Question

find remainder when 3x^3-4x^2+7x-5 is divided by (x-3)and(x+3)

Answer 1

hiii. ........friend

the answer is here,

$= > f(x) = \: 3 {x}^{3} - 4 {x}^{2} + 7x - 5$

If it is divided by (x-3). Then according to remainder theorem , the remainder will be f (3).

$= > \: f(3) = 3( {3})^{3} - 4 ({3)}^{2} + 7(3) - 5$

$= > \: 3(27) - 4(9) + 21 - 5$
=> 81+21-41.

=> 61.

So, The remainder is 61.

If we divide the the polynomial by (x+3).Then according to remainder thereom the remainder will be f (-3).

$= > \: f( - 3) = 3 ({ - 3})^{3} - 4 ({ - 3})^{2} + 7( - 3) - 5$

=> 3 (-27)-4 (9)-27-5.

=> -81-36-27-5

=> -143.

So, The remainder is -143 , When divided by (x+3).


:-)Hope it helps u.

Answer 2

Step-by-step explanation:

= > f(x) = \: 3 {x}^{3} - 4 {x}^{2} + 7x - 5=>f(x)=3x

3

−4x

2

+7x−5

If it is divided by (x-3). Then according to remainder theorem , the remainder will be f (3).

= > \: f(3) = 3( {3})^{3} - 4 ({3)}^{2} + 7(3) - 5=>f(3)=3(3)

3

−4(3)

2

+7(3)−5

= > \: 3(27) - 4(9) + 21 - 5=>3(27)−4(9)+21−5

=> 81+21-41.

=> 61.

So, The remainder is 61.

If we divide the the polynomial by (x+3).Then according to remainder thereom the remainder will be f (-3).

= > \: f( - 3) = 3 ({ - 3})^{3} - 4 ({ - 3})^{2} + 7( - 3) - 5=>f(−3)=3(−3)

3

−4(−3)

2

+7(−3)−5

=> 3 (-27)-4 (9)-27-5.

=> -81-36-27-5

=> -143.

So, The remainder is -143 , When divided by (x+3).