Question

find the remender when x³+3x²+3x+1 is divided by x+1 by remendering theorem​

Answer 1

p(x) = x³+3x²+3x+1

= x³+1 + 3x²+3x

= (x+1)(x²-x+1) + 3x(x+1)

= (x+1)(x²-x+1+3x)

= (x+1)(x²+2x+1)

= (x+1)³

So the remainder when x³+3x²+3x+1 is divided by x+1, is 0.

Answer 2

The remainder is 0.

ExplanaTion:-

Here we can find out the remainder by 2 methods.

But first let us find it by remainder theorem as it is told in question.

By remainder theorem:-

So by remainder theorem we get,

$\\ \\ \tt\mapsto x + 1 = 0. \\ \\ \tt\mapsto x = 1.$

So by putting the value of x we get,

$\\ \\ \tt\mapsto {x}^{3} + 3 {x}^{2} + 3x + 1. \\ \\ \tt = ( - 1) {}^{3} + 3 {( - 1)}^{2} + 3( - 1) + 1. \\ \\ \tt = - 1 + 3 - 3 + 1. \\ \\ \tt = 3 - 3 + 1 - 1. \\ \\ \tt = 0 + 0. \\ \\ \tt =0= Remainder.$

$\tt\therefore$ The Remainder is 0.

So when $\tt\mapsto {x}^{3} + 3 {x}^{2} + 3x + 1.$ is divided by (x+1) then the remainder would be equal to 0.

$\rule{200}2$

We can also find the remainder by long division method which is given below:-

Long Division Method:-

$\\ \\ \begin{array}{r|cccc} &x^2 + 2x + 1&\\ \cline{2-5} x + 1 & x^3-3x^2+3x + 1&\\ &\pm x^3\pm x^2& \\ \cline{2-5} & & 2x^2 &+3x& + 1\\ & & \pm 2x^2 & \pm 2x \\ \cline{2-5} & & & x & + 1\\& & & \pm x &\pm1\\ \cline{2-5}&&&0. \\ \cline{2-5} &\end{array}$

$\tt\therefore$ The remainder is 0.

$\rule{200}2$