Calculate: Polymomial equation with real coefficients taht has roots 3, 1 - i.

Question

Sadie Eaton · Accepted Answer

Polynomial equation with real coefficients that has the roots $3, 1 - i$
is $x^{3} - 5 x^{2} + 8 x - 6 = 0$ .
Given:
$3, 1 - i$
Formula Used: $(a + b) (a - b) = a^{2} - b^{2}$
Calculation:
If the polynomial has real coefficients, then it’s imaginary roots occur in conjugate pairs. So, a polynomial with the given root $1 - i$
must have another root as $1 + i$ .
Since each root of the equation corresponds to a factor of the polynomial, also, the roots indicate zeros of that polynomial, thus, the polynomial equation is written as,
$(x - 3) [x - (1 - i)] [x - (1 + i)] = 0$
$(x - 3) (x - 1 + i) (x - 1 - i) = 0$
Further use arithmetic rule,
$(a + b) (a - b) = a^{2} + b^{2}$
$and i^{2} = - 1$
Now, the polynomial equation is:
$(x - 3) (x^{2} - 2 x + 1 + 1) = 0$
$(x - 3) (x^{2} - 2 x + 2) = 0$
$x^{3} - 3 x^{2} - 2 x^{2} + 6 x + 2 x - 6 = 0$
$x^{3} - 5 x^{2} = 8 x - 6 = 0$
Hence, the polynomial equation of given roots $3, 1 - i is x^{3} - 5 x^{2} + 8 x - 6 = 0$

Calculate: Polymomial equation with real coefficients taht has roots 3, 1 - i.

Answered question

Answer & Explanation

New Questions in Algebra II