Question

X(x+1)(x+3)(x+4)=40​

Answer 1

(x²+4x)(x²+4x+3)=40;

x²+4x=t

t(t+3)=40⇒t²+3t-40=0, D=9+160=169,√D=13

t1=5,t2= -8

1)x²+4x=5,x²+4x-5=0, D=16+20=36,√D=6

X1=1,x2= -5.

2)x²+4x= -8,  x²+4x+8=0,D1=4-8<0--корней не имеет

ответ.1; -5.

Answer 2

To solve the equation X(x+1)(x+3)(x+4) = 40, we can follow these steps:

1. Expand the equation: X(x+1)(x+3)(x+4) = 40 (x^2 + 4x)(x+3)(x+4) = 40 (x^2 + 4x)(x^2 + 7x + 12) = 40

2. Distribute the terms: (x^4 + 7x^3 + 12x^2 + 4x^3 + 28x^2 + 48x) = 40 x^4 + 11x^3 + 40x^2 + 48x = 40

3. Simplify and move all terms to one side: x^4 + 11x^3 + 40x^2 + 48x - 40 = 0

4. Set the equation equal to zero and try to factor it or use numerical methods to find the solutions.

Unfortunately, the equation does not appear to have any simple factorization. You would need to use numerical methods, such as the Newton-Raphson method or a graphing calculator, to approximate the solutions.