Find the value of the expression $2 x^{3} + 2 x^{2} - 7 x + 72$ where $x = \frac{3 + 5 \sqrt{1}}{2}$ .

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Solution

It is given that $x = \frac{3 + 5 \sqrt{- 1}}{2}$ which is same as $x = \frac{3 + 5 i}{2}$ because $i^{2} = - 1$ .

Now, let us substitute $x = \frac{3 + 5 i}{2}$ in the given polynomial $2 x^{3} + 2 x^{2} - 7 x + 72$ as follows:

$2 x^{3} + 2 x^{2} - 7 x + 72 = 2 {(\frac{3 + 5 i}{2})}^{3} + 2 {(\frac{3 + 5 i}{2})}^{2} - 7 (\frac{3 + 5 i}{2}) + 72 = 2 (\frac{3^{3} + (5 i)^{3} + (3 \times 3^{2} \times 5 i) + (3 \times 3 \times (5 i)^{2})}{8}) + 2 (\frac{3^{2} + (5 i)^{2} + (2 \times 3 \times 5 i)}{2}) - (\frac{21 + 35 i}{2}) + 72 (∵ (a + b)^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}, (a + b)^{2} = a^{2} + b^{2} + 2 a b) = 2 (\frac{27 - 125 i + 135 i - 225}{8}) + 2 (\frac{9 - 25 + 30 i}{4}) - (\frac{21 + 35 i}{2}) + 72 (∵ i^{2} = - 1)$
$= (\frac{- 198 + 10 i}{4}) + (\frac{- 16 + 30 i}{2}) - (\frac{21 + 35 i}{2}) + 72 = (\frac{2 (- 99 + 5 i)}{4}) + (\frac{30 i - 16}{2}) - (\frac{35 i + 21}{2}) + 72 = \frac{5 i - 99}{2} + \frac{30 i - 16}{2} - (\frac{35 i + 21}{2}) + (\frac{72 \times 2}{2}) = \frac{5 i - 99 + 30 i - 16 - 35 i - 21 + 144}{2} = \frac{35 i - 35 i - 136 + 144}{2} = \frac{8}{2} = 4$

Hence, $2 x^{3} + 2 x^{2} - 7 x + 72 = 4$ when $x = \frac{3 + 5 \sqrt{- 1}}{2}$ .

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