Verify a quartic root and find the other roots

Evaluate the polynomial expression first; do not assume it equals zero at the start of a verification:

$\begin{aligned}f(2i)&=(2i)^4-2(2i)^3+7(2i)^2-8(2i)+12\\&=16i^4-16i^3+28i^2-16i+12\\&=16+16i-28-16i+12\\&=0.\end{aligned}$

This establishes the claim. The real coefficients then imply that $-2i$ is also a root. The two factors are $(z-2i)$ and $[z-(-2i)]$, whose product is $z^2+4$. The remaining factor is quadratic because the original degree is four.

Write

$\begin{aligned}f(z)&=(z-2i)[z-(-2i)]F(z)\\&=(z-2i)[z-(-2i)](z^2+Az+B)\\&=(z^2+4)(z^2+Az+B).\end{aligned}$

Compare constants: $4B=12$, so $B=3$. Compare coefficients of $z^3$: $A=-2$. Hence $f(z)=(z^2+4)(z^2-2z+3)$.

Divide by $z^2+4$. The quotient terms are $z^2$, $-2z$, then $3$. Each multiplication and subtraction is:

$\begin{aligned}z^2(z^2+4)&=z^4+4z^2,\\f(z)-(z^4+4z^2)&=-2z^3+3z^2-8z+12,\\(-2z)(z^2+4)&=-2z^3-8z,\\(-2z^3+3z^2-8z+12)-(-2z^3-8z)&=3z^2+12,\\3(z^2+4)&=3z^2+12,\\(3z^2+12)-(3z^2+12)&=0.\end{aligned}$

Thus the quotient is $z^2-2z+3$ with zero remainder.

Use the quadratic factor:

$\begin{aligned}z^2-2z+3&=0,\\z&=\frac{2\pm\sqrt{4-4(1)(3)}}2\\&=\frac{2\pm\sqrt{-8}}2\\&=1\pm\sqrt2\,i.\end{aligned}$

Include the conjugate root $-2i$ already found. The given root $2i$ is not listed again among the other roots.

On the illustrated calculator, press apps, select PlySmlt2, then POLYNOMIAL ROOT FINDER. Select Order 4 and a+bi. Enter coefficients $1,-2,7,-8,12$ and press GRAPH.

The screen lists $x_1=2i$, $x_2=-2i$, $x_3=1+1.414213562\ldots i$, $x_4=1-1.414213562\ldots i$. Use the right cursor to see values truncated by an ellipsis. These decimal outputs check the exact roots above.

The question says verify and hence, so show the verification and factorisation working; calculator roots alone do not explain the deduction.