Given,
The polynomial x²-2x-80
To find,
The zero of the given polynomial x²-2x-80 and verify the relationship between zeroes and the coefficients.
Solution,
We can simply solve for the zero of the polynomial x²-2x-80 by putting the polynomial equals to 0.
x²-2x-80 =0
Now factorize the given polynomial
x²-(10-8)x-80=0
Using Distributive property
x²-10x+8x-80=0
Taking x common from the first 2 terms and 8 from the last two terms
x(x-10)+8(x-10)=0
Taking (x-10) common from both sides
(x-10)(x+8)=0
This implies either the first part is zero or the second part is zero.
(x-10)=0 (x+8)=0
x=10 x=-8
So the zeroes of the polynomial x²-2x-80 are 10 and -8.
Sum of zeroes = -b/a where 'b' is the coefficient of x and 'a' is the coefficient of x².
= -(- 2)/1
= 2
Product of zeros = c/a where 'c' is the constant term and 'a' is the coefficient of x².
=80/1
=80
As we can see that the sum of 10 and -8 is 2 and their product is 80 which is verified by the relationship between zeroes and the coefficients.
Hence, the required zeroes of polynomial x²-2x-80 are 10 and -8 and the relationship between zeroes and coefficients is verified.
