# Math Help - help please :)

Identities Find values A, B and R

X^2 - 2X + 7 = (x+3) (Ax + B ) + R

thanks

2. Originally Posted by gracey
Identities Find values A, B and R

X^2 - 2X + 7 = (x+3) (Ax + B ) + R

thanks
$x^2 - 2x + 7 = (x+3) (Ax + B ) + R$

$x^2 - 2x + 7 = Ax^2 +Bx + 3Ax + 3B + R$

$(1-A)x^2 - (B+3A+2)x + (7-3B-R) =0$

Are you familiar with the quadratic formula?

That will get you values of x. I now realized that you want the constants...

$x^2 - 2x + 7 = (x+3) (Ax + B ) + R$

Set $x = -3$

$9 + 6 + 7 = R$

$R = 22$

Now, set $x=0$

$7 = 3B + R$

$7 = 3B + 22$

$B = -5$

I will leave A up to you.

3. Have you considered multiplication?

(x+3) (Ax + B ) + R = Ax^2 + (3A+B)x + (R+3B)

Now match up the coefficients on the terms of equal degree.

Don't use the quadratic formula. You are not solving for 'x'.

4. Simply divide both sides by (x+3).

$x^2 - 2x + 7 = (x+3) (Ax + B ) + R$

$\frac{x^2 - 2x + 7}{x+3}=\frac{(x+3) (Ax + B ) + R}{x+3}$

$x-5 + \frac{22}{x+3} = Ax+B + \frac{R}{x+3}$

5. Originally Posted by TKHunny
Have you considered multiplication?

(x+3) (Ax + B ) + R = Ax^2 + (3A+B)x + (R+3B)

Now match up the coefficients on the terms of equal degree.

Don't use the quadratic formula. You are not solving for 'x'.
This gives you $A = 1, B = -5, R=22$ matching what I gave.

6. Originally Posted by wingless
Simply divide both sides by (x+3).

$x-5 + \frac{22}{x+3}$
Notice that this requires a division algorithm. It isn't really magic if you put the complicated steps back in.

7. Originally Posted by TKHunny
Notice that this requires a division algorithm. It isn't really magic if you put the complicated steps back in.
I don't think it's complicated. I can do it even without writing, anyone who knows division can do it too..