Question

x2 - 11 x - 80 please answer this question

Answer 1

Heya !!!

This is the solution of your question.....

Given polynomial.
x² - 11x - 80

we have to find two numbers a and b such that a +b = -11 and ab = 80.

By looking at the prime factors of 80.

80= 2×2×2×2×5

We get they are 5 and 16...

So,
x²-11x-80
=> x² -16x + 5x -80 = 0
=> x(x -16) + 5 (x-16) = 0
=> (X+5)(x-16) = 0

Now, comparing with 0,
x +5 = 0. or. x-16=0
x = -5 or x = 16.

Hence, the zeroes of polynomial are 16 and -5.

Hope you got the answer

Answer 2

Heya,

x^2 - 11x - 80

=> Splitting the middle of term

=> x^2 - 16x + 5x - 80

=> x (x -16) 5 (x-16)

=> (x + 5) (x - 16) are the factors

=> -5 & +16 are the zeros of the given polynomial

Hope my answer helps you :)

Regards,
Shobana