How do you find all solutions to {3}+1=0?

Question

nebajcioz · Accepted Answer

Step 1
Using synthetic division and the fact that

x = - 1

is obviously a solution we find that we can expand this to:

(x + 1) (x^{2} - x + 1) = 0

In order to have

L H S = R H S

need one of the brackets to be equal to zero, ie
1)

(x + 1) = 0

2)

(x^{2} - x + 1) = 0

Fro m1 we note that

x = - 1

is a solution. We shall solve 2 using the quadratic formula:

x^{2} - x + 1 = 0

x = \frac{1 \pm \sqrt{{(- 1)}^{2} - 4 (1) (1)}}{2} = \frac{1 \pm \sqrt{- 3}}{2} = \frac{1 \pm \sqrt{3} i}{2}

Rosa Nicholson · Accepted Answer

Step 1 Given: x3+1=0 By Rational Root Theorem, all rational roots of a polynomial are in the form pq, where p divides the constant term 1 and q divides the leading coefficient 1. List all candidates pq ±1 Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions. x=−1 By Factor theorem, x−k is a factor of the polynomial for each root k. Divide x3+1 by z+1 to get x2−x+1. Solve the equation where the result equals to 0. x2−x+1=0 All equations of the form ax2+bx+c=0 can be solved using the quadratic formula: −b±b2−4ac2a Substitute 1 for a, -1 for b, and 1 for c in the quadratic formula. x=−(−1)±(−1)2−4×1×12 Do the calculations. x=1±−32 Since the square root of a negative number is not defined in the real field, there are no solutions. x∈⧸0 List all found solutions. x=−1

How do you find all solutions to ^{3}+1=0?

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