Question

12. Find the zeroes of the polynomial
x2 + 1/6x - 2 and verify the relationship
between the coefficients and the zeroes
of the polynomial​

Answer 1

Answer:

x²+1/6x-2=0

Multiply each term by 6

x²×6+1/6×6x-2×6=0

6x²+x-12=0

6x²+9x-8x-12=0

(6x²+9x)-(8x-12)=0

3x(2x+3)-4(2x+3)=0

(2x+3)(3x-4)=0

(2x+3=0)(3x-4=0)

x=(-3/2) and x=4/3

Therefore, the zeroes of the polynomial are (-3/2) and 4/3.

Sum of zeroes=(-3/2)+4/3=(-9+8)/6=(-1/6)= -(Coefficient of x)/Coefficient of x²=1/6

Product of zeroes=(-3/2)×(4/3)=(-2)=(-2)/1= Constant term/Coefficient of x²

Hence verified