Question

show that p(x)=x3-3x2+2x-6 has only one real zero​

Answer 1

Answer:

P(x)=x3−3x2+2x−6,P(3)=33−3(3)2+2(3)−6=0 So x−3 is a factor and x=3 is a zero of above polynomial.So, P(x)=(x−3)(x2+ax+b)=x3+(a−3)x2+(b−3a)x−3bComparing coefficient of x2,a−3=−3,(a=0)Comparing coefficient of x0,−3b=−6,(b=2)P(x)=(x−3)(x2+2)=0x=3,±i2–√So given polynomial has only one real zero x=3.