Question

find the reminder when x
${x}^{3} + 3x ^{2} + 3x + 1$
is divided by
$x + 1$
​

Answer 1

Answer:-

The remainder when x³ + 3x² + 3x + 1 is divided by (x+1) is 0.

Hence, f(x) is a multiple of g(x)

Solution:-

Let f(x) = x³ + 3x² + 3x + 1

and g(x) = x + 1

At first we will find out the zero of the linear polynomial = (x + 1)

=> (x + 1) = 0

=> x = -1

From the remainder theorem, we know that when f(x) = x³ + 3x² + 3x + 1 is divided by the linear polynomial g(x) = (x + 1),

The required value of x is f(-1)

Now, We will put the value of x :-

x³ + 3x² + 3x + 1

= (-1)³ + 3.(-1)² + 3.(-1) + 1

= -1 + 3.1 + (-3) + 1

= -1 + 3 - 3 + 1

= 0

Hence, The remainder is 0.

And f(x) = x³ + 3x² + 3x + 1 is the multiple of linear polynomial g(x) = (x + 1) respectively.

Answer 2

Answer:

0.

Step-by-step explanation:

Given :

$\large p(x)=x^3+3x^2+3x+1 \ and \ g(x)=x+1$

We have to find remainder here.

First zeroes of g ( x ) = x + 1 = 0

x = - 1

Now putting x = - 1  in p ( x )  to get remainder

$\large p(x)=x^3+3x^2+3x+1\\\\\\\large p(-1)=(-1)^3+3(-1)^2+3(-1)+1\\\\\\\large p(x)=-1+3-3+1\\\\\\\large p(x)=1-1+3-3\\\\\\\large p(x)=0$

Since remainder is zero x + 1 is factor of p ( x ) .