Question

If the polynomial ax3+3x2-13and 5x3-8x+a leave the same remainder when divided by (x+1). then find value of a​

Answer 1

Answer:

Let P(x)=ax^3+3x^2-13

F(x)=5x^3-8x+a

x+1=0

x=-1

P(x)=ax^3+3x^2-13

P(-1)=a×(-1)^3+3×(-1)^2-13

=a×-1+3×1-13

=-a+3-13

=-a-10

F(x)=5x^3-8x+a

x(-1)=5×(-1)^3-8×(-1)+a

=5×-1-8×-1+a

=-5+8+a

=3+a

-a-10=3+a

2a=-13

a=-13/2

Answer 2

Hey mate....

f(x) = ax3 + 3x2 -13,p(x) = 5x3 -8x + a

r(x) = (x+1)

solution:-

x+1 = 0

x = -1

f(-1) = a(-1) + 3(-1)^2 -13

f(-1) = - a + 3 -13

f(-1) = -a -10 [equation 1]

p(x) = 5x^3 -8x + a

p(-1) = 5(-1)^3 - 8(-1) + a

p(-1) = -5 + 8 + a [equation 2]

equating equation (1) & (2):-

-a -10 = -5 + 8 + a

-a -10 + 5 = 8 + a

-a -5 = 8 + a

-5 = 8 + a + a

-5 = 8 + 2a

-5 -8 = 2a

-13/2 = a

therefore, the value of a is -13/2....

Hope its help you.....