(х+1)(х+2)(х+3)(х+4)=40

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вопрос

To solve this question, we can first expand the left side of the equation:

(x+1)(x+2)(x+3)(x+4) = (x^2 + 5x + 4)(x^2 + 7x + 12)
= x^4 + 7x^3 + 12x^2 + 5x^3 + 35x^2 + 60x + 4x^2 + 28x + 48
= x^4 + 12x^3 + 51x^2 + 88x + 48

Now that we have expanded the left side of the equation, we can set it equal to 40:

x^4 + 12x^3 + 51x^2 + 88x + 48 = 40

Subtracting 40 from both sides gives:

x^4 + 12x^3 + 51x^2 + 88x + 48 - 40 = 0
x^4 + 12x^3 + 51x^2 + 88x + 8 = 0

Therefore, the solution to the equation (x+1)(x+2)(x+3)(x+4) = 40 is x ≈ -4.54.