Question

find the remainder when p(x)=x³-6x²+2x-4 is divided by g(x) =1-3/2x​

Answer 1

Answer:-

$p(x) =x {}^{3} - 6x + x - 4$

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

$= \frac{1}{8} - 6( \frac{1}{4} ) + 1 - 4$

$= \frac{1}{8} \frac{ - 6}{8} + 1 - 4$

$= \frac{1}{8} \frac{ - 3}{2} + 1 - 4$

Now take L.C.M. for both the denominators

$= (1 - 12 + 8 - \frac{32}{8}) = \frac{ - 35}{8}$

Remainder - 35/8

Now division - 6x² - 1

After division answer u get

Remainder is - 11x³ - 3

Checking the answer

$- 11 \times 3 - 3</p><p>$

$⇒ - 111/8 - 3</p><p>$

$⇒ (- 11 - \frac{24}{8} )</p><p>$

$⇒ \frac{ - 35}{8}$

Hence Answer proved

Answer 2

Given:

$p(x) = {x}^{3} - 6 {x}^{2} + 2x - 4$

$g(x) = 1 - \frac{3}{2}x$

To find the remainder when p(x) is divided by g(x), we are going to use the following theorem:-

Remainder theorem:

When a polynomial p(x) is divided by (x-a) then the remainder will be p(a).

In g(x):

$x = \frac{3}{2} - 1 \\ \\ x = \frac{3 - 2}{2} \\ \\ x = \frac{1}{2}$

Substituting the value of x in p(x):

${ (\frac{1}{2}) }^{3} - {6( \frac{1}{2}) }^{2} + 2( \frac{1}{2} ) - 4 \\ = \frac{1}{8} - \frac{3}{2} + 1 - 4 \\ = \frac{1 - 12 - 24}{8} \\ = - \frac{35}{8}$

Therefore, the remainder when p(x) is divided by g(x) is -35/8.