4 x^3 - 4x^2 - 11x + 6 = 0

Need to solve the equation algebraically

IF there is a rational real root, it is q/p, where p and q are integer factors of the x^3 term and the constant term. Once you have one root (by trial and error), the others can be obtained by solving the quadratic you get by factoring out the first root.

In your case, one of the roots is 2. Divide the cubic by x-2 to get the quadratic that will get you the other two roots.

If you are lazy like me and want to peek at the answers, use:
http://www.cubicsolver.com/

Angle QRS is a straight angle. If a=2c, c=2b, and b=d/3, what are the values of a,b,c,and d? a+b+c+d = 180

To solve the equation algebraically, we'll start by factoring the equation if possible. If we can't factor it, we'll use another method like the quadratic formula.

Step 1: Look for common factors.
In this equation, we don't have any common factors among the terms.

Step 2: Check if the equation can be factored.
Sometimes, equations can be factored into simpler expressions. However, in this case, factoring is not immediately apparent.

Step 3: Use the quadratic formula.
If the equation cannot be factored, we can use the quadratic formula to find the solution. The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 4, b = -4, and c = 6. Substituting these values into the quadratic formula, we get:
x = (-(-4) ± √((-4)^2 - 4(4)(6))) / (2(4))
x = (4 ± √(16 - 96)) / 8
x = (4 ± √(-80)) / 8

Step 4: Simplify the square root.
Since we have a negative value under the square root (√(-80)), we know that the equation has imaginary solutions. We can simplify it further by factoring out -1 from the square root:
x = (4 ± √(-1 * 16 * 5)) / 8
x = (4 ± 4i√5) / 8
x = (1 ± i√5) / 2

So the solutions to the equation 4x^3 - 4x^2 - 11x + 6 = 0 are:
x = (1 + i√5) / 2
x = (1 - i√5) / 2