1
Simplify \frac{1}{2}\imath n(x+3)21ın(x+3) to \frac{\imath n(x+3)}{2}2ın(x+3)
\frac{\imath n(x+3)}{2}-\imath nx=02ın(x+3)−ınx=0
2
Add \imath nxınx to both sides
\frac{\imath n(x+3)}{2}=\imath nx2ın(x+3)=ınx
3
Multiply both sides by 22
\imath n(x+3)=\imath nx\times 2ın(x+3)=ınx×2
4
Regroup terms
\imath n(x+3)=nx\times 2\imathın(x+3)=nx×2ı
5
Cancel \imathı on both sides
n(x+3)=nx\times 2n(x+3)=nx×2
6
Divide both sides by nn
x+3=\frac{nx\times 2}{n}x+3=nnx×2
7
Subtract 33 from both sides
x=\frac{nx\times 2}{n}-3x=nnx×2−3
Answer:
C. 
Step-by-step explanation:
The polynomial form that reveals most quickly the zeroes is the form of a product of binomials. That is:

Where
is the product function and
is the i-th root of the polynomial.
Hence,
resembles a form that is close to the form described above. The right option is C.
Look at the solution ( again me)