Question

What should be added to x^3-3x^2+6x-15 so that it is completely divisible bu x-3?

Answer 1

Answer:

-3

Step-by-step explanation:

Concept:

The remainder when f(x) is divided by (x-a)

is f(a)

Let f(x)= x³-3x² +6x-15

The remainder when f(x) is divided by (x-3)

=f(3)

=(3)³-3(3)² +6(3)-15

=27-27+18-15

=3

Therefore -3 should be added so that it is exactly divisible by (x-3)

Answer 2

Answer:

So it will be completely divisible if we add -3 to the polynomial

Step-by-step explanation:

According to remainder Theorem:

A polynomial f(x) is divided by a linear polynomial x-a then f(a) is the remainder of the polynomial divided by linear polynomial . And it is completely divisible when x-a is a factor of the f(x) which means that f(a) = 0

Now in our question

The given polynomial is

f(x) = x ³ - 3 x² + 6 x -15

Linear polynomial is x-3

To get the remainder we put liner polynomial equal to zero

so

x-3 =0

and x=3

Putting x = 3 in f(x)

it becomes

f(3) = 3 ³ - 3 (3)² + 6 (3) -15

      = 27 - 27 + 18 -15

       = 18 -15

       = 3

So the remainder is 3

To make it a factor we should put it equal to zero according to the definition

so f (3) should be zero

For this adding -3 to f(3)

f(3)+ (-3) = 3-3

f(3)+(-3) = 0

So it will be completely divisible if we add -3 to the polynomial