frac{3}{2}w^6 + w^2 + \frac{1}{4}

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