Question

show that 2 and -1/3 are zeroes of the polynomial 3x2-5x-2​

Answer 1

To prove :-

Here , We have to prove that 2 and -1/3 are the zeroes of polynomials 3x^2 - 5x - 2

Solution :-

The given zeroes are 2 and -1/3

Given equation :- 3x^2 - 5x - 2

Put the required values in the equation,

3( 2 )^2 - 5( 28 ) - 2

12 - 10 - 2

2 - 2 = 0

Now, Take zero -1/3

3( -1/3)^2 - 5( -1/3 ) - 2

3 * 1/9 + 5/3 - 2

1/3 + 5/3 - 2

By taking LCM,

1 + 5 - 6 / 3

6 - 6 / 3

0/3 = 0

Hence, 2 and -1/3 are zeroes

of 3x^2 - 5x - 2 $.$

Answer 2

Answer:

$f(x) = 3 {x}^{2} - 5x - 2 \\ f(2) = 3 \times {2}^{2} - 5 \times 2 - 2 \\ f(2) = 3 \times 4 - 5 \times 2 - 2 \\ f(2) = 0 \\ f( - \frac{ 1}{3} ) = 3 \times ( - \frac{ 1}{3} ) ^{2} - 5( - \frac{1}{3} ) - 2 \\ f( - \frac{1}{3} ) = 3 \times ( \frac{1}{9} ) + 5( \frac{1}{3} ) - 2 \\ f( - \frac{1}{3} ) = \frac{1}{3} + \frac{5}{3} - 2 \\ f( - \frac{1}{3} ) = \frac{1 + 5 - 6}{3} \\ f( - \frac{1}{3} ) = 0$

hope it helps:)