Question

find remainder using remainder theorem.
$p(x) \: = \: {x}^{3} - \: {6x \: }^{2} + \: 2x \: - \: 4 \: \: \: \\ be \: divide \: by \: g(x) \: = \: 1 - \frac{3}{2} x$
​

Answer 1

Step-by-step explanation:

Given =》

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

Solution =》

$p(x) = x {}^{3} - 6x + 2x - 4 \\ g(x) = 1 - \frac{3}{2} \\ = \frac{1}{1} - \frac{3}{2} \\ take \: lcm \: of \: 1and \: \\ = \frac{2 - 3}{2} \\ = \frac{ - 1}{2}$

Putting x= -1/ 2 in p(x)

$p(x) = x {}^{3} - 6x + 2x - 4 \\ = ( \frac{ - 1}{2} ) {}^{3} - 6 \times ( \frac{ - 1}{2} ) {}^{2} + 2 \times ( \frac{ - 1}{2} ) - 4 \\ = \frac{ - 1}{8} - \frac{6}{4} - \frac{2}{2} - 4 \\ = \frac{ - 1 - 12 - 8 - 32}{8} \\ = \frac{ - 53}{8}$

Answer 2

Question :-

Find remainder using remainder theorem when p(x) = x³ - 6x² + 2x - 4 is divided by g(x) = 1 - 3x/2

Explanation :-

Given :

• p(x) = x³ - 6x² + 2x - 4
• g(x) = 1 - 3x/2

To find :

The remaining when p(x) is divided by g(x)

Solution :

From the remainder theorem, Let's find the zero of polynomial g(x).

$\sf\rightarrow 1 - \: \dfrac{3x}{2} = \: 0$

$\sf\rightarrow 1 = \: \dfrac{3x}{2}$

$\sf\rightarrow 3x = 2$

$\sf\rightarrow x = \dfrac{2}{3}$

Now, Putting the value of x in polynomial p(x),

$\sf\longrightarrow { (\dfrac{2}{3}) }^{3} - 6 \times { (\dfrac{2}{3}) }^{2} + 2 \times \dfrac{2}{3} - 4$

$\sf\longrightarrow \dfrac{8}{27} - 6 \times \dfrac{4}{9} + \dfrac{4}{3} - 4$

$\sf\longrightarrow \dfrac{8}{27} - \dfrac{24}{9} + \dfrac{4}{3} - 4$

$\sf\longrightarrow \dfrac{8 - 72 + 36 - 108}{27}$

$\sf\longrightarrow \dfrac{44 - 180}{27}$

$\sf\huge\therefore - \dfrac{136}{27}$

Hence, the remainder is -136/27.