Write a polynomial of least degree with real coefficients and with the root

Question

Write a polynomial of least degree with real coefficients and with the root

–5+13𝑖

Answer 1

the other root is -5-13i, so

y = (x-(-5+3i))(x-(-5-13i))
y = ((x+5)-13i)((x+5)+13i)
y = (x+5)^2 + 13^2
y = x^2+10x+194

Answer 2

or

sum of roots = -5-13i + -5+13i = -10
product of roots = (-5-13i)(-5+13i)
= 25 - 169i^2 = 25+169 = 194

x^2 + 10x + 194

Answer 3

I like it!

Answer 4

To write a polynomial with the given complex root, we can make use of the conjugate pair theorem. The conjugate pair theorem states that if a polynomial has real coefficients, then complex roots must occur in conjugate pairs. In other words, if a+bi is a root of a polynomial with real coefficients, then a-bi must also be a root.

In this case, the given root is -5+13𝑖. Since we know that complex roots occur in conjugate pairs, the conjugate of -5+13𝑖 is -5-13𝑖. Therefore, -5-13𝑖 must also be a root of the polynomial.

To find the polynomial, we can set up the factors based on the roots:
(x - (-5 + 13𝑖))(x - (-5 - 13𝑖))

Simplifying this expression:
(x + 5 - 13𝑖)(x + 5 + 13𝑖)

Using the difference of squares formula (a^2 - b^2 = (a + b)(a - b)):
[(x + 5)^2 - (13𝑖)^2]

Squaring binomials:
(x^2 + 10x + 25 - 169𝑖^2)

The imaginary unit i^2 is defined as -1, so the expression simplifies further:
(x^2 + 10x + 25 + 169)

Combining like terms, the polynomial is:
x^2 + 10x + 194