According to the Rational Root Theorem, which could be a factor of the polynomial f(x) = 6x4 – 21x3 – 4x2 + 24x – 35?

a. 2x – 7
b. 2x + 7
c. 3x – 7
d. 3x + 7

Respuesta :

the answer is the first option, A) 2x-7

Answer with explanation:

The given Polynomial function is

[tex]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})[/tex]

By Rational root Theorem ,factors of the above expression can be

[tex]\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[/tex]

[tex],\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}[/tex]

That is , by observing at the options, the factors of the given expression could be

[tex]x=\pm \frac{7}{2}\\\\ \text{and}, x=\pm \frac{7}{3}[/tex]

Substituting these four  Values one by one  in the Polynomial function

  [tex]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[/tex]

As,we have to find single factor,you will find that,

[tex]f(\frac{-7}{2})\neq 0 \text{and} f(\frac{\pm7}{3})\neq 0.[/tex]

So, [tex]x=\frac{7}{2}\\\\ 2x-7[/tex]

is the factor of the polynomial.

Option A: 2 x -7