Divide: $81 x^{3} (50 x^{2} - 98)$ by $27 x^{2} (5 x + 7)$

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Solution

The given polynomial $81 x^{3} (50 x^{2} - 98)$ can be rewritten as:

$81 x^{3} (50 x^{2} - 98) = 162 x^{3} [(5 x)^{2} - (7)^{2}] = 162 x^{3} [(5 x + 7) (5 x - 7)] (∵ a^{2} - b^{2} = (a + b) (a - b))$

Now, we divide $162 x^{3} (5 x + 7) (5 x - 7)$ by the binomial $27 x^{2} (5 x + 7)$ as shown below:

$\frac{162 x^{3} (5 x + 7) (5 x - 7)}{27 x^{2} (5 x + 7)} = 6 x (5 x - 7)$

Hence, $\frac{81 x^{3} (50 x^{2} - 98)}{27 x^{2} (5 x + 7)} = 6 x (5 x - 7)$ .

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81 x^{3} (50 x^{2} - 98)

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Find the remainder when $7^{98}$ is divided by 5.

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