Analytically show that the equation below represents trigonometric
X^2-6x=13
(x-3)^2-9=13
(x-3)^2=22
(x-3)= +-sqrt22
x = sqrt22+3 or x = -sqrt22+3
Given:
The function is:
![y=3x^3-5x^2-11x-3](https://tex.z-dn.net/?f=y%3D3x%5E3-5x%5E2-11x-3)
It is given that -1 is a zero of the given function.
To find:
The other zeroes of the given function.
Solution:
If c is a zero of a polynomial P(x), then (x-c) is a factor of the polynomial.
It is given that -1 is a zero of the given function. So,
is a factor of the given function.
We have,
![y=3x^3-5x^2-11x-3](https://tex.z-dn.net/?f=y%3D3x%5E3-5x%5E2-11x-3)
Split the middle terms in such a way so that we get (x+1) as a factor.
![y=3x^3+3x^2-8x^2-8x-3x-3](https://tex.z-dn.net/?f=y%3D3x%5E3%2B3x%5E2-8x%5E2-8x-3x-3)
![y=3x^2(x+1)-8x(x+1)-3(x+1)](https://tex.z-dn.net/?f=y%3D3x%5E2%28x%2B1%29-8x%28x%2B1%29-3%28x%2B1%29)
![y=(x+1)(3x^2-8x-3)](https://tex.z-dn.net/?f=y%3D%28x%2B1%29%283x%5E2-8x-3%29)
Again splitting the middle term, we get
![y=(x+1)(3x^2-9x+x-3)](https://tex.z-dn.net/?f=y%3D%28x%2B1%29%283x%5E2-9x%2Bx-3%29)
![y=(x+1)(3x(x-3)+1(x-3))](https://tex.z-dn.net/?f=y%3D%28x%2B1%29%283x%28x-3%29%2B1%28x-3%29%29)
![y=(x+1)(3x+1)(x-3)](https://tex.z-dn.net/?f=y%3D%28x%2B1%29%283x%2B1%29%28x-3%29)
For zeroes,
.
![(x+1)(3x+1)(x-3)=0](https://tex.z-dn.net/?f=%28x%2B1%29%283x%2B1%29%28x-3%29%3D0)
and
and ![x-3=0](https://tex.z-dn.net/?f=x-3%3D0)
and
and ![x=3](https://tex.z-dn.net/?f=x%3D3)
Therefore, the other two zeroes of the given function are
and
.
Letter C
The - 3 at the end of the function means the function is shifted down 3 units. This also shifts the asymptote down 3 units.
Answer:
65
Step-by-step explanation:
You find the one equally in the center of the numbers after they are in order.