Question

Solve the question in aatachment​

Answer 1

Step-by-step explanation:

Tip:

Remainder Theorem:If a polynomial p(x) is divided by a linear polynomial g(x) whose zero is x = a, the remainder is given by r = p(a).

Given:

$\begin{gathered}i) \bold{ p(x) = {x}^{3} - 2 {x}^{2} - 4x - 1 }\\ \bold{g(x) = x + 1 }\\ \end{gathered}$

Solution:

To find remainder put value of x from g(x) into p(x)

$\begin{gathered}x + 1 = 0 \\ \\ x = - 1 \\ \\ p( - 1) = { (- 1)}^{3} - 2 {( - 1)}^{2} - 4( - 1) - 1 \\ \\ = - 1 - 2 + 4 - 1 \\ \\ = - 4 + 4 \\ \\ \bold{p( - 1)= 0} \\ \end{gathered}$

Thus,

Remainder is zero.

$\begin{gathered}ii) \:\bold{ p(x) = {x}^{3} - 3{x}^{2} + 4x + 50} \\ \bold{g(x) = x - 3} \\ \end{gathered}$

$\begin{gathered} g(x) = 0 \\ \\ x - 3 = 0 \\ \\ x = 3 \\ \\ p(3) = {(3)}^{3} - 3{(3)}^{2} + 4(3) + 50 \\ \\ = 27 - 27 + 12 + 50 \\ \\ \bold{p(3)= 62} \\ \end{gathered}$

Remainder is 62.

$\begin{gathered}iii) \: \bold{p(x) = 4 {x}^{3} -12 {x}^{2} + 14x - 3} \\ \bold{g(x) =2 x - 1} \\\end{gathered}$

Put g(x)=0 and from that put value of x in p(x)

$</p><p>\begin{gathered}2x - 1 = 0 \\ \\ 2x = 1 \\ \\ x = \frac{1}{2} \\ \end{gathered}$

$</p><p></p><p>\begin{gathered}\: p( \frac{1}{2} ) = 4 {( \frac{1}{2} )}^{3} -12 {( \frac{1}{2} )}^{2} + 14( \frac{1}{2} ) - 3 \\ \\ = \frac{1}{2} - 3 + 7 - 3 \\ \\ = \frac{1}{2} + 1 \\ \\ \bold{ p( \frac{1}{2} ) = \frac{3}{2}} \\ \end{gathered}$

Remainder is 3/2.

$\begin{gathered}iv) \: \bold{p(x) = {x}^{3} -6 {x}^{2} + 2x - 4 }\\ \bold{g(x) =1 - \frac{3}{2} x} \\\end{gathered}$

$\begin{gathered}1 - \frac{3}{2} x = 0 \\ \\ \frac{3}{2}x = 1 \\ \\ x = \frac{2}{3} \\ \\ \end{gathered}$

$\begin{gathered}p( \frac{2}{3} ) = {( \frac{2}{3})}^{3} -6 {( \frac{2}{3} )}^{2} + 2( \frac{2}{3} )- 4 \\ \\ = \frac{8}{27} - 6 \times \frac{4}{9} + \frac{4}{3} - 4 \\ \\ = \frac{8}{27} - \frac{8}{3} + \frac{4}{3} - 4 \\ \\ = \frac{8 - 72 + 36 -10 8}{27} \\ \\ = \frac{44 -1 80}{27} \\ \\ \bold{p( \frac{2}{3} )=\frac{-136}{27} }\\\end{gathered}$

Remainder is -136/27.

Hope it helps you.