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$\left(a^3+7a^2+14a+8\right)\left(a+3\right)$

Final answer to the exercise

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$16x^2-20x+100$

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$20+x=40$

$\left(10\left(9.4\right)^2+3\left(9.4\right)-4\right)$

$5m-5-m-8$
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