Given polynomial [tex]f(x)=4 x^{3}-12 x^{2}+2 x-6[/tex].
(a) To show that (x – 3) is a factor of f(x).
Factor theorem:
A polynomial f(x) has a factor of (x – a), if and only if f(a) = 0.
[tex]f(x)=4 x^{3}-12 x^{2}+2 x-6[/tex]
[tex]f(3)=4 (3)^{3}-12 (3)^{2}+2 (3)-6[/tex]
[tex]=4(27)-12(9)+6-6[/tex]
[tex]=108-108[/tex]
= 0
f(3) = 0
Hence (x – 3) is a factor of f(x).
(b) [tex]f(x)=0[/tex]
[tex]4 x^{3}-12 x^{2}+2 x-6=0[/tex]
Here [tex]4x^2[/tex] is common in first 2 terms and 2 is common in next 2 terms.
[tex]4x^2(x-3 )+2 (x-3)=0[/tex]
Now, (x – 3) is common in both terms.
[tex](x-3)(4x^2+2)=0[/tex]
[tex](x-3)=0[/tex] and [tex](4x^2+2)=0[/tex]
x = 3 and [tex]4x^2+2=0[/tex]
3 is the real root.
using quadratic formula solve [tex]4x^2+2=0[/tex] and find the other roots.
[tex]$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
[tex]$x=\frac{-0 \pm \sqrt{0^{2}-4 (4) (2)}}{2 (4)}[/tex]
[tex]$x=\pm\frac{\sqrt{-32}}{8}[/tex]
[tex]$x=\pm\frac{4\sqrt{2}i}{8}[/tex]
[tex]$x=\pm\frac{\sqrt{2}i}{2}[/tex]
These are complex roots.
Hence 3 is the only real root of the equation f(x) = 0.
