Question

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 Steven Zheng posted 3 weeks ago


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 Steven Zheng posted 3 weeks ago

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