Question

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

Answer 1

(x+1)(x+2)(x+3)(x+4)-3
=((x+1)(x+4))((x+2)(x+3))-3
=(x^2+5x+4)(x^2+5x+6)
Now let (x^2+5x+5) be t. Then,
=(t-1)(t+1)-3
=t^2-1-3
=t^2-4
=(t-2)(t+2)
=(x^2+5x+3)(x^2+5x+7)

PLZ MARK IT AS BRAINLIEST..

Answer 2

ⓢⓞⓛⓤⓣⓘⓞⓝ

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

it is a special case of factorization

here we used some general rules to solve the factorization

below the steps are given:-

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

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

=(x²+4x+x+4)(x²+3x+2x+6)-3

=(x²+5x+4)(x²+5x+6)-3

=(p+4)(p+6)-3[let p =x²+5x]

=(p²+6p+4p+24)-3

=P²+10p+24-3

=p²+10p+21

=p²+7p+3p+21

=p(p+7)+3(p+7)

=(p+3)(p+7)

=(x²+5x+3)(x²+5x+7)[putting the value of p]

so the final answer=>(x²+5x+3)(x²+5x+7)

=>here we replace the x²+5x by p and use the common terms of all terms . here also we use middle term factorization.