Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. 4, -14, and 5 + 8i
For an equation, if a+bi is a solution, a-bi is also a solution. So we have these four solutions; 4, -14, 5+8i, 5-8i Then, the polynomial equation with those 0's would be.. a(x-4)(x+14)(x-(5+8i))(x-(5-8i)) (a is constant that is not equal to 0) If we foil it, it becomes \[ax ^{4}-67ax ^{2}+1450ax-4984a\]. If a becomes negative, -4984a will be way too big, so it has to be positive. So, the smallest possible value for a is 1. So, the equation becomes... \[x ^{4}-67x ^{2}+1450x-4984\]
Thank you so much. I knew that the top part but then I started getting confused with all the signs. Again thank you
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