By Remainder Theorem find the remainder, when $p (x)$ is divided by $g (x)$ , where $p (x) = 4 x^{3} - 12 x^{2} + 14 x - 3, g (x) = 2 x - 1$

\frac{3}{2}

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\frac{5}{2}

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\frac{1}{2}

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\frac{3}{4}

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Solution

The correct option is A $\frac{3}{2}$
Here $p (x) = 4 x^{3} - 12 x^{2} + 14 x - 3$ and $g (x) = 2 x - 1$

Putting $g (x) = 0$
$\Rightarrow 2 x - 1 = 0$
$\Rightarrow x = \frac{1}{2}$

Here, $p (x)$ is divided by $g (x)$ ,

By remainder theorem, we can write,
The remainder $= p (\frac{1}{2})$

$= 4 {(\frac{1}{2})}^{3} - 12 {(\frac{1}{2})}^{2} + 14 (\frac{1}{2}) - 3$

$= \frac{4}{8} - \frac{12}{4} + \frac{14}{2} - 3$

$= \frac{1}{2} - 3 + 7 - 3$

$= \frac{1}{2} + 1$

$= \frac{3}{2}$

Thus, the remainder is $\frac{3}{2}$

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Similar questions

p (x) = x^{3} - 2 x^{2} - 4 x - 1, g (x) = x + 1

p (x) = x^{3} - 3 x^{2} + 4 x + 50, g (x) = x - 3

p (x) = 4 x^{3} - 12 x^{2} + 14 x - 3, g (x) = 2 x - 1

p (x) = x^{3} - 6 x^{2} + 2 x - 4, g (x) = 1 - \frac{3}{2} x

p (x) = x^{3} - 2 x^{2} - 4 x - 1, g (x) = x + 1

p (x) = x^{3} - 3 x^{2} + 4 x + 50, g (x) = x - 3

p (x) = 4 x^{3} - 12 x^{2} + 14 x - 3, g (x) = 2 x - 1

p (x) = x^{3} - 6 x^{2} + 2 x - 4, g (x) = 1 - \frac{3}{2} x

p (x)

g (x)

p (x) = 4 x^{3} - 12 x^{2} + 14 x - 3 g (x) = 2 x - 1

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$(p x) = 6 x^{3} + 13 x^{2} + 3, g (x) = 3 x + 2$

f (x) = 4 x^{3} - 12 x^{2} + 14 x - 3

g (x) = 2 x - 1

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