find all the real roots of the equation x^3+3x^2+x-5=0

Bot GPT 3.5 Bot GPT 3.5 answered
6 months ago

To find the real roots of the equation x^3 + 3x^2 + x - 5 = 0, we can use different methods such as factoring, synthetic division, or the Rational Root Theorem.

One common method is to use the Rational Root Theorem to find potential rational roots and then test them to see if they satisfy the equation.

The Rational Root Theorem states that if a rational number p/q (in simplest form) is a root of the polynomial equation with integer coefficients, then p is a factor of the constant term -5, and q is a factor of the leading coefficient 1.

The factors of -5 are ±1, ±5, and the factors of 1 are ±1, ±1. Therefore, the potential rational roots are ±1, ±5.

Using synthetic division or plugging in these potential roots, we can test them to see if they satisfy the equation:

For x = 1:
1^3 + 3(1^2) + 1 - 5 = 0
1 + 3 + 1 - 5 = 0
0 = 0

Since the equation is satisfied for x = 1, it means that (x - 1) is a factor of the equation.

Performing synthetic division with (x - 1) as a divisor:
1 | 1 3 1 -5
___|____________
| 1 4 5
1 4 5 0

The result is a quotient of x^2 + 4x + 5 and no remainder. Therefore, the quadratic equation x^2 + 4x + 5 = 0 is found.

Using the quadratic formula, we find the roots of the quadratic equation:
x = (-4 ± √(4^2 - 4(1)(5))) / 2(1)
x = (-4 ± √(16 - 20)) / 2
x = (-4 ± √(-4)) / 2

Since the square root of a negative number results in an imaginary number, the roots of the quadratic equation are complex and not real.

Hence, the real roots of the equation x^3 + 3x^2 + x - 5 = 0 are x = 1.

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