 Polynomial equation with real coefficients that has the roots 3, 1 -i texr{is} x^{3} - 5x^{2} + 8x - 6 = 0. SchachtN 2020-12-25 Answered

Polynomial equation with real coefficients that has the roots $$3, 1 -i\ \text{is}\ x^{3} - 5x^{2} + 8x - 6 = 0.$$

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Polynomial equation with real coefficients that has the roots $$3, 1 - i\ is\ x^{3} - 5x^{2} + 8x - 6 = 0$$.
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,
$$(a + b)(a - b) =a^{2} - b^{2}$$
$$\text{Here } a=x-1,b=i$$
Now, the polynomial equation is,
$$(x - 3)[(x - 1)^{2} - (i)^{2}] = 0$$
Use ariyhmetic rule.
$$(a - b)^{2} = a^{2} - 2ab + b^{2}$$
$$\text{and } i^{2} = -1$$.
Now the polynomial equation is:
$$(x - 3)(x^2 - 2x + 1 + 1) = 0$$
$$(x - 3)(x^{2} -2x + 2) = 0$$
$$x^{3} - 3x^{2} + 6x + 2x - 6 = 0$$
$$x^{3} - 5x^{2} + 8x - 6 = 0$$
Hence, the polynomial equation of given roots $$3, 1 - i \text{ is } x^{3} - 5x^{2} + 8x - 6 = 0$$