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## Answers

p(x) = x² + (a + 1)x + b

The two zeros are 2 and -3.

So p(2) = 0

∴2² + (a+1)*2 + b = 0

∴4 + 2a + 2 + b = 0

∴2a + b = -6 -------------------- (1)

Also, p(-3) = 0

∴(-3)² + (a+1) * (-3) + b = 0

∴9 -3a - 3 + b = 0

∴6 = 3a - b

∴3a - b = 6 ----------------- (2)

Solving the two equations by Elimination Method (You can use other methods also),

2a + b = -6

3a - b = 6 (Adding (1) and (2) )

----------------

∴ 5a = 0

∴

**a = 0**

Substituting a = 0 in equation (1),

2a + b = -6

∴2(0) + b = -6

∴0 + b = -6

∴

**b = -6**

Thus a = 0 and b = -6.

Quadratic polynomial is x² + (a+1)x + b

zeroes of the polynomial are 2 and -3

sum of the zeroes = = 2-3

a+1 = -1

a = -2

product of the zeroes = = 2(-3)

b = -6