which is a factor of the polynomial $f(x)=6x^{4}-21x^{3}-4x^{2}+24x - 35$?\n$2x - 7$\n$2x + 7$\n$3x…

Explanation:

Step1: Use the factor - theorem

The factor - theorem states that if (f(a)=0), then ((x - a)) is a factor of (f(x)). For a linear factor (ax + b), we find (x=-\frac{b}{a}) and check (f(-\frac{b}{a})). For the factor (2x - 7), let (2x-7 = 0), then (x=\frac{7}{2}). [ \begin{align*} f\left(\frac{7}{2}\right)&=6\left(\frac{7}{2}\right)^4-21\left(\frac{7}{2}\right)^3-4\left(\frac{7}{2}\right)^2 + 24\left(\frac{7}{2}\right)-35\ &=6\times\frac{2401}{16}-21\times\frac{343}{8}-4\times\frac{49}{4}+84 - 35\ &=\frac{7203}{8}-\frac{7203}{8}-49 + 84-35\ &=( \frac{7203 - 7203}{8})+(84-(49 + 35))\ &=0+0\ &=0 \end{align*} ]

Step2: Check other factors

For the factor (2x + 7), let (2x+7 = 0), then (x=-\frac{7}{2}). [ \begin{align*} f\left(-\frac{7}{2}\right)&=6\left(-\frac{7}{2}\right)^4-21\left(-\frac{7}{2}\right)^3-4\left(-\frac{7}{2}\right)^2+24\left(-\frac{7}{2}\right)-35\ &=6\times\frac{2401}{16}+21\times\frac{343}{8}-4\times\frac{49}{4}-84 - 35\ &\neq0 \end{align*} ] For the factor (3x - 7), let (3x - 7=0), then (x = \frac{7}{3}). [ \begin{align*} f\left(\frac{7}{3}\right)&=6\left(\frac{7}{3}\right)^4-21\left(\frac{7}{3}\right)^3-4\left(\frac{7}{3}\right)^2+24\left(\frac{7}{3}\right)-35\ &\neq0 \end{align*} ] For the factor (3x + 7), let (3x+7 = 0), then (x=-\frac{7}{3}). [ \begin{align*} f\left(-\frac{7}{3}\right)&=6\left(-\frac{7}{3}\right)^4-21\left(-\frac{7}{3}\right)^3-4\left(-\frac{7}{3}\right)^2+24\left(-\frac{7}{3}\right)-35\ &\neq0 \end{align*} ]

Answer:

A. (2x - 7)