Odpowiedź:
[tex]\huge\boxed{e)\ w(x)=-3x(x-\sqrt5)^2(x+\sqrt5)^2}\\\boxed{f)\ w(x)=2x(2x-1)^2(2x+1)^2}[/tex]
Szczegółowe wyjaśnienie:
[tex]e)\ w(x)=-3x^5+30x^3-75x=-3x(x^4-10x^2+25)\\\\=-3x\underbrace{\bigg[(x^2)^2-2\cdot x^2\cdot5+5^2\bigg]}_{(a-b)^2=a^2-2ab+b^2}=-3x(x^2-5)^2\\\\=-3x\bigg[x^2-(\sqrt5)^2\bigg]^2=-3x\bigg[\underbrace{(x-\sqrt5)(x+\sqrt5)}_{a^2-b^2=(a-b)(a+b)}\bigg]^2\\\\=-3x(x-\sqrt5)^2(x+\sqrt5)^2[/tex]
[tex]f)\ w(x)=32x^6-16x^4+2x^2=2x^2(16x^4-8x^2+1)\\\\=2x^2\bigg[\underbrace{(4x^2)^2-2\cdot4x^2\cdot1+1^2}_{(a-b)^2=a^2-2ab+b^2}\bigg]=2x^2(4x^2-1)^2\\\\=2x^2\bigg[(2x)^2-1^2\bigg]^2=2x^2\bigg[\underbrace{(2x-1)(2x+1)}_{a^2-b^2=(a-b)(a+b)}\bigg]^2\\\\=2x(2x-1)^2(2x+1)^2[/tex]
