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
ANSWERED
asked 2021-04-06
Find all rational zeros of the polynomial, and write the polynomial in factored form.
\(\displaystyle{P}{\left({x}\right)}={6}{x}^{{{4}}}-{23}{x}^{{{3}}}-{13}{x}^{{{2}}}+{32}{x}+{16}\)

Answers (1)

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).
0
 
Best answer

expert advice

Need a better answer?
...