Question

\((2x − 1)\) is a factor of the polynomial \(\displaystyle{6}{x}^{{6}}+{x}^{{5}}-{92}{x}^{{4}}+{45}{x}^{{3}}+{184}{x}^{{2}}+{4}{x}-{48}\). Determine whether the statement is true or false. Justify your answer.

Answer 1

The factors of a polynomial decompose the polynomial into two or more polynomials, so the product of the factors gives the polynomial. In other words, a factor of a polynomial is a polynomial that divides the original polynomial with the remainder, 0.
To check, whether \(2x−1\) is the factor of \(\displaystyle{6}{x}^{{6}}+{x}^{{5}}-{92}{x}^{{4}}+{45}{x}^{{3}}+{184}{x}^{{2}}+{4}{x}-{48}\) or not, substitute the value of x from \(2x−1\) and if this results in zero, then \(2x−1\) is the factor of \(f(x)\). Substitute x with \(\displaystyle{\frac{{{1}}}{{{2}}}}\) in \(f(x)\) and simplify each power.
\(\displaystyle{2}{x}-{1}={0}\)
\(\displaystyle{2}{x}={1}\)
\(\displaystyle{x}={\frac{{{1}}}{{{2}}}}\)
\(\displaystyle{f{{\left({x}\right)}}}={6}{x}^{{6}}+{x}^{{5}}-{92}{x}^{{4}}+{45}{x}^{{3}}+{184}{x}^{{2}}+{4}{x}-{48}\)
\(\displaystyle{f{{\left({\frac{{{1}}}{{{2}}}}\right)}}}={0}\)
\(\displaystyle{6}{\left({\frac{{{1}}}{{{2}}}}\right)}^{{6}}+{\left({\frac{{{1}}}{{{2}}}}\right)}^{{5}}-{92}{\left({\frac{{{1}}}{{{2}}}}\right)}^{{4}}+{45}{\left({\frac{{{1}}}{{{2}}}}\right)}^{{3}}+{184}{\left({\frac{{{1}}}{{{2}}}}\right)}^{{2}}+{4}{\left({\frac{{{1}}}{{{2}}}}\right)}-{48}={0}\)
Cancel the common factors from the numerator and the denominator of each term and combine the terms with the same signs. Perform the addition and then subtraction to obtain 0. Hence, \(2x-1\) is the factor of
\(\displaystyle{6}{x}^{{6}}+{x}^{{5}}-{92}{x}^{{4}}+{45}{x}^{{3}}+{184}{x}^{{2}}+{4}{x}-{48}\)
\(\displaystyle{6}{\left({\frac{{{1}}}{{{2}}}}\right)}^{{6}}+{\left({\frac{{{1}}}{{{2}}}}\right)}^{{5}}-{92}{\left({\frac{{{1}}}{{{2}}}}\right)}^{{4}}+{45}{\left({\frac{{{1}}}{{{2}}}}\right)}^{{3}}+{184}{\left({\frac{{{1}}}{{{2}}}}\right)}^{{2}}+{4}{\left({\frac{{{1}}}{{{2}}}}\right)}-{48}={0}\)
\(\displaystyle{6}{\left({\frac{{{1}}}{{{64}}}}\right)}+{\left({\frac{{{1}}}{{{32}}}}\right)}-{92}{\left({\frac{{{1}}}{{{16}}}}\right)}+{45}{\left({\frac{{{1}}}{{{2}}}}\right)}^{{3}}+{184}{\left({\frac{{{1}}}{{{2}}}}\right)}^{{2}}+{4}{\left({\frac{{{1}}}{{{2}}}}\right)}-{48}={0}\)
\(\displaystyle{\frac{{{3}}}{{{32}}}}+{\frac{{{1}}}{{{32}}}}-{\frac{{{23}}}{{{4}}}}+{\frac{{{45}}}{{{8}}}}+{46}-{46}={0}\)
\(\displaystyle{\left({\frac{{{3}}}{{{32}}}}+{\frac{{{1}}}{{{32}}}}+{\frac{{{45}}}{{{8}}}}+{46}\right)}+{\left(-{\frac{{{23}}}{{{4}}}}-{46}\right)}={0}\)
\(\displaystyle{\frac{{{207}}}{{{4}}}}-{\frac{{{207}}}{{{4}}}}={0}\)
\(\displaystyle{0}={0}\)