1 - The given polynomial has degree 4 and positive leading coefficient and the graph should rise on the left and right sides.
2-p(0) = 1, graph shows negative value.
3 - The equation x - x² + 1 = 0 has no solution which suggests that the polynomial p(x) = x¹ - x² + 1 has no zeros. The graph shows x intercepts.
4- The graph has a zero of multiplicity 2, a zero of multiplicity 3 and and a zero of multiplicity 1. So t degree of the graphed polynomial should be at least 6. The given polynomial has degree 4.
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Answer:
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Solution
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1 - The given polynomial has degree 4 and positive leading coefficient and the graph should rise on the left and right sides.
2-p(0) = 1, graph shows negative value.
3 - The equation x - x² + 1 = 0 has no solution which suggests that the polynomial p(x) = x¹ - x² + 1 has no zeros. The graph shows x intercepts.
4- The graph has a zero of multiplicity 2, a zero of multiplicity 3 and and a zero of multiplicity 1. So t degree of the graphed polynomial should be at least 6. The given polynomial has degree 4.
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