Answer:
![2x^2(6x - 1) + 3(6x - 1)](https://tex.z-dn.net/?f=2x%5E2%286x%20-%201%29%20%2B%203%286x%20-%201%29)
Step-by-step explanation:
Given
![12x^3 - 2x^2 + 18x - 3](https://tex.z-dn.net/?f=12x%5E3%20-%202x%5E2%20%2B%2018x%20-%203)
Required
Factorize
To start with; we need to group the expression into two
![(12x^3 - 2x^2) + (18x - 3)](https://tex.z-dn.net/?f=%2812x%5E3%20-%202x%5E2%29%20%2B%20%2818x%20-%203%29)
Factorize each grouped expression, using their common factor
![2x^2(6x - 1) + 3(6x - 1)](https://tex.z-dn.net/?f=2x%5E2%286x%20-%201%29%20%2B%203%286x%20-%201%29)
From the list of given options;
Option A is correct
The correct answer is x = 1/28y^2
Answer:
Step-by-step explanation:
a) Find the gradient
![\nabla f(x,y) =f_x(x,y)\bold{i} +f_y (x,y)\bold{j}](https://tex.z-dn.net/?f=%5Cnabla%20f%28x%2Cy%29%20%3Df_x%28x%2Cy%29%5Cbold%7Bi%7D%20%2Bf_y%20%28x%2Cy%29%5Cbold%7Bj%7D)
![Also, sin(4x+3y)= sin(4x)cos(3y)+sin(3y)cos (4x)](https://tex.z-dn.net/?f=Also%2C%20sin%284x%2B3y%29%3D%20sin%284x%29cos%283y%29%2Bsin%283y%29cos%20%284x%29)
![f_x(x,y) = 4cos(4x)cos(3y)-4sin(3y)sin(4x) = 4 cos(4x-3y)](https://tex.z-dn.net/?f=f_x%28x%2Cy%29%20%3D%204cos%284x%29cos%283y%29-4sin%283y%29sin%284x%29%20%3D%204%20cos%284x-3y%29)
![f_y(x,y) = -3sin(4x)sin(3y)+3cos(3y)cos(4x)=3cos(4x-3y)](https://tex.z-dn.net/?f=f_y%28x%2Cy%29%20%3D%20-3sin%284x%29sin%283y%29%2B3cos%283y%29cos%284x%29%3D3cos%284x-3y%29)
Hence
![\nabla f(x,y) = 4cos(4x-3y)\bold{i}+3cos(4x-3y)\bold{j}](https://tex.z-dn.net/?f=%5Cnabla%20f%28x%2Cy%29%20%3D%204cos%284x-3y%29%5Cbold%7Bi%7D%2B3cos%284x-3y%29%5Cbold%7Bj%7D)
b) At point (-6, 8) just replace the values in gradient to find it out. You can do it.
c) directional derivative in the direction u.
![D_uf(x,y)=\nabla f(x,y). u](https://tex.z-dn.net/?f=D_uf%28x%2Cy%29%3D%5Cnabla%20f%28x%2Cy%29.%20u)
I stuck with your 1 2 3 i -j . What does 1 2 3 mean? is it not 123 or 1 2/3 or else?
When you move the decimal to the left, it adds 1 to your exponent. So if you move it to the left 4 times, you will get 0.00018.