A) No factor of degree between 2 and 5
B) \[{{x}^{2}}-3x+5\] as a factor
C) \[x+1\] as a factor
D) \[x-125\] as a factor
Correct Answer: B
Solution :
Putting \[{{x}^{2}}=3x-5\], we get \[{{x}^{6}}+18{{x}^{3}}+125\] \[={{(3x-5)}^{3}}+18x(3x-5)+125\] \[=27{{x}^{3}}-125-45x(3x-5)+54{{x}^{2}}-90x+125\] \[=27{{x}^{3}}-135{{x}^{2}}+225x+54{{x}^{2}}-90x\] \[=27{{x}^{3}}-81{{x}^{2}}+135x\] \[=27x({{x}^{2}}-3x+5)\] \[=27x(3x-5-3x+5)\] \[=0\] Hence \[{{x}^{2}}-3x+5\] is factor.
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