# Find a cubic polynomial function f with real coefficients that has the given complex zeros and x-intercept. Complex Zeroes. x=2\pm\sqrt{2}i

Find a cubic polynomial function f with real coefficients that has the given complex zeros and x-intercept. (There are many correct answers. Expand your answer completely.)
Complex Zeroes
$$\displaystyle{x}={2}\pm\sqrt{{{2}}}{i}$$
X-Intercept (5,0)

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tabuordg
zeros of cubic polynomial is given.
complex zeros are in pair and one real.
complex zeros are $$\displaystyle{2}+\sqrt{{{2}}}{i},{2}-\sqrt{{{2}}}{i}$$
real zeros is 5.
then factors of the polynomial will be
$$\displaystyle{x}-{5},{x}-{\left({2}+\sqrt{{{2}}}{i}\right)},{x}-{\left({2}-\sqrt{{{2}}}{i}\right)}$$
equation will be
$$\displaystyle{\left({x}-{5}\right)}{\left\lbrace{x}-{\left({2}+\sqrt{{{2}}}{i}\right)}\right\rbrace}{\left\lbrace{x}-{\left({2}-\sqrt{{{2}}}{i}\right)}\right\rbrace}={0}$$
$$\displaystyle{\left({x}-{5}\right)}{\left({x}^{{2}}-{x}{\left({2}-\sqrt{{{2}}}{i}\right)}-{x}{\left({2}+\sqrt{{2}}{i}\right)}+{\left({2}+\sqrt{{2}}{i}\right)}{\left({2}-\sqrt{{2}}{i}\right)}\right)}={0}$$
$$\displaystyle{\left({x}-{5}\right)}{\left({x}^{{2}}-{x}{\left({2}+\sqrt{{2}}{i}+{2}-\sqrt{{2}}{i}\right)}+{2}^{{2}}-{\left(\sqrt{{2}}{i}\right)}^{{2}}\right)}={0}$$
$$\displaystyle{\left({x}-{5}\right)}{\left({x}^{{2}}-{4}{x}+{4}+{2}\right)}={0}$$
$$\displaystyle{\left({x}-{5}\right)}{\left({x}^{{2}}-{4}{x}+{4}+{2}\right)}={0}$$
$$\displaystyle{x}^{{3}}-{4}{x}^{{2}}+{6}{x}-{5}{x}^{{2}}+{20}{x}-{30}={0}$$
$$\displaystyle{x}^{{3}}-{9}{x}^{{2}}+{26}{x}-{30}={0}$$
polynomial will be
$$\displaystyle{y}={a}{\left({x}^{{3}}-{9}{x}^{{2}}+{26}{x}-{30}\right)}.\ {a}\ne{0}$$
there are infinite number of polynomials.