Question

Verify that 3, -1 and −1 3 , are zeroes of the polynomial p(x) = 3x3 – 5x2 – 11x – 3. Then, verify the relationships between the zeroes and its coefficients

Answer 1

Answer:

p(x)= 3(3)³- 5(3)² - 11(3) - 3

p(x)= 27 - 45- 33 -3

p(x)= 0

||| ly check for other two values

Answer 2

SOLUTION:-

Given

p(x) =3x³-5x²-11x-3

Explanation

p(x) = 3x³-5x²-11x-3

putting x= 3

p(3)= 3×(3)³-5×(3)²-11×3-3

= 3×27-5×9-33-3

= 81-45-33-3

=81-81

=0

p(x) = 3x³-5x²-11x-3

Putting x= -1

p(-1)= 3×(-1)³-5×(-1)²-11×(-1)-3

= 3×(-1)-5×1+11-3

=-3-5+11-3

=-11+11

=0

p(x) = 3x³-5x²-11x-3

putting x= -1/3

p(-1/3)= 3×(-1/3)³-5×(-1/3)²-11×(-1/3)-3

= 3×(-1/27)-5×1/9+11/3-3

= -1/9-5/9+11/3-3

= -1-5+33-27 / 9

= -33+33 / 9

= 0

Relation between zeros and coefficients

$\alpha + \beta + \gamma = \frac{ - b}{a} \\ \\ = 3 + ( - 1) + ( \frac{ - 1}{3} ) = \frac{ - ( - 5)}{3} \\ \\ = 3 - 1 - \frac{1}{3} = \frac{5}{3} \\ \\ = \frac{9 - 3 - 1}{3} = \frac{5}{3} \\ \\ = \frac{9 - 4}{3} = \frac{5}{3} \\ \\ = \frac{5}{3} = \frac{5}{3}$

$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} \\ \\ = 3 \times ( - 1) + ( - 1) \times \frac{ - 1}{3} + ( \frac{ - 1}{3} \times 3) = \frac{ - 11}{3} \\ \\ = - 3 + \frac{1}{3} - \frac{3}{3} = \frac{ - 11}{3} \\ \\ = \frac{ - 9 + 1 - 3}{3} = \frac{ - 11}{3} \\ \\ = \frac{ - 11}{3} = \frac{ - 11}{3}$

$\alpha \beta \gamma = \frac{ - d}{a} \\ \\ = 3 \times ( - 1) \times \frac{1}{3} = \frac{ - 3}{3} \\ \\ = \frac{ - 3}{3} = \frac{ - 3}{3}$

Hence verified!!!