Answer:
(D) 3x−4
Step-by-step explanation:
Factor Theorem
Given a polynomial P(x) and a linear function x-a, If P(a)=0, then the linear function x-a is a factor of P(a).
In Option A:
[tex]L$inear Function =2x+3\\Set 2x+3=0$\\x=-\frac{3}{2} \\p(x)=6x^4+x^3-45x^2+26x+24\\p(-\frac{3}{2})=6(-\frac{3}{2})^4+(-\frac{3}{2})^3-45(-\frac{3}{2})^2+26(-\frac{3}{2})+24\\\\p(-\frac{3}{2})=-89.25[/tex]
In Option B
[tex]L$inear Function =3x-2\\Set 3x-2=0$\\x=\frac{2}{3} \\p(x)=6x^4+x^3-45x^2+26x+24\\p(\frac{2}{3})=6(\frac{2}{3})^4+(\frac{2}{3})^3-45(\frac{2}{3})^2+26(\frac{2}{3})+24\\\\p(\frac{2}{3})=22.8[/tex]
In Option C
[tex]L$inear Function =2x-1\\Set 2x-1=0$\\x=\frac{1}{2} \\p(x)=6x^4+x^3-45x^2+26x+24\\p(\frac{1}{2} )=6(\frac{1}{2} )^4+(\frac{1}{2} )^3-45(\frac{1}{2} )^2+26(\frac{1}{2} )+24\\\\p(\frac{1}{2} )=26.25[/tex]
In Option D
[tex]L$inear Function =3x-4\\Set 3x-4=0$\\x=\frac{4}{3} \\p(x)=6x^4+x^3-45x^2+26x+24\\p(\frac{4}{3} )=6(\frac{4}{3})^4+(\frac{4}{3} )^3-45(\frac{4}{3} )^2+26(\frac{4}{3})+24\\\\p(\frac{4}{3} )=0[/tex]
We can see that only Option D: 3x−4 gives a result of 0. Therefore, by the factor theorem, it is a factor of the polynomial.
