Respuesta :
According to factor theorem, if f(x) is a polynomial of degree n ≥ 1 and 'a' is any real number, then, (x-a) is a factor of f(x), if f(a)=0.
The factor theorem states
[tex]$(x-a)$ is a factor of $f(x) \Leftrightarrow f(a)=0$[/tex]
f(x)=4 x^3-3 x^2-8 x+4
we have (x-2)[tex]\Rightarrow a=2[/tex]
& [tex]f(2)=4 \times 2^3-3 \times 2^2-8 \times 2+4 \\& f(2)=4 \times 8-3 \times 4-16+4 \\& f(2)=32-12-16+4 \\& f(2)=8 \neq 0\end{aligned}[/tex]
[tex]$\therefore x-2$ is not a factor of $4 x^3-3 x^2-8 x+4$[/tex]
however by the remainder theorem when
[tex]$f(x)=4 x^3-3 x^2-8 x+4$[/tex] is divided by [tex]$(x-2)$[/tex] the remainder is 8
the factor theorem being a special case of the remainder theorem
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