Polynomial equation with real coefficients that has roots -1, 4 - 2i is x^3 - 7x^2 + 12x + 20 = 0.

Question

Polynomial equation with real coefficients that has roots $- 1, 4 - 2 i i s x^{3} - 7 x^{2} + 12 x + 20 = 0.$

Answer 1

Given:)

- 1, 4 - 2 i

Used Formula:

(a + b) (a - b) = a^{2} - b^{2}

Calculation:
If the polynomial has real coefficients, then it’s imaginary roots occur in conjugate pairs. So, a polynomial with the given root 4 — 2i must have another root as 4 + 2i.
Since each root of the equation corresponds to a factor of the polynomial also the roots indicate zeros of that polynomial.
Hence, below mentioned equation is written as.

[x - (- 1)] [x - (4 - 2 i)] [x - (4 + 2 i)] = 0

(x + 1) (x - 4 + 2 i) (x - 4 - 2 i) = 0

Further use arithmetic rule.

(a + b) (a - b) = a^{2} - b^{2}

Here,

a = x - 4, b = 2 i

Now, the polynomial equation is,

(x + 1) [(x - 4)^{2} - (2 i)^{2}] = 0

Use arithmetic rule.

(a - b)^{2} = a^{2} - 2 a b + b^{2}

And

i^{2} = - 1

Now, the polynomial equation is,

(x + 1) (x^{2} - 8 x + 16 + 4) = 0

(x + 1) (x^{2} - 8 x + 20) = 0

(x^{3} + x^{2} - 8 x^{2} - 8 x + 20 x + 20) = 0

x^{3} - 7 x^{2} + 12 x + 20 = 0

Hence, the polynomial equation of giveb roots

- 1, 4 - 2 i i s x^{3} - 7 x^{2} + 12 x + 20 = 0

Polynomial equation with real coefficients that has roots -1, 4 - 2i is x^3 - 7x^2 + 12x + 20 = 0.

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