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Factor by applying the difference of cubes.
Roots: 2, - 1 + isqrt(3), - 1 - isqrt(3)
Number and Type of Roots: one real root, two imaginary roots
To solve the given equation, we will apply the difference of two cubes. a^3-b^3 ⇔ (a-b)(a^2+ab+b^2) Let's start by rewriting 8 as 2^3 and applying the difference of two cubes.
Use the Zero Product Property
(I): LHS+2=RHS+2
From Equation (I), we found that one root is x=2. To find other roots, we will solve Equation (II). Note that this is a quadratic equation. Thus, we will use the Quadratic Formula. ax^2+bx+c=0 ⇕ x=- b±sqrt(b^2-4ac)/2a To do so, we first need to identify a, b, and c. x^2+2x+4 = 0 ⇕ 1x^2+ 2x+ 4=0 We can see above that a= 1, b= 2, and c= 4. Let's substitute these values into the formula and evaluate the right-hand side.
Substitute values
Calculate power
Identity Property of Multiplication
Multiply
Subtract term
sqrt(- a)= sqrt(a)* i
Split into factors
sqrt(a* b)=sqrt(a)*sqrt(b)
Calculate root
Factor out 2
Cancel out common factors
Commutative Property of Multiplication
We found that the roots of the quadratic equation, and thus of the given equation, are x=- 1 ± isqrt(3). Roots x=2, x=- 1 + isqrt(3), x=- 1 - isqrt(3) We can see that there are three roots. One of these roots is real and two of them are imaginary.