Posted by Acezar on Monday, November 1, 2010 at 8:21pm.
Nevermind, I figured it out:
The quotient:
(10x^6 - 15x^5 - 5x^3)/(5x^3 )
(10x^6)/(5x^3 ) - (15x^5)/(5x^3 ) - (5x^3)/(5x^3 )
2x3 - (15x^5)/(5x^3 ) - (5x^3)/(5x^3 )
2x3 - 3x2 - (5x^3)/(5x^3 )
2x3 - 3x2 - 1
The product using the quotient:
(x2 - 3x + 5)(2x3 - 3x2 - 1)
x2 ∙ 2x3 + x2 ∙ -3x2 + x2 ∙ - 1 - 3x ∙ 2x3 - 3x ∙ -3x2 - 3x ∙ -1 + 5 ∙ 2x3 + 5 ∙ -3x2 + 5 ∙ -1
2x5 - 9x4 + 19x3 - 16x2 + 3x - 5
Here's a better format. Good luck to anyone else who has this problem.
The quotient:
(10x^6 - 15x^5 - 5x^3)/(5x^3 )
(10x^6)/(5x^3 ) - (15x^5)/(5x^3 ) - (5x^3)/(5x^3 )
2x^3 - (15x^5)/(5x^3 ) - (5x^3)/(5x^3 )
2x^3 - 3x^2 - (5x^3)/(5x^3 )
2x^3 - 3x^2 - 1
The product using the quotient:
(x^2 - 3x + 5)(2x^3 - 3x^2 - 1)
x^2 * 2x^3 + x^2 * -3x^2 + x^2 * - 1 - 3x * 2x^3 - 3x * -3x^2 - 3x * -1 + 5 * 2x^3 + 5 * -3x^2 + 5 * -1
2x^5 - 9x^4 + 19x^3 - 16x^2 + 3x - 5
The idea is to solve the quotient first, by dividing each term by 5x^3.
Once the solution is found: 2x^3 - 3x^2 - 1
Then use that quotient and multiply it by (x^2 - 3x + 5).
Thus: (x^2 - 3x + 5)(2x^3 - 3x^2 - 1)
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