Find the polynomial equation of lowest degree, with real coefficients and leading coefficient 1 1 that has the given roots.

x^4 + 2x^3 + 3x^2 + 2x + 2 = 0

Step-by-step explanation:

The computation of the polynomial equation of the lowest degree is shown below

As we know that the complex roots would always arise in the conjucate pairs

As -i is a root, i is also a root

As  -1 + i is a root

And, -1 is also a root

Now the polynomial equation would be

(x + i)(x - i)(x + 1 - i)(x + 1 - i) = 0

(x^2 - i^2)[(x + 1)^2 - i^2] = 0

(x^2 + 1)[(x + 1)^2 + 1] = 0

(x^2 + 1)(x^2 + 2x + 2) = 0

x^4 + 2x^3 + 3x^2 + 2x + 2 = 0

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°