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Chapter 3: Problem 32

Find a polynomial equation with real coefficients that has the given roots. $$-2, i \sqrt{2}$$

Short Answer

Expert verified
The polynomial equation is x虏+ 22.

Step by step solution

01

Note the Roots

Given roots are -2 and i Therefore the roots are i and -i because they're conjugates of one another.For the remaining root of i 鈭�2, we must also take the conjugate -i鈭�2 because coefficients must be real numbers.So they must all be the conjugates of on another.
02

Form Factors for Each Pair of Roots

For the roots -2, (x + 2) (x- i鈭�2), (x + i鈭� 2) as follows because the conjugate coincides -They give a factor of:
03

Combine Like Terms

Combining like terms, we get polynomial (x + 2)*(x - i 鈭�2)*( x+ i鈭�2) in one polynomial where for practical reasons where we know used 2 square roots to terms that we've gotten egative (-2 ), cases as always equalsSo our polynomial equation will then be:f x) x^2 + 2= 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Roots of Polynomials
Polynomials can have several types of roots (solutions), which are values of x for which the polynomial equals zero. These roots can be real numbers, complex numbers, or a combination of both.

When given specific roots, like -2 and i鈭�2 in this problem, you can create a polynomial by setting up factors for each of these roots.
For example, if x = r is a root, then (x - r) is a factor.
  • For the root -2, the factor is (x + 2) because -2 is the root.
  • For the root i鈭�2, the factor is (x - i鈭�2).

When the polynomial has real coefficients, any complex roots must come in conjugate pairs. This means that if i鈭�2 is a root, then its conjugate -i鈭�2 is also a root. Thus, the factor (x + i鈭�2) must also be included.
Complex Conjugates
Complex conjugates are pairs of numbers of the form a + bi and a - bi, where 'a' and 'b' are real numbers, and 'i' is the imaginary unit, which satisfies i虏 = -1.

In polynomial equations with real coefficients, complex roots occur in conjugate pairs to ensure that the polynomial remains real.

In our example, the given root i鈭�2 pairs with -i鈭�2 because:
  • Their sum and product are both real numbers.
  • They ensure that the polynomial's coefficients stay real.
  • When multiplying the factors of these roots, the imaginary parts cancel out, which keeps the coefficients real.

So, the full polynomial formed includes both (x - i鈭�2) and (x + i鈭�2) as factors.
Real Coefficients
A polynomial has real coefficients when all of its numerical coefficients are real numbers. To form a polynomial with real coefficients, even when you have complex roots, you need to include both the root and its conjugate.

Using our example:
  • The factors of -2, i鈭�2, and -i鈭�2 lead to (x + 2), (x - i鈭�2), and (x + i鈭�2).
  • To confirm the polynomial has real coefficients, notice how loomsout in multiping the conjugates (x - i鈭�2) and (x + i鈭�2) where the imaginary parts cancel: (x - i鈭�2)(x + i鈭�2) = x虏 - (i鈭�2)虏 = x虏 + 2.

So, combining factors from -2 and the conjugate pairs results in (x + 2)(x虏 + 2) = 0. This is a polynomial with real coefficients: x鲁 + 2x + 4 = 0.

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Most popular questions from this chapter

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