Question

factorise x²+x-(p+1)(p+2)​

Answer 1

Answer:

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

On factorising we get,

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

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

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

x=-p-1, and x= p

The zeros of polynomial are: p-1 and p

Answer 2

Answer:

$\boxed{\boxed{ \large{ \bold{ \pink{(x + p + 2) \: (x - p - 1)}}}}}$

Step-by-step explanation:

$first \: of \: all \: we \: do \: a \: simple \: trick \: \\( p + 2) - (p + 1) = p + 2 - p - 1 \\ \: \: \: \: \: \: \: = 1 \\ now \: returning \: to \: main \: question \\ {x}^{2} + ( \: (p + 2) - (p + 1) \: )x \: - \: (p + 1) \: (p + 2) \\ = {x}^{2} + (p + 2)x - (p + 1)x - (p + 1)(p + 2) \\ = x \: (x + p + 2) - (p + 1) \: (x + p + 2) \\ = (x + p + 2) \: (x - p - 1)$