Write a polynomial f(x) that satisfies the given conditions. Polynomial of lowest degree with zeros of 35 (multiplicity 2 ) and −32 (multiplicity 1 ) and with f(0)=−100.
Q. Write a polynomial f(x) that satisfies the given conditions. Polynomial of lowest degree with zeros of 35 (multiplicity 2 ) and −32 (multiplicity 1 ) and with f(0)=−100.
Identify Zeros and Factors: Identify the linear factors associated with the given zeros, considering their multiplicities. The zero (35) with multiplicity 2 gives us the factor (x−35)2, and the zero −(32) with multiplicity 1 gives us the factor (x+32).
Write Factored Form: Write the polynomial in its factored form using the identified factors. The polynomial is f(x)=(x−35)2⋅(x+32).
Expand and Simplify: Expand the polynomial to find the standard form. First, expand (x−35)2 to get x2−(2⋅35)x+(35)2=x2−(310)x+925.
Distribute Terms: Next, expand the entire polynomial f(x)=(x2−310x+925)∗(x+32).
Combine Like Terms: Distribute the terms to get f(x)=x3+(32)x2−(310)x2−(920)x+(925)x+(2750).
Adjust Constant Term: Combine like terms to simplify the polynomial. We get f(x)=x3−38x2+95x+2750.
Calculate New Constant: Now, we need to adjust the polynomial so that f(0)=−100. Substitute x=0 into the polynomial to find the current value of f(0). We get f(0)=03−(8/3)⋅02+(5/9)⋅0+(50/27)=2750.
Write Final Polynomial: To make f(0)=−100, we need to adjust the constant term. Since f(0) is currently 2750, we need to subtract (2750+100) from the constant term to get the desired value. Calculate the new constant term: (2750)−(2750+100)=−100.
Write Final Polynomial: To make f(0)=−100, we need to adjust the constant term. Since f(0) is currently 2750, we need to subtract (2750+100) from the constant term to get the desired value. Calculate the new constant term: (2750)−(2750+100)=−100.Write the final polynomial with the adjusted constant term. The polynomial is f(x)=x3−(38)x2+(95)x−100.
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