Question

$(x+1)(x+2)(x+3)(x+4)=24$

• A. 0, 1, 2, 3
• B. 0, 5
• C. 0, 3, 2, -1
• D. 1, 2, -3, -4
• E. 0, -5

Answer 1

The correct option is E 0, -5

Soln:

$\left[\right(x+1\left)\right(x+4\left)\right]\left[\right(x+2\left)\right(x+3\left)\right]=24$

$(x^2+5x+4)(x^2+5x+6)=24$

Let$x^2+5x=k$

$(k+4)(k+6)=24$

$k^2+10k=0$

$k(k+10)=0$

$K=0,k=-10$

$x^2+5x=0,x^2+5x=-10$

$x(x+5)=0,x^2+5x+10=0$

$x=0,x=-5;D=25-40=-15$ No real roots.

So, roots are 0, -5. Hence option (e)

2nd Method:- By factorising 24 into four factors then only possibilities are-

$24=1\times 2\times 3\times 4or-1\times -2\times -3\times -4$

thus we can find roots easily that is 0 and -5. choice (e) .