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Find a polynomial function of degree 3 with -1, 3, 5 as zeros. A) f(x) = x^3 - 7x^2 + 7x + 15 B) f(x) = 3x^3 + 7x^2 + 7x - 15 C) f(x) = x^3 + 9x^2 + 7x - 3 D) f(x) = x^3 + 7x^2 + 7x - 15
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Ok if it has the zeros, -1, 3 and 5, that means it will be (x+1)(x-3)(x-5)
\(\large { x=\{ -1, 3, 5\}\to \begin{cases} x=-1\to {\color{brown}{ x+1}}=0\\ x=3\to {\color{brown}{ x-3}}=0\\ x=5\to {\color{brown}{ x-5}}=0 \end{cases} \\ \quad \\ ({\color{brown}{ x+1}})({\color{brown}{ x-3}})({\color{brown}{ x-5}})=0\\ \quad \\\implies ({\color{brown}{ x+1}})({\color{brown}{ x-3}})({\color{brown}{ x-5}})=\textit{original polynomial} }\)
Now you multiply them and you have the function
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