|=0 ⇒x(x2−56)−7(4x−40)+5(28−5x)=0 ⇒x3−56x−28x+280+140−25x=0 ⇒x3−109x+420=0 Since 5 and 7 are the roots of the given equation so that x2−12x+35 is a factor of the given equation, when we divide the given equation by x2−12x+35 it gives ⇒x+12=0 x=−12.