Question

$\frac{3}{2}-\frac{x - 4}{2x^{2}-4x + 2}$
$\frac{ax^{2}+bx + c}{2(x - 1)^{2}}$
$a=$
$b=$
$c=$

Explanation:

Step1: Factor the denominator

First, factor \(2x^{2}-4x + 2=2(x^{2}-2x + 1)=2(x - 1)^{2}\). Then \(\frac{3}{2}-\frac{x - 4}{2x^{2}-4x + 2}=\frac{3}{2}-\frac{x - 4}{2(x - 1)^{2}}\).
Find a common - denominator, which is \(2(x - 1)^{2}\). So \(\frac{3}{2}\times\frac{(x - 1)^{2}}{(x - 1)^{2}}-\frac{x - 4}{2(x - 1)^{2}}=\frac{3(x^{2}-2x + 1)-(x - 4)}{2(x - 1)^{2}}\).

Step2: Expand the numerator

Expand \(3(x^{2}-2x + 1)-(x - 4)\):
\[

$$\begin{align*} 3(x^{2}-2x + 1)-(x - 4)&=3x^{2}-6x+3 - x + 4\\ &=3x^{2}-7x + 7 \end{align*}$$

\]
Since \(\frac{3x^{2}-7x + 7}{2(x - 1)^{2}}=\frac{Ax^{2}+Bx + C}{2(x - 1)^{2}}\), by comparing the numerators, we have \(A = 3\), \(B=-7\), \(C = 7\).

Answer:

A = 3
B = -7
C = 7