Yes
factor out the 2z^2 in each term
(2z^2)(z^2-5z+4)
factor some more
z^2-5z+4
find what 2 numbers multiply to get 4 and add to get -5
the numbers are -1 and -4
(z-1)(z-4)
the factored form is
(2z^2)(z-1)(z-4)
Answer:
1+i
Step-by-step explanation:
To find the 8th roots of unity, you have to find the trigonometric form of unity.
1. Since
then

and

This gives you 
Thus,

2. The 8th roots can be calculated using following formula:
![\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.](https://tex.z-dn.net/?f=%5Csqrt%5B8%5D%7Bz%7D%3D%5C%7B%5Csqrt%5B8%5D%7B%7Cz%7C%7D%20%28%5Ccos%5Cdfrac%7B%5Cvarphi%2B2%5Cpi%20k%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B%5Cvarphi%2B2%5Cpi%20k%7D%7B8%7D%29%2C%20k%3D0%2C%5C%201%2C%5Cdots%2C7%5C%7D.)
Now
at k=0, ![z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;](https://tex.z-dn.net/?f=z_0%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%200%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%200%7D%7B8%7D%29%3D1%5Ccdot%20%281%2B0%5Ccdot%20i%29%3D1%3B)
at k=1, ![z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};](https://tex.z-dn.net/?f=z_1%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%201%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%201%7D%7B8%7D%29%3D1%5Ccdot%20%28%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%3D%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3B)
at k=2, ![z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;](https://tex.z-dn.net/?f=z_2%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%202%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%202%7D%7B8%7D%29%3D1%5Ccdot%20%280%2B1%5Ccdot%20i%29%3Di%3B)
at k=3, ![z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};](https://tex.z-dn.net/?f=z_3%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%203%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%203%7D%7B8%7D%29%3D1%5Ccdot%20%28-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%3D-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3B)
at k=4, ![z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;](https://tex.z-dn.net/?f=z_4%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%204%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%204%7D%7B8%7D%29%3D1%5Ccdot%20%28-1%2B0%5Ccdot%20i%29%3D-1%3B)
at k=5, ![z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};](https://tex.z-dn.net/?f=z_5%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%205%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%205%7D%7B8%7D%29%3D1%5Ccdot%20%28-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%3D-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3B)
at k=6, ![z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;](https://tex.z-dn.net/?f=z_6%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%206%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%206%7D%7B8%7D%29%3D1%5Ccdot%20%280-1%5Ccdot%20i%29%3D-i%3B)
at k=7, ![z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};](https://tex.z-dn.net/?f=z_7%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%207%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%207%7D%7B8%7D%29%3D1%5Ccdot%20%28%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%3D%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3B)
The 8th roots are

Option C is icncorrect.
The equation for which square method is possible is x²-8=1
Step-by-step explanation:
For checking which of the equation satisfies the complete square condition, we proceed by checking each of the available options
1). x²+20x=52
Rewriting it as x²+20x-52
This binomial expression is not a perfect square since the product of the coefficient of x²(i.e. 1) and independent constant (i.e. 52) is not a perfect square.
2). 5x² + 3x = 9
This equation can be rearranged as 5x²+3x-9=0
This binomial expression is not a perfect square since the product of the coefficient of x²(i.e. 5) and independent constant(i.e. 9) is not a perfect square.
3.) x² −8=1
This equation can be rearranged as x²=9
Hence x= ±3
This binomial expression is a perfect square and can be done by the square method.
4). 3x² −x+17=0
This binomial expression is not a perfect square since the product of the coefficient of x²(i.e. 3) and independent constant(i.e. 17) is not a perfect square.
<span>
(1/6)/ 1.5 =
(1/6)/(15/10)=
1/6*10/15=10/90=1/9 width</span>