Question

Solve by factoring and simplify all irrational and complex solutions :2×^3-5×^2+40x-100=0

Answer 1

First, we need to factor the polynomial 2x^3 - 5x^2 + 40x - 100.

We can start by factoring out the greatest common factor of the coefficients, which is 2:

2(x^3 - (5/2)x^2 + 20x - 50) = 0

Next, we can try to factor the cubic polynomial inside the parentheses. One possible way to do this is to use synthetic division to test possible roots. We can start by testing x = 2, which is a root if and only if the remainder of the division is zero:

2 | 1 -5/2 20 -50
| 2 -5 30
|----------------
1 -1/2 15 -20

Since the remainder is not zero, x = 2 is not a root. We can try another value, such as x = 5:

5 | 1 -5/2 20 -50
| 5 12.5 162.5
|----------------
1 2.5 32.5 112.5

Since the remainder is not zero, x = 5 is not a root either. However, we can see that the quotient is a quadratic polynomial with rational coefficients, which we can factor using the quadratic formula:

x^2 + 2.5x + 32.5 = 0

x = (-2.5 ± sqrt(2.5^2 - 4(1)(32.5))) / (2(1))
x = (-2.5 ± sqrt(-115)) / 2
x = (-2.5 ± 5sqrt(23)i) / 2

Therefore, the solutions to the equation 2x^3 - 5x^2 + 40x - 100 = 0 are:

x = 2.5
x = (-2.5 + 5sqrt(23)i) / 2
x = (-2.5 - 5sqrt(23)i) / 2

Note that the first solution is rational, while the other two solutions are complex (because they involve the imaginary unit i). Also, note that we simplified the complex solutions by dividing both the numerator and denominator by 2.

Answer 2

To solve this equation by factoring, we can first try to factor out a common factor of 2:

2x^3 - 5x^2 + 40x - 100 = 0

2(x^3 - (5/2)x^2 + 20x - 50) = 0

Next, we can use the rational root theorem to find possible rational roots of the polynomial x^3 - (5/2)x^2 + 20x - 50. The rational root theorem states that any rational root of a polynomial with integer coefficients must be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.

The constant term of our polynomial is -50, which has factors of ±1, ±2, ±5, ±10, ±25, and ±50. The leading coefficient is 1, which has factors of ±1. So, the possible rational roots are:

±1, ±2, ±5, ±10, ±25, and ±50

We can try plugging in each of these values and see if any of them make the polynomial equal to 0. After some trial and error, we find that x = 5 is a root:

x^3 - (5/2)x^2 + 20x - 50 = 0

(5)^3 - (5/2)(5)^2 + 20(5) - 50 = 0

125 - (5/2)25 + 100 - 50 = 0

125 - 62.5 + 50 = 0

112.5 ≠ 0

Since x = 5 is a root, we can factor the polynomial as:

2(x - 5)(x^2 + 5x + 10) = 0

Using the quadratic formula, we can find the roots of the quadratic factor:

x = (-5 ± sqrt(5^2 - 4(1)(10))) / 2(1)

x = (-5 ± sqrt(-15)) / 2

x = (-5 ± i sqrt(15)) / 2

So, the solutions to the equation 2x^3 - 5x^2 + 40x - 100 = 0 are:

x = 5, (-5 + i sqrt(15)) / 2, (-5 - i sqrt(15)) / 2