Question

Find the zeros of the polynomial 6x2 - 3 - 7x and verify the relationship between the zeros and the coefficients of the polynomial.

Answer 1

=6x²-7x-3
=6x²+2x-9x-3
=2x(3x+1)-3(3x+1)
=(2x-3)(3x+1)
⇒2x-3=0                                    or                            ⇒3x+1=0
⇒x=3/2                                       or                           ⇒ x= - 1/3
α=3/2 ,β= - 1/3
⇒α+β= -b/a
⇒3/2+(-1/3)= - (-3)/6
⇒3/2-1/3=1/2
⇒7/6 =1/2
 ⇒αβ=c/a
 ⇒ 3/2(-1/3)= -7/6
⇒-1/2= -7/6
 ⇒1/2=7/6      

Answer 2

given,
6x²-7x - 3
to find the zeroes of the polynomial,
6x²-7x - 3 = 0
6x²-9x +2x -3 = 0
3x(2x-3) + 1(2x -3) = 0
(3x+1) (2x-3) = 0
x = -1/3 or 3/2
sum of the zeroes = 3/2 - 1/3 = 7/6 = - ( co-efficient of x) / co-efficient of x²
product of the zeroes = -1/3 × 3/2 = -1/2  = constant / co-efficient of x²