1. your leading coefficient has to be 1 (nothing before the x^2). If there is you have to divide that out before you start.
2. Move your constant (the number without any x attached) to the other side of the equation
3. take 1/2 of the b term (the one with the x attached) and then square it and then add it to both sides
4. Factor the left side
5. Set each factor equal to 0 and solve
Here is an example:
4x^2-24x+20=0
The first term is not a 1 so we have to divide it out by 4 first
x^2-6x+5=0
Move the 5 to the other side. It becomes negative.
x^2-6x=-5
Take 1/2 of 6 (3) then square it (9) and add it to both sides.
x^2-6x+9=-5+9
Factor the left side
(x-3)(x-3)=4
(x-3)^2=4
To solve you need to square root both sides
x-3=+/-
x-3=+/-2
x=3+2=5
x=3-2=1
Those would be your two answers.
<span>Hope that helps</span>
f(x) = 5x − 1 and g(x) = 2x^2 + 1
(f × g)(x) = (5x − 1)(2x^2 + 1)
(f × g)(x) = 10x^3 - 2x^2 + 5x - 1
Substitute x = - 3
(f × g)(-3) = 10(-3)^3 - 2(-3)^2 + 5(-3) - 1
(f × g)(-3) = 10(-27) - 2(9) -15 - 1
(f × g)(-3) =-270 - 18 - 16
(f × g)(-3) = -236
Answer
- 236
The answer is going to be D
We first obtain the equation of the lines bounding R.
For the line with points (0, 0) and (8, 1), the equation is given by:

For the line with points (0, 0) and (1, 8), the equation is given by:

For the line with points (8, 1) and (1, 8), the equation is given by:

The Jacobian determinant is given by

The integrand x - 3y is transformed as 8u + v - 3(u + 8v) = 8u + v - 3u - 24v = 5u - 23v
Therefore, the integration is given by: