BenoguigoliB

2021-03-09

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

Polynomial equation with real coefficients that has the roots
is ${x}^{3}-5{x}^{2}+8x-6=0$.
Given:
$3,1-i$
Formula Used: $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$
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,
$\left(x-3\right)\left[x-\left(1-i\right)\right]\left[x-\left(1+i\right)\right]=0$
$\left(x-3\right)\left(x-1+i\right)\left(x-1-i\right)=0$
Further use arithmetic rule,
$\left(a+b\right)\left(a-b\right)={a}^{2}+{b}^{2}$

Now, the polynomial equation is:
$\left(x-3\right)\left({x}^{2}-2x+1+1\right)=0$
$\left(x-3\right)\left({x}^{2}-2x+2\right)=0$
${x}^{3}-3{x}^{2}-2{x}^{2}+6x+2x-6=0$
${x}^{3}-5{x}^{2}=8x-6=0$
Hence, the polynomial equation of given roots

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