Now we will find the zeros of the polynomial x2+x-p(p+1) by factorization method. And find the roots of the equation x2+x-p(p+1)=0 where p is a constant

We have a polynomial p(x) = x2+x-p(p+1) ….equ 1

let’s compare p(x) = 0 with standard polynomial p(x) = ax^{2} +bx +c …..equ 2

while comparing equation 1 with equation 2 we get

a = 1, b = 1 , c = p(p+1)

x2+x-p(p+1) = 0

## The Zeros of The Polynomial x2+x-p(p+1) by middle term split

x2+x-p(p+1) = 0

x2+(p+1)x -px -p(p+1) = 0

x{x+(p+1)} -p{x+(p+1)} = 0

{x-p}{x+(p+1)} = 0

now we got two factors

(x-p) = 0

x = p

{x+(p+1)} = 0

x = -(p+1)

Hence (p) and -(p+1) are the zeros of the Polynomial x2+x-p(p+1).

## Find the roots of the equation x2+x-p(p+1)=0 where p is a constant

If In the equation x2+x-p(p+1)=0, p is constant then we will compare this with standard quadratic equation like ax2+bx+c = 0.

now solve this equation be middle term split method to find the values of x.

Hence:

x2+x-p(p+1) = 0

x2+(p+1)x -px -p(p+1) = 0

x{x+(p+1)} -p{x+(p+1)} = 0

{x-p}{x+(p+1)} = 0

now we got two factors

(x-p) = 0

x = p

{x+(p+1)} = 0

x = -(p+1)

So, the root of the equation x2+x-p(p+1)=0 are x = p, and x = -(p+1)

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