Question

when a polynomial 2xcube +3x square +ax+b is divided by (x-2) leaves remainder 2, and (x+2) leaves remainder-2. find a and b​

Answer 1

Step-by-step explanation:

let x-2=0

x=2

f(x)=2x^3+3x^2+ax+b

f(2)=2(8)+3(4)+a(2)+b

2=16+12+2a+b

-26=2a+b

let x+2=0

x= -2

f(x)=2x^3+3x^2+ax+b

f(-2)=2(-8)+3(4)+a(-2)+b

-2=-16+12-2a+b

-2=-4-2a+b

2=-2a +b

2a + b = -26

-2a+ b = 2

2b=-24

b= -12

-26=2a+b

-26=2a-12

-14=2a

-7=a

Answer 2

Answer:

Thus a=-7 and b=-6

Step-by-step explanation:

Let p(x)=2x³+3x²+ax+b

When p(x) is divided by (x-2) leaves remainder 2

so p(2)=2

Putting x=2 in p(x) we get

2=2*8+3*4+2a+b

2a+b=2-16-12=-26

2a+b=-26---------(1)

When p(x) is divided by (x+2) leaves remainder -2

so p(-2) = -2

Putting x= -2 in p(x) we get

2.(-2)³+3(-2)²-2a+b=-2

-2*8+3*4-2a+b=-2

-16+12-2a+b=-2

-2a+b =-2+16-12

-2a+b=2--------(2)

adding (1) and (2)

2b=-24, b=-12

from(1)

2a-12=-26

2a=-26+12=-14

a=-7

Thus a=-7 and b=-6