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Please solve
(2x^2 +x+4) (x+7) = 6x^3+29x^2+47x+28
with explanation
ASAP!!!
On a time crunch!!!!!

Sagot :

Answer: x = 0

Step-by-step explanation:

Subtract 6x^3+29x^2+47x+28 from both sides:

(2^2 + x + 4) (x - 7) - 6x^3+29x^2+47x+28

(6x^3+29x^2+47x+28)-6x^3+29x^2+47x+28

= -4x^3 - 14x^2 - 36x = 0

Now factor the left side:

-4x^3 - 14x^2 - 36x = 0

(-2x) (2x^2 + 7x + 18) = 0

Now set the factors equal to 0

-2x = 0 or 2x^2 + 7x + 18 = 0

x = 0

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