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Write a polynomial f(x) that satisfies the given conditions. Polynomial of lowest degree with zeros of -(1)/(6) (multiplicity 2 ) and (3)/(4) (multiplicity 1 ) and with f(0)=3. f(x)=

Write a polynomial f(x) f(x) that satisfies the given conditions.\newlinePolynomial of lowest degree with zeros of −16 -\frac{1}{6} (multiplicity 22 ) and 34 \frac{3}{4} (multiplicity 11 ) and with f(0)=3 f(0)=3 .\newlinef(x)= f(x)=

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Q. Write a polynomial f(x) f(x) that satisfies the given conditions.\newlinePolynomial of lowest degree with zeros of −16 -\frac{1}{6} (multiplicity 22 ) and 34 \frac{3}{4} (multiplicity 11 ) and with f(0)=3 f(0)=3 .\newlinef(x)= f(x)=
  1. Identify Zeros: To find the polynomial of the lowest degree with the given zeros, we need to write down the factors corresponding to each zero. The zero at −(1/6)-(1/6) with multiplicity 22 means we will have a factor of (x+1/6)2(x + 1/6)^2. The zero at (3/4)(3/4) with multiplicity 11 means we will have a factor of (x−3/4)(x - 3/4).
  2. Write Factors: Now we write down the factors as part of the polynomial: f(x)=(x+16)2⋅(x−34)f(x) = (x + \frac{1}{6})^2 \cdot (x - \frac{3}{4})
  3. Expand Factors: Next, we need to expand the factors to write the polynomial in standard form: f(x)=[(x+16)(x+16)]⋅(x−34)f(x) = [(x + \frac{1}{6})(x + \frac{1}{6})] \cdot (x - \frac{3}{4})
  4. Combine Terms: Expanding the squared factor:\newlinef(x)=[x2+(16)x+(16)x+(136)]⋅(x−34)f(x) = [x^2 + (\frac{1}{6})x + (\frac{1}{6})x + (\frac{1}{36})] \cdot (x - \frac{3}{4})\newlinef(x)=[x2+(13)x+(136)]⋅(x−34)f(x) = [x^2 + (\frac{1}{3})x + (\frac{1}{36})] \cdot (x - \frac{3}{4})
  5. Adjust Constant Term: Now we expand the remaining factors:\newlinef(x)=x3−34x2+13x2−14x+136x−136⋅34f(x) = x^3 - \frac{3}{4}x^2 + \frac{1}{3}x^2 - \frac{1}{4}x + \frac{1}{36}x - \frac{1}{36}\cdot\frac{3}{4}\newlinef(x)=x3−34x2+13x2−14x+136x−3144f(x) = x^3 - \frac{3}{4}x^2 + \frac{1}{3}x^2 - \frac{1}{4}x + \frac{1}{36}x - \frac{3}{144}\newlinef(x)=x3−912x2+412x2−312x+136x−148f(x) = x^3 - \frac{9}{12}x^2 + \frac{4}{12}x^2 - \frac{3}{12}x + \frac{1}{36}x - \frac{1}{48}
  6. Adjust Constant Term: Now we expand the remaining factors:\newlinef(x) = x3−34x2+13x2−14x+136x−136×34x^3 - \frac{3}{4}x^2 + \frac{1}{3}x^2 - \frac{1}{4}x + \frac{1}{36}x - \frac{1}{36}\times\frac{3}{4}\newlinef(x) = x3−34x2+13x2−14x+136x−3144x^3 - \frac{3}{4}x^2 + \frac{1}{3}x^2 - \frac{1}{4}x + \frac{1}{36}x - \frac{3}{144}\newlinef(x) = x3−912x2+412x2−312x+136x−148x^3 - \frac{9}{12}x^2 + \frac{4}{12}x^2 - \frac{3}{12}x + \frac{1}{36}x - \frac{1}{48} Combine like terms:\newlinef(x) = x3−512x2−212x−148x^3 - \frac{5}{12}x^2 - \frac{2}{12}x - \frac{1}{48}
  7. Adjust Constant Term: Now we expand the remaining factors:\newlinef(x)=x3−34x2+13x2−14x+136x−136⋅34f(x) = x^3 - \frac{3}{4}x^2 + \frac{1}{3}x^2 - \frac{1}{4}x + \frac{1}{36}x - \frac{1}{36}\cdot\frac{3}{4}\newlinef(x)=x3−34x2+13x2−14x+136x−3144f(x) = x^3 - \frac{3}{4}x^2 + \frac{1}{3}x^2 - \frac{1}{4}x + \frac{1}{36}x - \frac{3}{144}\newlinef(x)=x3−912x2+412x2−312x+136x−148f(x) = x^3 - \frac{9}{12}x^2 + \frac{4}{12}x^2 - \frac{3}{12}x + \frac{1}{36}x - \frac{1}{48}Combine like terms:\newlinef(x)=x3−512x2−212x−148f(x) = x^3 - \frac{5}{12}x^2 - \frac{2}{12}x - \frac{1}{48}To ensure that f(0)=3f(0) = 3, we need to adjust the constant term. Since the constant term is currently −148-\frac{1}{48}, we need to add 3+1483 + \frac{1}{48} to it to make f(0)f(0) equal to 33.
  8. Adjust Constant Term: Now we expand the remaining factors:\newlinef(x)=x3−34x2+13x2−14x+136x−136⋅34f(x) = x^3 - \frac{3}{4}x^2 + \frac{1}{3}x^2 - \frac{1}{4}x + \frac{1}{36}x - \frac{1}{36}\cdot\frac{3}{4}\newlinef(x)=x3−34x2+13x2−14x+136x−3144f(x) = x^3 - \frac{3}{4}x^2 + \frac{1}{3}x^2 - \frac{1}{4}x + \frac{1}{36}x - \frac{3}{144}\newlinef(x)=x3−912x2+412x2−312x+136x−148f(x) = x^3 - \frac{9}{12}x^2 + \frac{4}{12}x^2 - \frac{3}{12}x + \frac{1}{36}x - \frac{1}{48} Combine like terms:\newlinef(x)=x3−512x2−212x−148f(x) = x^3 - \frac{5}{12}x^2 - \frac{2}{12}x - \frac{1}{48} To ensure that f(0)=3f(0) = 3, we need to adjust the constant term. Since the constant term is currently −148-\frac{1}{48}, we need to add 3+1483 + \frac{1}{48} to it to make f(0)f(0) equal to 33.Calculate the necessary adjustment to the constant term:\newline3+148=14448+148=145483 + \frac{1}{48} = \frac{144}{48} + \frac{1}{48} = \frac{145}{48}
  9. Adjust Constant Term: Now we expand the remaining factors:\newlinef(x)=x3−34x2+13x2−14x+136x−136⋅34f(x) = x^3 - \frac{3}{4}x^2 + \frac{1}{3}x^2 - \frac{1}{4}x + \frac{1}{36}x - \frac{1}{36}\cdot\frac{3}{4}\newlinef(x)=x3−34x2+13x2−14x+136x−3144f(x) = x^3 - \frac{3}{4}x^2 + \frac{1}{3}x^2 - \frac{1}{4}x + \frac{1}{36}x - \frac{3}{144}\newlinef(x)=x3−912x2+412x2−312x+136x−148f(x) = x^3 - \frac{9}{12}x^2 + \frac{4}{12}x^2 - \frac{3}{12}x + \frac{1}{36}x - \frac{1}{48} Combine like terms:\newlinef(x)=x3−512x2−212x−148f(x) = x^3 - \frac{5}{12}x^2 - \frac{2}{12}x - \frac{1}{48} To ensure that f(0)=3f(0) = 3, we need to adjust the constant term. Since the constant term is currently −148-\frac{1}{48}, we need to add 3+1483 + \frac{1}{48} to it to make f(0)f(0) equal to 33. Calculate the necessary adjustment to the constant term:\newline3+148=14448+148=145483 + \frac{1}{48} = \frac{144}{48} + \frac{1}{48} = \frac{145}{48} Now we write the final polynomial with the adjusted constant term:\newlinef(x)=x3−34x2+13x2−14x+136x−3144f(x) = x^3 - \frac{3}{4}x^2 + \frac{1}{3}x^2 - \frac{1}{4}x + \frac{1}{36}x - \frac{3}{144}00

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