Question

Can someone please help me solve 91

Answer 1

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

Answer 2

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}