I don't know how to factorise the first polynomial but I found its root : \(\dfrac{1}{3}\left(\sqrt[3]{28+3\sqrt{87}}+\sqrt[3]{28-3\sqrt{87}}-2\right)\)

The seocnd polynomial can't be factorised without using the imaginary unit \(i\):

\(x^2-2x+5=x^2-2x+1-\left(-4\right)=\left(x-1\right)^2-\left(2i\right)^2\)

\(=\left(x-1-2i\right)\left(x-1+2i\right)\)

\(a^3+4a^2+4a+3=a^3+a^2+a+3a^2+3a+3\)

\(=a\left(a^2+a+1\right)+3\left(a^2+a+1\right)=\left(a+3\right)\left(a^2+a+1\right)\)

\(4a^2b^2-\left(a^2-b^2\right)^2=\left(2ab-a^2+b^2\right)\left(2ab+a^2-b^2\right)\)

\(=\left(a^2+2ab+b^2-2a^2\right)\left(a^2+2ab+b^2-2b^2\right)\)

\(=\left[\left(a+b\right)^2-\left(\sqrt{2}a\right)^2\right]\left[\left(a+b\right)^2-\left(\sqrt{2}b\right)^2\right]\)

\(=\left(a+b-\sqrt{2}a\right)\left(a+b+\sqrt{2}a\right)\left(a+b-\sqrt{2}b\right)\left(a+b+\sqrt{2}b\right)\)

\(a^3+6a^2+8+12a=\left(a+2\right)^3\)

\(4+4x-x^2=-\left(x^2-4x-4\right)=-\left(x^2-4x+4-8\right)\)

\(=-\left[\left(x-2\right)^2-\left(2\sqrt{2}\right)^2\right]=-\left(x-2-2\sqrt{2}\right)\left(x-2+2\sqrt{2}\right)\)

\(\left(2x-5\right)^2-\left(5x-3x\right)=4x^2-20x+25-2x\)

\(=4\left(x^2-\dfrac{11}{2}x+\dfrac{25}{4}\right)=4\left(x^2-2x.\dfrac{11}{4}+\dfrac{121}{16}-\dfrac{21}{16}\right)\)

\(=4\left[\left(x-\dfrac{11}{4}\right)^2-\left(\dfrac{\sqrt{21}}{4}\right)^2\right]=4\left(x-\dfrac{11+\sqrt{21}}{4}\right)\left(x-\dfrac{11-\sqrt{21}}{4}\right)\)