### Home > PC > Chapter 6 > Lesson 6.2.3 > Problem6-100

6-100.

We know that $\frac { x ^ { 2 } - 3 x + A } { x - 1 }= x + B$ for some numbers $A$ and $B$. Find $A$ and $B$.

Multiply both sides by $\left(x − 1\right)$.

$x^{2} − 3x + A = \left(x + B\right)\left(x − 1\right)$

Multiply out on the right side.

$x^{2} − 3x + A = x^{2} + Bx − x − B$

Subtract $x^{2}$ from both sides.

$−3x + A = Bx − x − B$

Subtract $x$ from both sides of the equation.

$−2x + A = Bx − B$

Set the $x$-terms $=$ to each other.
Set the constant terms $=$ to each other.

$∴ − 2 = B$ and $A = − B$
$∴ A = 2$