Question

If the polynomial P(x)= ax^3+4x^2+3x-4 and g(x)= x^3-4x+a leaves the same remainder when divided by (x-3) , find the value of a​

Answer 1

Step-by-step explanation:

$\huge \pink{ \boxed{\bf \blue{Answer}}}$

A] given ax3+4x2 +3x−4=0 &

x 3 −4x+a=0 leave same

A] given ax

3

+4x

2

+3x−4=0 &

x

3

−4x+a=0 leave same

remainder when divided by x-3

P(x) = ax

3

+4x

2

+3x−4

q(x)= x

3

−4x+a

Remainder theorem,

P(3)=q(3)

a(3)

3

+4(3)

2

+3(3)−4=3

3

−4(3)+a

27a+36+9−4=27−12+a

26a=15−41

26a=−26

∴a=−1

P(x)=−x

3

+4x

2

+3x−4

when divide by (x-2)

P(2)=−(2)

3

+4(2)

2

+3(2)−4

=−8+16+6−4

=8+2

=10

Answer 2

Step-by-step explanation:

Step-by-step explanation:

$\huge \pink{ \boxed{\bf \blue{Ashish}}}$

A] given ax3+4x2 +3x−4=0 &

x 3 −4x+a=0 leave same

A] given ax

3

+4x

2

+3x−4=0 &

x

3

−4x+a=0 leave same

remainder when divided by x-3

P(x) = ax

3

+4x

2

+3x−4

q(x)= x

3

−4x+a

Remainder theorem,

P(3)=q(3)

a(3)

3

+4(3)

2

+3(3)−4=3

3

−4(3)+a

27a+36+9−4=27−12+a

26a=15−41

26a=−26

∴a=−1

P(x)=−x

3

+4x

2

+3x−4

when divide by (x-2)

P(2)=−(2)

3

+4(2)

2

+3(2)−4

=−8+16+6−4

=8+2

=10