If a^2-3a+1=0 , find the value of the algebraic expression 3a^3-8a^2+a+\dfrac{3}{a^2+1}
Steven Zheng
posted 3 weeks ago
Question
Answer
3a^3-8a^2+a+\dfrac{3}{a^2+1}
= 3a^3-9a^2+3a+a^2-2a+\dfrac{3}{a^2+1}
=3a(a^2-3a+1) +a^2-2a+\dfrac{3}{a^2+1}
= a^2-2a+\dfrac{3}{a^2+1}
=a^2-3a+1+a-1+\dfrac{3}{a^2-3a+1+3a}
=a-1+\dfrac{1}{a}
=\dfrac{a^2-a+1}{a}
=\dfrac{a^2-3a+1+2a}{a}
=2
Steven Zheng
posted 3 weeks ago