# Find all rational zeros of the polynomial, and write the polynomial in factored form. P(x)=4x^{3}+4x^{2}-x-1

Find all rational zeros of the polynomial, and write the polynomial in factored form.
$$\displaystyle{P}{\left({x}\right)}={4}{x}^{{{3}}}+{4}{x}^{{{2}}}-{x}-{1}$$

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Pohanginah
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.