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24x^3=10x^2+4x I know the roots for this equation but I cannot figure out the check.

Sagot :

Konrad509idn
[tex]24x^3=10x^2+4x \\ 24x^3-10x^2-4x=0\\ 2x(12x^2-5x-2)=0\\ 2x(12x^2+3x-8x-2)=0\\ 2x(3x(4x+1)-2(4x+1))=0\\ 2x(3x-2)(4x+1)=0\\ x=0 \vee x=\frac{2}{3} \vee x=-\frac{1}{4}[/tex]
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