$y^{2} + 2 y - 3 = ? (a) (y - 1) (y + 3) (b) (y + 1) (y - 3) (c) (y - 1) (y - 3) (d) (y + 2) (y - 3)$

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Solution

$y^{2} + 2 y - 3 = y^{2} + 3 y - y - 3 {\begin{matrix} ∵ - 3 = 3 \times (- 1) 2 = 3 - 1 \end{matrix}} = y (y + 3) - 1 (y + 3) = (y + 3) (y - 1)$

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Using factor theorem the factors of $y^{3} - 2 y^{2} - y + 2$ is

\frac{y}{y - 2} + \frac{1}{y + 2} + \frac{3 y}{4 - y^{2}}

f (x, y) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} y - x^{2} & x - y^{2} & x y - 1 x - y^{2} & x y - 1 & y - x^{2} x y - 1 & y - x^{2} & x - y^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

f (2, 2)

f : A \to B

f (x) = \frac{x - 1}{x - 2}

A = R - {2}

B = R - {1}

f

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