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

$$(2x − 1)$$ 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.

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

yagombyeR

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
$$\displaystyle{6}{x}^{{6}}+{x}^{{5}}-{92}{x}^{{4}}+{45}{x}^{{3}}+{184}{x}^{{2}}+{4}{x}-{48}$$
$$\displaystyle{6}{\left({\frac{{{1}}}{{{2}}}}\right)}^{{6}}+{\left({\frac{{{1}}}{{{2}}}}\right)}^{{5}}-{92}{\left({\frac{{{1}}}{{{2}}}}\right)}^{{4}}+{45}{\left({\frac{{{1}}}{{{2}}}}\right)}^{{3}}+{184}{\left({\frac{{{1}}}{{{2}}}}\right)}^{{2}}+{4}{\left({\frac{{{1}}}{{{2}}}}\right)}-{48}={0}$$
$$\displaystyle{6}{\left({\frac{{{1}}}{{{64}}}}\right)}+{\left({\frac{{{1}}}{{{32}}}}\right)}-{92}{\left({\frac{{{1}}}{{{16}}}}\right)}+{45}{\left({\frac{{{1}}}{{{2}}}}\right)}^{{3}}+{184}{\left({\frac{{{1}}}{{{2}}}}\right)}^{{2}}+{4}{\left({\frac{{{1}}}{{{2}}}}\right)}-{48}={0}$$
$$\displaystyle{\frac{{{3}}}{{{32}}}}+{\frac{{{1}}}{{{32}}}}-{\frac{{{23}}}{{{4}}}}+{\frac{{{45}}}{{{8}}}}+{46}-{46}={0}$$
$$\displaystyle{\left({\frac{{{3}}}{{{32}}}}+{\frac{{{1}}}{{{32}}}}+{\frac{{{45}}}{{{8}}}}+{46}\right)}+{\left(-{\frac{{{23}}}{{{4}}}}-{46}\right)}={0}$$
$$\displaystyle{\frac{{{207}}}{{{4}}}}-{\frac{{{207}}}{{{4}}}}={0}$$
$$\displaystyle{0}={0}$$