$26 z^{3} (32 z^{2} - 18) \div 13 z^{2} (4 z - 3)$

z (4 z + 3)

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(4 z + 3)

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4 z (4 z + 3)

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None

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Solution

The correct option is B $4 z (4 z + 3)$
Let us first factorize $(32 z^{2} - 18)$ as follows:

$32 z^{2} - 18 = 2 (16 z^{2} - 9) = 2 [(4 z)^{2} - 3^{2})] = 2 (4 z + 3) (4 z - 3) (∵ a^{2} - b^{2} = (a + b) (a - b))$

Since we have to divide $\frac{26 z^{3} (32 z^{2} - 18)}{13 z^{2} (4 z - 3)}$ , therefore, we will divide $\frac{52 z^{3} (4 z + 3) (4 z - 3)}{13 z^{2} (4 z - 3)}$ as shown below:

$\frac{52 z^{3} (4 z + 3) (4 z - 3)}{13 z^{2} (4 z - 3)} = \frac{52}{13} \times \frac{z^{3}}{z^{2}} \times \frac{(4 z + 3) (4 z - 3)}{(4 z - 3)} = \frac{2 \times 2 \times 13}{13} \times \frac{z^{3} \times z^{- 2}}{1} \times (4 z + 3) = 4 \times z^{3 - 2} \times (4 z + 3) (∵ a^{x} + a^{y} = a^{x + y}) = 4 z (4 z + 3)$

Hence, $\frac{26 z^{3} (32 z^{2} - 18)}{13 z^{2} (4 z - 3)} = 4 z (4 z + 3)$ .

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