Question

If 16x^4-8x^3+x^2-8x+4 is divided by 2x-1,then the remainder is I want the answer in remainder theorem. ​

Answer 1

Question :-

Find the Remainder when the Polynomial $p (x) = 16x^4 - 8x^3 + x^2 - 8x + 4$ is Divided by $g (x) = 2x - 1$

Solution :-

Let g (x) = 0 ⇔ 2x - 1 = 0 ⇒ 2x = 1

∴ x = $\bold {\frac{1}{2}}$

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By the Remainder Theorem, We know that when p (x) is Divided by (2x - 1), then the Remainder is $\bold {p (\frac{1}{2})}$

Now, $p (\frac{1}{2}) = \bold {\frac{1}{4}}$

$\begin {aligned} \therefore \bold {p (\frac{1}{2})} & = 16(\frac{1}{2})^4 - 8(\frac{1}{2})^3 + (\frac{1}{2}) ^2 - 8(\frac{1}{2}) + 4\\\\ & \Rightarrow 16 (\frac{1}{16}) - 8(\frac{1}{8}) + \frac{1}{4} - 4 + 4\\\\ & \Rightarrow \frac{16}{16} - \frac{8}{8} + \frac{1}{4} - 4 + 4\\\\ & \Rightarrow \cancel {1} \cancel {- 1} + \frac{1}{4} \cancel {- 4} + \cancel {4} \\\\ & \therefore \boxed {\bold {x = \frac{1}{4}}} \end {aligned}$

$\mathfrak {Hence, \: \red {The\: Required\: Remainder\: is\: \bold {\frac{1}{4}}}}$

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