Question

If - 2 and 3 are the zeros of the quadratic polynomial x2 +(a + 1)x + b, then

Answer 1

Answer:

a=10/3 b=-46/3

Step-by-step explanation:

-2&3 are roots of equation

put x=-2 in above equation you will get

-2a+b=-2

put x=3

a+b=-12

solving both equation you get

a=10/3

b=-46/3

Answer 2

Answer: a = -2, b = -6

Step-by-step explanation:

p(x) = x² + (a + 1)x + b

Zeroes of this polynomial are (-2) and 3. This means if we put (-2) and 3 as x in the above equation, we should get 0.

p(-2) = (-2)² + (a + 1)(-2) + b = 0

4 - 2a - 2 + b = 0

-2a + b + 2 = 0 (First Equation)

p(3) = 3² + (a + 1)3 + b = 0

9 + 3a + 3 + b = 0

3a + b + 12 = 0 (Second Equation)

Now, we have two equations and two variables. So we can find the values of a and b.

First Equation: -2a + b + 2 = 0

b = 2a - 2

Put the value of b obtained here in the second equation.

Second Equation: 3a + b + 12 = 0

3a + (2a - 2) + 12 = 0

3a + 2a - 2 + 12 = 0

5a + 10 = 0

5a = -10

a = -2

Now, find the value of b.

b = 2a - 2

b = 2(-2) - 2

b = - 4 - 2

b = -6