Question

if zero of x3 -3x2+x+1 are a-b,a,a+b, then find the value of a2​

Answer 1

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

Answer 2

Step-by-step explanation:

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

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