Answer with explanation:
The given Polynomial function is
\(f(x)= 6x^4 - 21 x^3 - 4 x^2 + 24 x - 35\\\\6(x^4-\frac{21}{6}\times x^3-\frac{4}{6}\times x^2+4 x-\frac{35}{6})\)
By Rational root Theorem ,factors of the above expression can be
\(\text{Factors of 35,factors of 6,and factors of }\frac{35}{6}=\pm 1,\pm 5,\pm 7, \pm 35,\pm 2,\pm 3, \pm 6\)
\(,\pm \frac{1}{2},\pm\frac{1}{3},\pm \frac{1}{6},\pm\frac{5}{2},\pm\frac{5}{3},\pm\frac{5}{6},\pm\frac{7}{2},\pm\frac{7}{3},\pm\frac{7}{6},\pm\frac{35}{2},\pm\frac{35}{3},\pm\frac{35}{6}\)
That is , by observing at the options, the factors of the given expression could be
\(x=\pm \frac{7}{2}\\\\ \text{and}, x=\pm \frac{7}{3}\)
Substituting these four Values one by one in the Polynomial function
\(f(\frac{7}{2})=6 \times (\frac{7}{2})^4 -21 \times (\frac{7}{2})^3 -4 \times (\frac{7}{2})^2 +24 \times (\frac{7}{2})-35\\\\ f(\frac{7}{2})=900.375 -900.375 - 49+84-35\\\\f(\frac{7}{2})=0\)
As,we have to find single factor,you will find that,
\(f(\frac{-7}{2})
eq 0 \text{and} f(\frac{\pm7}{3})
eq 0.\)
So, \(x=\frac{7}{2}\\\\ 2x-7\)
is the factor of the polynomial.
Option A: 2 x -7
