Question

Find the zeroes of polynomial and verify rhe relationship between zeroes and coefficient of x2-x-6

Answer 1

We have,

${x}^{2} - x - 6 \\ = {x}^{2} - 2x + 3x - 6 \\ = x(x - 2) + 3(x - 2) \\ = (x + 3)(x - 2)$

Hence, the zeroes of the polynomial are (-3) and (2).

Thus, verification:

-3 + 2 = -1 = -1/1 = coefficient of x/coefficient of x^2

And, -3×2 = -6 = -6/1 = constant term/coefficient of x^2

HOPE THIS COULD HELP!!!!

Answer 2

Answer:

Step-by-step explanation:

Let p(x)=x2-x-6

To find zeroes of p(x)

Let p(x) =0

=> x2-x-6=0

=> x2-3x+2x-6=0

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

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

=> x-3=0 & x+2=0

=> x= 3 & x= -2

Therefore the zeroes of p(X) are 3 & -2

Now sum of zeroes = -b/a

= -(-1)/1

=1

And 3+(-2)= 1

Also product of zeroes = c/a

=-6/1

=-6

And 2*(-3)=-6

Hence verified