• # question_answer The HCF of ${{X}^{4}}-1$and${{X}^{4}}-2{{X}^{3}}-2{{X}^{2}}-2X-3$is A)  $({{x}^{2}}+1)\,(x-1)$      B)         $({{x}^{2}}+1)$               C)         $({{x}^{2}}+1)\,(x+1)$    D)         $(x+1)$

${{x}^{4}}-1=({{x}^{2}}-1)\,({{x}^{2}}+1)=(x-1)\,(x+1)$ $({{x}^{2}}+1)$ Now ${{x}^{4}}-2{{x}^{3}}-2{{x}^{2}}-2x-3$ Putting $x=-1$in this equation gives 0, so $(x+1)$ is a factor, divide ${{x}^{4}}-2{{x}^{3}}-2{{x}^{2}}-2x-3$by $(x+1)$ gives ${{x}^{3}}-3{{x}^{2}}+x-3$ Now put $x=3,$gives 0, so another factor is $(x-3),$ divide $(x-3)$ gives ${{x}^{2}}+1$which cannot be farther divided So${{x}^{4}}-2{{x}^{3}}-2{{x}^{2}}-2x-3=({{x}^{2}}+1)(x+1)(x-3)$ Now common factors in both expressions are $({{x}^{2}}+1)\,(x+1)$ which is the HCF.