(x-1)(x-3)(x+2)(x+6)=72x^2

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вопрос

To solve this equation, we first distribute the terms on the left side:

(x-1)(x-3)(x+2)(x+6) = 72x^2
=> (x^2 - 3x - x + 3)(x^2 + 6x + 2x + 12) = 72x^2
=> (x^2 - 4x + 3)(x^2 + 8x + 12) = 72x^2
=> (x^2 + 8x - 4x - 32)(x^2 + 8x + 12) = 72x^2
=> (x^2 + 4x - 32)(x^2 + 8x + 12) = 72x^2

Now, we expand the left side further:

=> x^4 + 8x^3 + 12x^2 + 4x^3 + 32x^2 + 48x - 32x^2 - 256x - 384 = 72x^2
=> x^4 + 12x^3 + 12x^2 + 48x - 256x - 384 = 72x^2
=> x^4 + 12x^3 - 60x^2 - 208x - 384 = 72x^2
=> x^4 + 12x^3 - 132x^2 - 208x - 384 = 0

Now that we have the polynomial equation set equal to zero, we can try factoring it further or using another method to solve for x.