Question

Find zeroes of each of the following polynomials and verify the relationship between the zeroes and its coefficients.
1. 3y^{2}-10y-8

Answer 1

Answer:

Hello mate

here is your answer.

$3 {y}^{2} - 10y - 8$

$= 3 {y}^{2} - 12y + 2y - 8$

$= 3y(y - 4) + 2(y - 4)$

Therefore,

$\alpha = 4$

$\beta = - \frac{2}{3}$

Verification,

$\alpha + \beta = 4 + ( - \frac{2}{3})$

By taking LCM

$\alpha + \beta = \frac{12 - 2}{3}$

$\alpha + \beta = \frac{10}{3}$

$- \frac{b}{a} = - ( \frac{ - 10}{3} )$

$- \frac{b}{a} = \frac{10}{3}$

$\alpha \times \beta = 4 \times \frac{ - 2}{3}$

$\frac{c}{a} = \frac{ - 8}{3}$

$\frac{ - 8}{3} = \frac{ - 8}{3}$

Hence Verified.

hope it helps.

Answer 2

3y^2 - 10y - 8 [a=3;b=(-10);c=(-8)]
3y^2 - 12y + 2y -8 =
3y (y-4) + 2(y-4) =
(3y+2)(y-4)

Zero of (3y+2). Zero of (y-4)
3y+2=0. y-4=0
3y=(-2). y=4
y=(-2)/3

Sum of zeroes
-2/3+4 = (-2+12)/3 = (10/3) = -b/a
Product of zeroes
-2/3 * 4 = (-8/3). = c/a