How would you solve for x with an equation like this?
0.2x^3 - 2.84x^2 + 8.532x - 2.4 = 0
I thought you would use the quadratic formula somehow, but I'm not sure
To solve for x in this equation, you would not use the quadratic formula since the equation is not quadratic. Instead, you would use a method called "factoring" or "solving by factoring." Here's how you can do it:
Step 1: Look for any common factors among the coefficients. In this case, all the coefficients are divisible by 0.2, so you can simplify the equation by dividing every term by 0.2:
x^3 - 14.2x^2 + 42.66x - 12 = 0
Step 2: Check for any possible rational roots using the rational root theorem. The rational root theorem states that any rational root (if it exists) of the equation will be of the form p/q, where p is a factor of the constant term (in this case, 12) and q is a factor of the leading coefficient (in this case, 1).
The factors of 12 are ±1, ±2, ±3, ±4, ±6, and ±12. The factors of 1 are ±1. So the possible rational roots are ±1, ±2, ±3, ±4, ±6, and ±12.
Step 3: Begin testing the possible rational roots using synthetic division. Synthetic division allows you to quickly check if the possible root is indeed a root of the equation. Start by dividing the polynomial by one of the possible roots.
For example, let's test if x = 1 is a root:
1 | 1 - 14.2 42.66 -12
- 1 + 1.8 44.46 32.46
--------------
0 -12.4 87.12 20.46
Since the remainder is not zero, x = 1 is not a root.
Repeat the process for the other possible rational roots until you find one that gives a remainder of zero. The root that gives a remainder of zero is an actual root of the equation. Once you find one root, you can divide the polynomial by the binomial (x - root) to obtain a quadratic equation, which you can then solve using the quadratic formula.
Please note that solving polynomial equations can sometimes be complex, and there might not always be rational roots or straightforward factoring. In such cases, alternative techniques like graphing or numerical methods may be necessary to find the solutions.