1. ## f(x) again :)

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

divide f(x) by x=6

2. Originally Posted by coyoteflare
$\displaystyle f(x)=x^3+9x^2+20x+12$

divide f(x) by x + 6

Hello coyote,

Use synthetic division:

Code:


-6 | 1  9  20  12
-6 -18 -12
--------------
1  3   2   0
$\displaystyle (x^3+9x^2+20x+12) \div (x+6)=x^2+3x+2$

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

3. is this the only way of doing it because i dont know synthetic division

4. 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.

5. Originally Posted by coyoteflare
is this the only way of doing it because i dont know synthetic division
Coyote, I've edited my post to show the long division.

6. ## no problem

Originally Posted by masters
Coyote, I've edited my post to show the long division.
i thanked you also as yours was explained clearer although ive got the hang of it now anyway thanks though, i would like to learn synthetic division but not until after my exams if i can get by without it xD
thanks again