Express f (x) as a product of linear and quadratic polynomials with real coefficients that are irreducible over R. -1,0,3+i. degree 4

Expert Answer

Viktor Wiley

Answered 2021-09-04 Author has 8343 answers

Given Roots of the polynomial of degree 4 is -1, 0, 3+i
To find the polynomial f (x) with real coefficients and leading coefficient 1 has the given zero(s) and degree. Also, to Express f (x) as a product of linear and quadratic polynomials with real coefficients that are irreducible over R.
Definition Used We know if a polynomial having a complex root then its complex conjugate will also be the root of the given polynomial.
Formation of the polynomial f(x)
Since by the definition if 3+i is one root then 3-i will also be the root of the given polynomial
In this way, we have a total number of 4 zeroes i.e.
-1,0,3+i and 3-i
Then the polynomial f(x) is
\(\displaystyle{f{{\left({x}\right)}}}={\left({x}-{\left(-{1}\right)}\right)}{\left({x}-{0}\right)}{\left({x}-{\left({3}+{i}\right)}{\left({x}-{\left({3}-{i}\right)}\right)}\right.}\)
\(\displaystyle={x}{\left({x}+{1}\right)}{\left({x}-{3}-{i}\right)}{\left({x}-{3}+{i}\right)}\)
\(\displaystyle={x}{\left({x}+{i}\right)}{\left\lbrace{\left({x}-{3}\right)}^{{2}}-{\left({i}\right)}^{{2}}\right\rbrace}\)
\(\displaystyle={x}{\left({x}+{1}\right)}{\left\lbrace{x}^{{2}}-{6}{x}+{9}+{1}\right\rbrace}\)
\(\displaystyle={\left({x}^{{2}}+{x}\right)}{\left({x}^{{2}}-{6}{x}+{10}\right)}\)
\(\displaystyle={x}^{{4}}-{6}{x}^{{3}}+{10}{x}^{{2}}+{x}^{{3}}-{6}{x}^{{2}}+{10}{x}\)
\(\displaystyle={x}^{{4}}-{5}{x}^{{3}}+{4}{x}^{{2}}+{10}{x}\)

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