Question

# Find all rational zeros of the polynomial, and write the polynomial in factored form. P(x)=6x^{4}-23x^{3}-13x^{2}+32x+16

Polynomial factorization
Find all rational zeros of the polynomial, and write the polynomial in factored form.
$$\displaystyle{P}{\left({x}\right)}={6}{x}^{{{4}}}-{23}{x}^{{{3}}}-{13}{x}^{{{2}}}+{32}{x}+{16}$$

2021-04-08
Step 1
Given polynomial is, $$\displaystyle{P}{\left({x}\right)}={6}{x}^{{{4}}}-{23}{x}^{{{3}}}-{13}{x}^{{{2}}}+{32}{x}+{16}$$.
The zeroes are found as follows:
$$\displaystyle{6}{x}^{{{4}}}-{23}{x}^{{{3}}}-{13}{x}^{{{2}}}+{32}{x}+{16}$$
$$\displaystyle{6}{x}^{{{3}}}{\left({x}+{1}\right)}-{29}{x}^{{{2}}}{\left({x}+{1}\right)}+{16}{x}{\left({x}+{1}\right)}+{16}{\left({x}+{1}\right)}={0}$$
$$\displaystyle{\left({x}+{1}\right)}{\left({6}{x}^{{{3}}}-{29}{x}^{{{2}}}+{16}{x}+{16}\right)}={0}$$
$$\displaystyle{\left({x}+{1}\right)}{\left({2}{x}+{1}\right)}{\left({3}{x}^{{{2}}}-{16}{x}+{16}\right)}={0}$$
(x+1)(x-4)(2x+1)(3x-4)=0
Step 2
On solving further,
x+1=0 or x-4=0 or 2x+1=0 or 3x-4=0
$$\displaystyle{x}=-{1},{x}={4},{x}=-{\frac{{{1}}}{{{2}}}},{x}={\frac{{{4}}}{{{3}}}}$$
Thus, rational zeroes are $$\displaystyle-{\frac{{{1}}}{{{2}}}}$$ and $$\displaystyle{\frac{{{4}}}{{{3}}}}$$. And, the factored form of the polynomial is (x+1)(x-4)(2x+1)(3x-4).