# Find the cubic equation that has -1 and 2i as roots

• MHB
• pan90
In summary, the conversation discusses finding a cubic polynomial with given roots and explains that for polynomials with real coefficients, a complex root also has its complex conjugate as a root. The solution process involves substituting the given roots into different expressions and eliminating options until the correct polynomial is found.
pan90
Answer is given, but no explanation or logic for it.

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pan90 said:
Answer is given, but no explanation or logic for it.
From HiSet free practice test
You need to show what you have tried, even if it is wrong. That will enable us to help you better.

Note that, for a polynomial with real coefficients, if we have a complex root then we also have the complex conjugate as a root as well. So your cubic polynomial is $$\displaystyle f(x) = a( x - (-1))(x - (2i))(x - (-2i))$$.

(The "a" is there because the polynomial will have the same roots no matter what constant is multiplying the whole thing. Your answer key is setting a = 1.)

-Dan

Edit: The method worked to give the right answer, but the signs were wrong. The terms for the given roots are of the form x - r, not x + r as I had originally written.

Substitute $x=-1$ and $x=2i$ in each of the given expressions. Do you get $0$? If the same expression gives $0$ for both these two values of $x$, then the equation is the one you’re looking for; otherwise, it isn’t.

Right, let’s do it one at a time. Start with $x=-1$. Substitute this in each of the five expressions A–E. You’ll find that B and D give you $0$ whereas A, C and E don’t. Therefore the correct answer is either B or D (you have eliminated A, C and E).

Now substitute $x=2i$ in B and in D. One will give you $0$ while the other won’t. The one that gives you $0$ is the one you’re looking for.

First, there exist an infinite number of different cubic equations that have "-1" and "2i" as roots, Notice that all of the proposed solutions have real coefficients. A polynomial equation with real coefficients but a non-real root must also have the complex conjugate of that non-real root as a root. The complex conjugate of "2i" is "-2i" so the polynomial must be $$(x+ 1)(x- 2i)(x+ 2i)= (x- 1)(x^2+ 4)= x^3- x^2+ 4x- 4$$.

## 1. What is a cubic equation?

A cubic equation is a polynomial equation of the form ax^3 + bx^2 + cx + d = 0, where a, b, c, and d are constants and x is the variable. It is called a cubic equation because the highest power of x is 3.

## 2. How do you find the roots of a cubic equation?

The roots of a cubic equation can be found by factoring, using the quadratic formula, or by using synthetic division. In this case, we can use the quadratic formula since we know that the roots are complex numbers.

## 3. What are the steps to find the cubic equation with -1 and 2i as roots?

The steps to find the cubic equation with -1 and 2i as roots are:

1. Write the equation in the form (x - r)(x - s)(x - t) = 0, where r, s, and t are the roots.
2. Substitute -1 and 2i for r and s, respectively.
3. Use the conjugate of 2i, -2i, as the third root.
4. Expand the equation and simplify to get the cubic equation.

## 4. Can a cubic equation have complex roots?

Yes, a cubic equation can have complex roots. In fact, every cubic equation has either three real roots or one real root and two complex roots.

## 5. How can I check if my cubic equation is correct?

You can check if your cubic equation is correct by substituting the roots into the equation and verifying that it equals 0. You can also graph the equation and see if the x-intercepts match the given roots.

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