Question

Verify that 1, 2 and 3/2are the zeroes of the cubic polynomial p (x) = 2x3 - 9x2 + 13x - 6.
Then, verify the relationship between the zeroes and the coefficients of the polynomial.​

Answer 1

Answer:

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Answer 2

p(x) = 2x³ - 9x² + 13x - 6

x= 1, 2, 3/2

putting x= 1, we get

= 2(1)³ - 9(1)² + 13(1) - 6

= 2 - 9 + 13 - 6

= -7 + 7

= 0

putting x= 2

= 2(2)³ - 9(2)² + 13(2) - 6

= 16 - 36 + 26 - 6

= -20 + 20

= 0

putting x= 3/2

= 2(3/2)³ - 9(3/2)² + 13(3/2) - 6

= 2× 27/8 - 9× 9/4 + 39/2 - 6

= 27/4 - 81/4 + 39/2 - 6

taking LCM

= 27 - 81 + 78 - 24/4

= -54 + 54/4

= 0/4

= 0

Sum of the zeroes= -(coefficient of x²)/coefficient of x³

1 + 2 + 3/2= -(-9)/2

3 + 3/2 = 9/2

9/2 = 9/2

Product of the sum of the zeroes = coefficient of x/coefficient of x²

1(2) + 2(3/2) + (1)3/2 = 13/2

2 + 3 + 3/2 = 13/2

13/2 = 13/2

products of the zeroes= -(constant term)/coefficient of x²

1 × 2 × 3/2 = -(-6)/2

3 = 3