If $P = {x / x^{2} + 14 x + 40 = 0}$
$Q = {x / x^{2} - 5 x + 6 = 0}$
$R = {x / x^{2} + 17 x - 60 = 0}$ and the universal set $X = {- 20, - 10, - 4, 2, 3, 4}$ .

Find $P \cup Q$ .

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Solution

$P = {x / x^{2} + 14 x + 40 = 0}$

$∴ P = {- 10, - 4}$

Similarly, $Q = {3, 2}$ , $R = {- 20, 3}$ and $X = {- 20, - 10, - 4, 2, 3, 4}$

$P \cup Q = {- 10, - 4, 3, 2}$ .

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If $({5 x}^{2} + 14 x + 2)^{2}$ – $({4 x}^{2} - 5 x + 7)^{2}$ is divided by $(x^{2} + x + 1)$ , then quotient q and remainder r are given by:

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= \frac{x + 6}{x + 7}

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