Question

find the remainder when p(x)=x^3-6x^2+12x-5 is divided by(x-2) by long division method and remainder theorem class 9
ch 2

Answer 1

$\large\underline{\sf{Solution-}}$

↝  Given polynomial is

$\rm :\longmapsto\:p(x) = {x}^{3} - {6x}^{2} + 12x - 5$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Remainder Theorem states that when a polynomial f(x) is divisible by linear polynomial (x - a), then remainder is f(a).

So,

$\rm :\longmapsto\:p(x) \: is \: divided \: by \: x - 2$

Then,

$\rm :\longmapsto\:Remainder = p(2)$

$\rm \: \: = \: {(2)}^{3} - {6(2)}^{2} + 12(2) - 5$

$\rm \: = \:8 - 24 + 24 - 5$

$\rm \: = \: 3$

So,

$\rm :\longmapsto\:Remainder \: when \: p(x) \: is \: divided \: by \: x - 2 \: is \: 3$

Second Part

↝ Given polynomial is

$\rm :\longmapsto\:p(x) = {x}^{3} - {6x}^{2} + 12x - 5$

↝ Using Long Division,

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\: {x}^{2} - 4x + 4\:\:}}}\\ {\underline{\sf{x - 2}}}& {\sf{\: {x}^{3}-{6x}^{2} + 12x - 5 \:\:}} \\{\sf{}}& \underline{\sf{\:-  {x}^{3} +2{x}^{2}   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:\:}} \\ {{\sf{}}}& {\sf{\: \:  \:  \:  \:  \:  \:  \:  \:  \:  -4{x}^{2} + 12x - 5  \:  \:  \:  \:   \:  \:  \:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \:  \:  \:  \:  \: 4{x}^{2} - 8x  \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \:  \:  \:  \:  \: \: \: \:4x - 5  \:\:}} \\{\sf{}}& \underline{\sf{\: \:  \:  \:  \:  \:  - 4x + 8\:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \: 3\:\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered}$

So,

$\rm :\longmapsto\:Remainder \: when \: p(x) \: is \: divided \: by \: x - 2 \: is \: 3$

Additional Information :-

Factor Theorem states that if (x - a) is a factor of polynomial p(x), then p(a) = 0.