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

Polynomial equation with real coefficients that has roots

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broliY
Given:) $-1,4-2i$
Used Formula:
$\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$
Calculation:
If the polynomial has real coefficients, then it’s imaginary roots occur in conjugate pairs. So, a polynomial with the given root 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.
$\left[x-\left(-1\right)\right]\left[x-\left(4-2i\right)\right]\left[x-\left(4+2i\right)\right]=0$
$\left(x+1\right)\left(x-4+2i\right)\left(x-4-2i\right)=0$
Further use arithmetic rule.
$\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$
Here, $a=x-4,b=2i$
Now, the polynomial equation is,
$\left(x+1\right)\left[\left(x-4{\right)}^{2}-\left(2i{\right)}^{2}\right]=0$
Use arithmetic rule.
$\left(a-b{\right)}^{2}={a}^{2}-2ab+{b}^{2}$
And ${i}^{2}=-1$
Now, the polynomial equation is,
$\left(x+1\right)\left({x}^{2}-8x+16+4\right)=0$
$\left(x+1\right)\left({x}^{2}-8x+20\right)=0$
$\left({x}^{3}+{x}^{2}-8{x}^{2}-8x+20x+20\right)=0$
${x}^{3}-7{x}^{2}+12x+20=0$
Hence, the polynomial equation of giveb roots $-1,4-2iis{x}^{3}-7{x}^{2}+12x+20=0$