I think the best answer is 3 to 9 but thats not a answer so 3 to 6
Degree is the value of the highest power of the variable, and hence in this case is 8
Answer:
(
0
,
2
)
Step-by-step explanation:
To find the x-intercept, substitute in 0 for y and solve for x . To find the y-intercept, substitute in 0 for x and solve for y .
x-intercept(s): None
y-intercept(s): ( 0 , 2 )
Answer:
The factored form of x^3 -1 will be:
![x^3-1^3=\left(x-1\right)\left(x^2+x+1\right)](https://tex.z-dn.net/?f=x%5E3-1%5E3%3D%5Cleft%28x-1%5Cright%29%5Cleft%28x%5E2%2Bx%2B1%5Cright%29)
Step-by-step explanation:
Given the expression
![x^3-1](https://tex.z-dn.net/?f=x%5E3-1)
Rewrite 1 as 1³
![=x^3-1^3](https://tex.z-dn.net/?f=%3Dx%5E3-1%5E3)
![\mathrm{Apply\:Difference\:of\:Cubes\:Formula:\:}x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3ADifference%5C%3Aof%5C%3ACubes%5C%3AFormula%3A%5C%3A%7Dx%5E3-y%5E3%3D%5Cleft%28x-y%5Cright%29%5Cleft%28x%5E2%2Bxy%2By%5E2%5Cright%29)
![x^3-1^3=\left(x-1\right)\left(x^2+x+1\right)](https://tex.z-dn.net/?f=x%5E3-1%5E3%3D%5Cleft%28x-1%5Cright%29%5Cleft%28x%5E2%2Bx%2B1%5Cright%29)
![=\left(x-1\right)\left(x^2+x+1\right)](https://tex.z-dn.net/?f=%3D%5Cleft%28x-1%5Cright%29%5Cleft%28x%5E2%2Bx%2B1%5Cright%29)
Thus, the factored form of x^3 -1 will be:
![x^3-1^3=\left(x-1\right)\left(x^2+x+1\right)](https://tex.z-dn.net/?f=x%5E3-1%5E3%3D%5Cleft%28x-1%5Cright%29%5Cleft%28x%5E2%2Bx%2B1%5Cright%29)