For the equation, determine the constants A and B that make the equation an identity.
-$$\frac { x - 3 } { x ^ { 2 } } = \frac { A } { x } + \frac { B } { x ^ { 2 } }$$
A) $A = - 3 , B = 1$
B) $\mathrm { A } = 3 , \mathrm {~B} = 1$
C) $\mathrm { A } = 1 , \mathrm {~B} = - 3$
D) $A = 1 , B = 3$

Question

Quizplus · Accepted Answer

To make the equation an identity, equate the numerators: $x - 3 = A \cdot x + B$. Solving gives $A = 1$ and $B = -3$.

For the Equation, Determine the Constants a and B That x−3x2=Ax+Bx2\frac { x - 3 } { x ^ { 2 } } = \frac { A } { x } + \frac { B } { x ^ { 2 } }x2x−3​=xA​+x2B​

For the Equation, Determine the Constants a and B That $\frac { x - 3 } { x ^ { 2 } } = \frac { A } { x } + \frac { B } { x ^ { 2 } }$