# Question 3 and 4

Dear Student,

**Answer 3:**

When 2 and -3 are the zeroes of the polynomial p(x) = x^{2} + (a + 1) x + b

Therefore,

p(2) = 2^{2}+ (a + 1) 2 + b = 0

4 + 2a + 2 + b = 0

6 + 2a + b = 0

2a + b = -6 - - - - - - - - - - (1)

Now,

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

9 - 3a - 3 + b = 0

6 - 3a + b = 0

-3a + b = -6

3a - b = 6 - - - - - - - - -- (2)

Adding (1) and (2),

2a + b = -6

3a - b = 6

- - - - - - - - -

5a = 0

$a=\frac{0}{5}$

**a = 0**

Now, Substituting the value of a in 2

3 x 0 - b = 6

-b = 6

**b = -6**

Therefore,

a = 0 and b = -6

Please post the remaining queries in separate threads so that we can help you as soon as possible.Regards

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