Question

when a polynomial 2x³+3x²+ax+b is divided by x-2 leaves remainder 2 and x+2 leaves remainder -2 find a and b

Answer 1

In 2x^3+3x^2+ax+b ,x-2=0 then x=2 so we have to add the value of 2 in place of x

2*8+3*4+2a+b=28+2a+b=2

x+2=0 so x=-2 
now we will add the value of -2 in place of x

2*-8+3*4-2a+b=4-2a+b=-2 
-4+2a-b=2
now adding both equations 
-4+2a-b+28+2a+b=4
24+4a=4
a=-5
as 4-2a+b=-2
adding value of a
14+b=-2
b=-16

Answer 2

P(X) = 2x³ + 3x² + ax + b

P(2) = 0 and p(-2) = 0

if P(2) = (2)³ + 3(2)² + a(2) + b = 0

= 16 + 12 + 2a + b = 0

= 2a + b + 28 = 0

2a + b = -28 -------------(1)

if P(-2) = 0 = 2(-2)³ + 3(-2)² + a(-2) + b = 0

= 2(-8) + 3(4) -2a + b = 0

= -16 + 12 - 2a + b = 0

= -2a + b - 4 = 0

= -2a + b = 4 -----------------(2)

2a + b = -28

- 2a + b = 4

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0 + 2B = -24

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b =$\frac{-24}{2}$

b = -12

therefore 2a + b = -28

2a = -28 - b

2a -12 = -28

2a = -28 + 12

2a = -16

a =$\frac{-16}{2}$

a = -8