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caroash09 · Jul 3, 2009, 11:13 AM

can some one help me fast in this question

Using long division or synthetic division calculate: x³ - 7x + 6 divided by x+3

I used synthetic division.. in end of the division I am not getting 0.. its coming 12

Unknown008 · Jul 3, 2009, 11:26 AM

You must have made some mistake...

Here it is:

$x^3 - 7x +6 = (x+3) (ax^2 + bx + c)$

There is no remainder since putting x as -3 gives 0 (as per the factor theorem)

So, expand;

$= ax^3 +(3a+b)x^2+(c+3b)x+3c$

Now equate the coefficients;

a = 1
3a+b = 0
c+3b = -7
3c = 6

(This is actually a way to check your work.)

Then, replace the values of a, b and c in your quotient.

caroash09 · Jul 3, 2009, 11:34 AM

Well sorry but I don't understand your method to check my ans. Anyway thanks for your help

Unknown008 · Jul 3, 2009, 11:44 AM

I'll try explain better. When you expand (x+3)(ax^2+bx+c), you should have your initial x^3 -7x+6. That's where I'm trying to guide you.

If you don't like that, I'll go for the long division. That'll be a little messy to be typed though...

OOO x^2__-3x___+2____
x+3 | x^3 OOOOO- 7x + 6
OOO x^3 + 3x^2
OOO___________
OOOOO0 - 3x^2 - 7x + 6
OOOOOiO-3x^2 - 9x
OOOOOO_________
OOOOOOOiO0 + 2x + 6
OOOOOOOOO + 2x + 6
OOOOOOOOO _______
OOOOOOOOOOO0 + 0