Question

If the zeroes of polynomial x^3+3x^2+x+1 are a-b, a, a+b, find a and b please answer me fast ♥♥♥.

Answer 1

sum of zeroes = ( -coefficient Of x^2)/ coefficient of x^3

So

a-b + a + a + b = -3

3a = -3

a= -1

product of zeroes = ( - constant)/ coefficient of x^3

(a-b)( a)( a+b) = -1

put a = -1

( -1-b) ( -1)( -1+ b) = -1

1 - b^2 = 1

b^2 = 0
b= 0

Answer 2

Answer:

a = -1, b = 0

Step-by-step explanation:

Given Equation is x³ + 3x² + x + 1.

Given Zeroes are a-b,a,a+b.

(i)

Sum of zeroes = -b/a

a - b + a + a + b = -3/1

3a = -3/1

3a = -3

a = -1.

(ii)

Product of zeroes : -d/a

(a - b) * a * (a + b) = -1

a³ - ab² = -1

(-1)³ - (-1)(b)² = -1

-1 + b² = -1

b² = 0

b = 0.

Therefore:

a = -1

b = 0.

Hope it helps!