-3,0,2 are the zeros of the polynomial p(x)=x3+(a-1) x2+bx+c find a and c
Answers
To find the answer, we will use the concept of relationship between zeros and coefficients of a cubic equation.
Answer:
The value of a is 2 and the value of c is 0.
Step-by-step explanation:
Since -3 , 0 , 2 are the zeroes of the polynomial
p(x) = x³ + (a - 1)x² + bx + c
It means that the values -3 , 0 , 2 will satisfy the given polynomial.
=> p(-3) = (-3)³ + (a - 1)(-3)² + b(-3) + c
0 = - 27 + 9 (a - 1) - 3b + c --(i)
=> p(0) = (0)³ + (a - 1)(0)² + b(0) + c
0 = 0 + 0 (a - 1) + 0 + c
Therefore, c = 0
=> p(2) = (2)³ + (a - 1)(2)² + b(2) + c
0 = 8 + 4 (a - 1) + 2b + c --(ii)
Putting c = 0 in equation (i) and (ii),
=> 0 = 8 + 4a - 4 + 2b + 0 and 0 = - 27 + 9 (a - 1) - 3b + 0
=> 0 = 4 + 4a + 2b + 0 and 0 = - 27 + 9a - 9 - 3b + 0
=> 0 = 2 + 2a + b and 0 = - 12 + 3a - b + 0
=> 2a + b = -2 and 3a - b = 12
Solving these two equations,
=> 5a = 10
=> a = 2
Therefore, the value of a is 2 and the value of c is 0.