Question
By how much does $8 x^3-6 x^2+9 x-10$ exceed $4 x^3+2 x^2+7 x-3$ ?
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Answer
$8x^3 – 6x^2 + 9x – 10$ exceeds $4x^3 + 2x^2 + 7x -3$
$= (8x^3 - 6x^2 + 9x - 10) - (4x^3 + 2x^2 + 7x - 3)$
$= 8x^3 - 6x^2 + 9x - 10 - 4x^3 + 2x^2 + 7x - 3$
$= 8x^3- 4x^3 - 6x^2 - 2x^2 + 9x - 7x - 10 + 3$
$= 4x^3 - 8x^2 + 2x - 7$
$= (8x^3 - 6x^2 + 9x - 10) - (4x^3 + 2x^2 + 7x - 3)$
$= 8x^3 - 6x^2 + 9x - 10 - 4x^3 + 2x^2 + 7x - 3$
$= 8x^3- 4x^3 - 6x^2 - 2x^2 + 9x - 7x - 10 + 3$
$= 4x^3 - 8x^2 + 2x - 7$
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