Question

If the zeroes of the polynomial x^2 + (a+1)x + b are 2 and -3. Then find a and b. ( please answer with explanation )

Answer 1

x = 2

Then,

x^2 + (a+1)x + b

= 2^2 + (a+1)2 + b (substituting x)

= 4 + 2a + 2 + b

Now, when zeroes of a polynomial are given, the polynomial is always = 0.

So,

4 + 2a + 2 + b = 0

6 + 2a + b = 0

2a + b = 6

a + b = 3

a = 3 - b

x = -3

x^2 + (a+1)x + b

-3^2 + (a+1)3 + b = 0

9 + 3a + 3 + b = 0

9 + 3 (3 - b) + 3 + b = 0

9 + 9 -3b + 3 + b = 0

21 - 2b = 0

2b = 21

b = 21/2

Now to find a:

a = 3 - b

a = 3 - 21/2

a = -15/2