Find a polynomial equation with real coefficients that has the given roots.

5 and -7i

Write a polynomial equation with roots 5 and -7i

x^3 - __ x^2 + __ x -__ = 0

Don't know if they still teach that, but ....

for any quadratic equation of the form
ax^2 + bx + c = 0
the sum of the roots is -b/a , and the product of the roots is c/a

complex roots always appear as conjugate pairs, so they must be
5-7i and 5+7i
sum = 10
product = (5-7u)(5+7i) = 25 - 49i^2 = 25+49 = 74

equation must be
x^2 - 10x + 74 = 0

or .... if the roots are 5-7i and 5+7i
then (x - (5-7i))(x - (5+7i)) = 0
(x - 5 + 7i)(x - 5 - 7i) = 0
x^2 - 5x - 7ix - 5x + 25 + 35i + 7ix - 35i - 49i^2 = 0
x^2 - 10x + 25 - 49i^2 = 0 , but i^2 = -1

x^2 - 10x + 74 = 0

Well, those are some interesting roots you've got there.

To construct a polynomial equation with real coefficients that has the given roots, you'll need the conjugate of -7i, which is 7i. This is because complex roots always come in conjugate pairs.

So, the polynomial equation you're looking for is:

(x - 5)(x + 7i)(x - 7i) = 0

Now, let's expand that equation and put it in standard form:

(x - 5)(x^2 - (7i)^2) = 0

(x - 5)(x^2 - 49i^2) = 0

(x - 5)(x^2 + 49) = 0

Now you have a polynomial equation with real coefficients:

x^3 - 5x^2 + 49x - 245 = 0

Voila! You've got your polynomial equation with the given roots.

To find a polynomial equation with the given roots, we need to use the fact that complex roots come in conjugate pairs. If -7i is a root, then its conjugate, 7i, is also a root.

Using the roots 5, -7i, and 7i, we can write the polynomial equation as follows:

(x - 5)(x - (-7i))(x - 7i) = 0

Now, let's simplify this equation.

(x - 5)(x + 7i)(x - 7i) = 0

Expanding the equation:

(x - 5)((x^2 - (7i)^2) = 0

Simplifying further:

(x - 5)(x^2 + 49) = 0

Finally, let's multiply the terms:

x^3 + 49x - 5x^2 - 245 = 0

Thus, the polynomial equation with real coefficients is:

x^3 - 5x^2 + 49x - 245 = 0

To find a polynomial equation with the given roots 5 and -7i, you need to use the fact that complex roots always occur in pairs. Since -7i is a root, its conjugate, 7i, must also be a root.

Now, let's build the polynomial equation by using the roots. We can start with the quadratic equation:

(x - 5)(x - 7i)(x - 7i) = 0

Expanding this equation:

(x - 5)(x^2 - 7ix - 7ix + 49) = 0
(x - 5)(x^2 - 14ix + 49) = 0

Now, we can simplify it further:

x(x^2 - 14ix + 49) - 5(x^2 - 14ix + 49) = 0
x^3 - 14ix^2 + 49x - 5x^2 + 70ix - 245 = 0

Finally, let's rearrange the terms in the equation to obtain the polynomial equation with real coefficients:

x^3 - (14i + 5)x^2 + (70i + 49)x - 245 = 0

Therefore, the polynomial equation with real coefficients that has the given roots is:

x^3 - (14i + 5)x^2 + (70i + 49)x - 245 = 0