Question

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

Answer 1

Solution:-

given by:-

$(x + 1)(x + 2)(x + 3)(x + 4) - 3 \\ ( {x}^{2} + 5x + 4)( {x}^{2} + 5x + 6) - 3 \\ ( {x}^{4} + 5 {x}^{3} + 6 {x}^{2} + 5 {x}^{3} + 25 {x}^{2} + 30x + 4 {x}^{2} + 20x + 24 - 3) \\ {x}^{4} + 10 {x}^{3} + 33 {x}^{2} + 50x + 21)ans$

Answer 2

Answer:

x^4+10x^3+35x^2+50x+21

Step-by-step explanation:

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

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

= (x^2+2x+x+2)(x^2+4x+3x+12)-3

= (x^2+3x+2)(x^2+7x+12)-3

= (x^4+7x^3+12x^2+3x^3+21x^2+36x+2x^2+14x+24)-3

= (x^4+10x^3+35x^2+50x+24)-3

= x^4+10x^3+35x^2+50x+24-3

= x^4+10x^3+35x^2+50x+21