if zero of x3 -3×2+x+1 are a-b,a,a+b, then find the value of a2​

2 thoughts on “if zero of x3 -3×2+x+1 are a-b,a,a+b, then find the value of a2​”

1. Step-by-step explanation:

   [tex] {x}^{3} – 3 {x}^{2} + x + 1 \\ \alpha = a – b \: , \: \beta =a \: , \: \gamma =a+b \\ sum \: of \: zeroes: \\ \alpha + \beta + \gamma = \frac{-b}{a} \\ (a-b) + a + (a+b)=\frac{-b}{a} \\ 3a = \frac{ – b}{a} \\ {a}^{2} = \frac{ – b}{3} \\ {a}^{2} = \frac{ – ( – 3)}{3} \\ {a}^{2} = 1[/tex]

   HOPE THIS HELPS

   MARK ME BRAINLIEST

   Reply

2. Step-by-step explanation:

   Correct option is

   D

   a=0,b=−6

   x

   2

   +(a+1)x+b is the quadratic polynomial.

   2 and −3 are the zeros of the quadratic polynomial.

   Thus, 2+(−3)=

   1

   −(a+1)

   =>

   1

   (a+1)

   =1

   =>a+1=1

   =>a=0

   Also, 2×(−3)=b

   =>b=−6

   Reply