We have, f(x)=4x4+4x3−23x2−12x+36=0 Using concept a polynomial f(x) has a multiple root r, then f(r)=0 and ‌f′(r)=0 ‌f′(x)=16x3+12x2−46x−12 Now, using hit and trial By taking x=−2 f(−2)=4(−2)4+4(−2)3−23(−2)2 −12(−2)+36 =64−32−92+24+36=0 ⇒f′(−2)=16(−2)3+12(−2)2 −46(−2)−12 =−128+48+92−12=0 ∵x=−2 is a multiple root. By dividing the equation by (x+2)2, we get another multiple root x=‌