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Given that K is a polynomial of degree 3, K has a root of multiplicity 2 at v = 5, K(-3) = 0, and K(0) = -50.25. Find an algebraic equation for K.

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Final answer:

To find an algebraic equation for the polynomial K of degree 3, with a given root multiplicity and specific function values, we determine the coefficient 'a' after forming the general polynomial expression and then substituting the given function value at K(0).

Step-by-step explanation:

Given that K is a polynomial of degree 3 with a root of multiplicity 2 at v = 5, another root at v = -3 (since K(-3) = 0), and K(0) = -50.25, we can start forming the algebraic equation for K.

Since the root at v = 5 has multiplicity 2, we know that (v - 5)2 will be a factor of K. The fact that K(-3) = 0 shows that (v + 3) is another factor of K. As a cubic polynomial, we can now express K(v) as: K(v) = a(v + 3)(v - 5)2. To find the value of the coefficient a, we use the point K(0) = -50.25: a(0 + 3)(0 - 5)2 = -50.25, which simplifies to 75a = -50.25.

Solving for a, we divide both sides by 75 to get a = -50.25 / 75 which simplifies to a = -0.67. Therefore, the polynomial equation for K is K(v) = -0.67(v + 3)(v - 5)2, which can be expanded if necessary to find the standard form of the cubic equation.

User Pavel Katiushyn
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