Question

$\frac{2}{3} \frac{3}{2}$
Find the polynomial by the given zeroes. ​

Answer 1

Given :

Zeroes of the polynomial are 2/3 and 3/2

To Find :

Find the polynomial by the given zeroes.

Solution:

As we know that :-

α + β = -b/a

→ 2/3 + 3/2 = -b/a

→ 4 + 9/6 = -b/a

→ 13/6 = -b/a ....... (i)

And

αβ = c/a

→ 2/3 × 3/2 = c/a

→ 6/6 = c/a ..... (ii)

Now, From (i) and (ii), we get :

a = 6

b = -13

c = 6

As we know that :-

For a quadratic polynomial :

ax² + bx + c

→ 6x² + (-13)x + 6

→ 6x² - 13x + 6

Verification :-

→ 6x² - 13x + 6

→ 6x² - 9x - 4x + 6

→ 3x(2x - 3) -2(2x - 3)

→ (3x - 2)(2x - 3)

Zeroes are -

3x - 2 = 0 and 2x - 3 = 0

x = 2/3 and x = 3/2

Verified.

Answer 2

$\large\underline\mathrm{Given:-}$

Zeroes of the polynomial are 2/3 and 3/2

$\large\underline\mathrm{To \: find}$

Find the polynomial by the given zeroes.

$\large\underline\mathrm{Solution}$

$\alpha + \beta = - b \div a$

$\implies$ (4 + 9)/6 = -b/a

$\implies$ 13/6 = -b/a. .....(1)

and

$\alpha \beta = c \div a$

$\implies$ 2/3 × 3/2 = c/a

$\implies$ 6/6 = c/a. ....(2)

from (1) and (2).. we get

$\implies$ a = 6

$\implies$ b = -13

$\implies$ c = 6

For a quadratic polynomial:

$\implies$ ax² + bx + c

$\implies$ 6x² + (-13)x + 6

$\implies$ 6x² - 13x + 6

$\implies$ 6x² - 9x - 4x + 6

$\implies$ 3x(2x - 3) - 2(2x - 3)

$\implies$ (3x - 2) (2x - 3)

$\implies$ x = 2/3, or 3/2