Question

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

Polynomial arithmetic
ANSWERED
asked 2021-01-19
Polynomial equation with real coefficients that has roots \(-1, 4 - 2i is x^3 - 7x^2 + 12x + 20 = 0.\)

Answers (1)

2021-01-20
Given:) \(-1, 4 - 2i\)
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.
\([x - (-1)] [x - (4 - 2i)][x-(4 + 2i)] = 0\)
\((x + 1) (x - 4 + 2i)(x - 4 - 2i) = 0\)
Further use arithmetic rule.
\((a + b)(a - b) = a^2 - b^2\)
Here, \(a = x - 4, b = 2i\)
Now, the polynomial equation is,
\((x + 1)[(x - 4)^2 - (2i)^2] = 0\)
Use arithmetic rule.
\((a - b)^2 = a^2 - 2ab + b^2\)
And \(i^2 = -1\)
Now, the polynomial equation is,
\((x + 1) (x^2 - 8x + 16 + 4) = 0\)
\((x + 1) (x^2 - 8x + 20) = 0\)
\((x^3 + x^2 - 8x^2 - 8x + 20x + 20) = 0\)
\(x^3 - 7x^2 + 12x + 20 = 0\)
Hence, the polynomial equation of giveb roots \(-1, 4 - 2i is x^3 - 7x^2 + 12x + 20 = 0\)
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