Question

(x-1)(x-2)(x+3)(x+4)+4 factorization​

Answer 1

Step-by-step explanation:

(x+1) (x+2)(x+3) (x+4) --15

( x+1)(x+4) = x^2 +5x +4:

(x+2) (x+3) = x^2 +5x+6

Now let us put x^2 + 5x+4 = m

The given problem is in terms of m

m (m+2) --15

m^2 +2m --15

Factoriese

m^2 +5m --3m --15

m (m+5) -3 ( m+5)

(m +5) (m--3)

Now factoriise

m+5 = x^2+5x+4+5 = x^2+5x+9

m-3 = x^2+5x+4 --3 = x^2+5x+1

(x^2+5x+9) (x^2 +5x +1) are the factors

Unless each is expressed as quadratic equation

The given problem should have 4 factors

The rest you can do

4 roots will be the answer

Answer 2

sᴏʟᴜᴛɪᴏɴ

Given:-(x-1)(x-2)(x+3)(x+4)+4 factorization

it is a special case of factorization

to solve the sum we use FOLLOWING steps as below:-

(x-1)(x-2)(x+3)(x+4)+4

=(x-1)(x+3)(x-2)(x+4)+4

=(x²+3x-x-3)(x²+4x-2x-8)+4

=(x²+2x-3)(x²+2x-8)+4

=(a-3)(a-8)+4[let a =x²+2x]

=(a²-8a-3a+24)+4

=a²-11a+24+4

=a²-11a+28

=a²-7a-4a+28

=a(a-7)-4(a-7)

=(a-4)(a-7)

=(x²+2x-4)(x²+2x-7)[putting the value of a]

so the final answer =(x²+2x-4)(x²+2x-7)

here we used common method means we find out here the common terms and in their places we place the a and then we use middle term factorization