Can someone please help me solve 91

asked Sep 3, 2018 48.4k views

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$3x^3+7x^2-20x=0\\ x(3x^2+7x-20)=0\\ x(3x^2-5x+12x-20)=0\\ x(x(3x-5)+4(3x-5))=0\\ x(x+4)(3x-5)=0\\ x=0 \vee x=-4 \vee x=(5)/(3)$

Panwen Wang answered Sep 5, 2018

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3x^3+7x^2-20x=0 factor out x first

x(3x^2+7x-20)=0

x(3x^2+12x-5x-20)=0

x(3x(x+4)-5(x+4))=0

x(3x-5)(x+4)=0

x={-4, 0, 5/3}

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