Question

p(p-1)(p-2)(p-3)-120 factorisation​

Answer 1

Answer:

Answer  = (p+2)(p-5)(p^2-3p+12)

Step-by-step explanation:

Hey mate!

Here is your answer!

(p^2-3p)(p^2-2p-p+2)-120

(p^2-3p)(p^2-3p+2)-120

let , p^2-3p = x

(x)(x+2) - 120

x^2 + 2x - 120

x^2+12x-10x - 120 = 0

x(x+12) - 10(x+12) = 0

(i)(p^2-3p-10)(p^2-3p+12) = 0

(p^2-5p+2p-10)(p^2-3p+12) = 0

{p(p-5)+2(p-5)}(p^2-3p+12) = 0

= {(p+2)(p-5)(p^2-3p+12) = 0

Ans = (p+2)(p-5)(p^2-3p+12)

hope it helps!

please mark me as brainliest!