find the zero of the polynomial x^2-2x-80 and verify the relationship between zeroes and the coefficients
Answers
Answer:
Zeroes are 10 and -8.
Step-by-step explanation:
Given,
x² - 2x - 80 = 0
x² - 10x + 8x - 80 = 0
x(x - 10) + 8(x - 10) = 0
(x - 10)(x + 8) = 0
x - 10 = 0 and x + 8 = 0
x = 10 and x = -8.
Therefore x = 10, -8
Let these zeroes be m and n.
Sum of zeroes = m + n = -b/a
Here a and b are coefficient of x² and x respectively.
Sum of zeroes = 10 - 8 = -(-2)/1
Sum of zeroes = 2 = 2
Hence verified.
Product of zeroes = m*n = c/a
Here c is constant term of equation.
Product of zeroes = 10*(-8) = -80/1
Product of zeroes = -80 = -80
Hence verified.
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.