Step 1

Zeros of the function are those values of x for which the value of the function is 0.

To find the zeros of the polynomial, first we need to factor it.

First rewrite the polynomial as

\(\displaystyle{4}{x}^{{{3}}}-{x}+{4}{x}^{{{2}}}-{1}\) and then factor out x and 1 from the first two term and the last two terms respectively, then factor out \(\displaystyle{\left({4}{x}^{{{2}}}-{1}\right)}\).

\(\displaystyle{4}{x}^{{{3}}}-{x}+{4}{x}^{{{2}}}-{1}={x}{\left({4}{x}^{{{2}}}-{1}\right)}+{1}{\left({4}{x}^{{{2}}}-{1}\right)}\)

\(\displaystyle={\left({4}{x}^{{{2}}}-{1}\right)}{\left({x}+{1}\right)}\)

Step 2

Use the identity \(\displaystyle{\left({a}^{{{2}}}-{b}^{{{2}}}\right)}={\left({a}-{b}\right)}{\left({a}+{b}\right)}\) to further factor the polynomial as follows:

\(\displaystyle{\left({4}{x}^{{{2}}}-{1}\right)}{\left({x}+{1}\right)}={\left({\left({2}{x}\right)}^{{{2}}}-{1}^{{{2}}}\right)}{\left({x}+{1}\right)}\)

=(2x-1)(2x+1)(x+1)

Therefore, the factored form of the polynomial is (2x−1)(2x+1)(x+1).

Step 3

Equate (2x−1)(2x+1)(x+1) to 0 and solve for x to find the zeros of the polynomial.

(2x-1)(2x+1)(x+1)=0

2x-1=0

\(\displaystyle{x}={\frac{{{1}}}{{{2}}}}\)

2x+1=0

\(\displaystyle{x}=-{\frac{{{1}}}{{{2}}}}\)

x+1=0

x=-1

Therefore, the zeros are \(\displaystyle{\frac{{{1}}}{{{2}}}},-{\frac{{{1}}}{{{2}}}}\) and -1.

Zeros of the function are those values of x for which the value of the function is 0.

To find the zeros of the polynomial, first we need to factor it.

First rewrite the polynomial as

\(\displaystyle{4}{x}^{{{3}}}-{x}+{4}{x}^{{{2}}}-{1}\) and then factor out x and 1 from the first two term and the last two terms respectively, then factor out \(\displaystyle{\left({4}{x}^{{{2}}}-{1}\right)}\).

\(\displaystyle{4}{x}^{{{3}}}-{x}+{4}{x}^{{{2}}}-{1}={x}{\left({4}{x}^{{{2}}}-{1}\right)}+{1}{\left({4}{x}^{{{2}}}-{1}\right)}\)

\(\displaystyle={\left({4}{x}^{{{2}}}-{1}\right)}{\left({x}+{1}\right)}\)

Step 2

Use the identity \(\displaystyle{\left({a}^{{{2}}}-{b}^{{{2}}}\right)}={\left({a}-{b}\right)}{\left({a}+{b}\right)}\) to further factor the polynomial as follows:

\(\displaystyle{\left({4}{x}^{{{2}}}-{1}\right)}{\left({x}+{1}\right)}={\left({\left({2}{x}\right)}^{{{2}}}-{1}^{{{2}}}\right)}{\left({x}+{1}\right)}\)

=(2x-1)(2x+1)(x+1)

Therefore, the factored form of the polynomial is (2x−1)(2x+1)(x+1).

Step 3

Equate (2x−1)(2x+1)(x+1) to 0 and solve for x to find the zeros of the polynomial.

(2x-1)(2x+1)(x+1)=0

2x-1=0

\(\displaystyle{x}={\frac{{{1}}}{{{2}}}}\)

2x+1=0

\(\displaystyle{x}=-{\frac{{{1}}}{{{2}}}}\)

x+1=0

x=-1

Therefore, the zeros are \(\displaystyle{\frac{{{1}}}{{{2}}}},-{\frac{{{1}}}{{{2}}}}\) and -1.