Question

complex numbers online practice
complete this assessment to review what you’ve learned. it will not count toward your grade.
given the equation ( x^3 + 6x^2 - 10x - 60 = 0 ), which of the following describes a possible solution set? (1 point)

• the equation has four non - real solutions.
• the equation has all real solutions.
• the equation has two real solutions and one non - real solution.
• the equation has all non - real solutions.

check answer remaining attempts : 3

Explanation:

Step1: Analyze the degree of the polynomial

The given equation is a cubic (degree 3) polynomial: \(x^3 + 6x^2 - 10x - 60 = 0\). By the Fundamental Theorem of Algebra, a degree \(n\) polynomial has \(n\) roots (real or complex), counting multiplicities. So a cubic has 3 roots in total.

Step2: Recall properties of complex roots

For polynomials with real coefficients, complex roots (non - real) come in conjugate pairs. That is, if \(a + bi\) is a root, then \(a - bi\) is also a root.

Step3: Analyze each option

• Option 1: The equation is cubic (degree 3), so it can have at most 3 roots. So "four non - real solutions" is impossible.
• Option 2: Let's try to factor the polynomial. We can factor by grouping:

Group the terms as \((x^3+6x^2)+(- 10x - 60)=x^2(x + 6)-10(x + 6)=(x^2 - 10)(x + 6)\).
Set each factor equal to zero:

• \(x+6 = 0\) gives \(x=-6\) (a real root).
• \(x^2-10=0\) gives \(x=\pm\sqrt{10}\) (both real roots, since \(\sqrt{10}\) and \(-\sqrt{10}\) are real numbers).

So the roots are \(x=-6,x = \sqrt{10},x=-\sqrt{10}\), all real.

• Option 3: Since complex roots come in pairs, if there was 1 non - real root, there would be 2 non - real roots (a pair), but the total number of roots is 3. So having 2 real and 1 non - real is impossible (because 1 non - real would imply at least 2 non - real, and \(2 + 2>3\) which is not possible for a cubic).
• Option 4: We found real roots, so "all non - real solutions" is false.

Answer:

The equation has all real solutions.