If the zeros of the qudratic polynomial x^2+(a+1)x+b are 2and -3then
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Answer:
a= -6
b= 6
Step-by-step explanation:
x^2 + (a+1)x +b
putting the value of x = 2
2^2 + (a+1)2 + b =0
4 + 2a + 2 + b =0
6 + 2a + b=0 ______eq1
putting the value of x = -3
3^2 + (a+1)3 +b =0
9 + 3a + 3 + b =0
12 + 3a + b =0 _______eq2
from eq 1 and 2
12 + 3a + b - ( 6 + 2a + b ) =0
12 + 3a + b - 6 - 2a - b =0
6 + a = 0
a= -6
from ep 1
6 + 2 (-6) + b = 0
6 - 12 + b = 0
-6 + b = 0
b= 6
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