$\displaystyle f(x)=x^3+9x^2+20x+12$

divide f(x) by x=6

thanks in advance

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- Jan 8th 2009, 10:06 AM #1

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- Jan 8th 2009, 10:15 AM #2

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Hello coyote,

Use synthetic division:

Code:-6 | 1 9 20 12 -6 -18 -12 -------------- 1 3 2 0

Sorry coyote. I should've asked if you knew synthetic division.

Code:x^2 + 3x + 2 __________________________ x + 6 | x^3 + 9x^2 + 20x + 12 -(x^3 + 6x^2) -------------- 3x^2 + 20x -(3x^2 + 18x) ------------ 2x + 12 -(2x + 12) ======= 0

- Jan 8th 2009, 10:35 AM #3

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- Jan 8th 2009, 10:50 AM #4

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No, you can just use "regular" division:

x+6) x^3+ 9x^2+ 20x+ 12

x divides into x^3 x^2 times so the quotient starts "x^2+ ..."

Multiplying x^2 by x+ 6 gives

x^3+ 9x^2+ 20x+ 12

x^3+ 6x^2

subtracting: 3x^2+ 20x+ 12

Now x divides into 3x^2 3x times so we have "x^2+ 3x+ ..."

Multiplying 3x by x+ 6 gives

3x^2+ 20x+ 12

3x^2+ 18x

subtracting: 2x+ 12

Finally, x divides into 2x twice so we have "x^2+ 15x+ 2"

Multiplying 2 by x+ 6,

2x+ 12

2x+ 12

which has remainder 0.

- Jan 8th 2009, 11:44 AM #5

- Jan 8th 2009, 01:35 PM #6

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## no problem