Verified Answer :

konrad509

\(x^3 + x^2 - 20x=0\\ x(x^2+x-20)=0\\ x(x^2+5x-4x-20)=0\\ x(x(x+5)-4(x+5))=0\\ x(x-4)(x+5)=0\\ x=0 \vee x=4 \vee x=-5\)

Answer:

B. 0, −5, 4

Step-by-step explanation:

Given :


\(x^{3} +x^{2} -20x\)


Solution:


\(x^{3} +x^{2} -20x\)


taking x common from the given polynomial


⇒\(x(x^{2} +x-20)=0\)


⇒\(x(x(x+5)-4(x+5))=0\)


⇒\(x(x+5)(x-4)=0\)


⇒ x=0 , x+5=0 , x-4=0


⇒x = 0 , x = -5 , x = 4


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