The expression f(x)=x³+ax²+bx+c leaves the same remainder,r, when it is divided by x+2 and when it is divided by x-2. evalute b. i need this as soon as possible (today )

b = -4

Step-by-step explanation:

f(2) = R  and f(-2) = R

Therefore f(2) = f(-2):-

(2)^3 + a(2)^2 + 2b + c =  (-2)^3 +a(-2)^2 - 2b + c

8 + 4a + 2b+ c  = -8 + 4a - 2b + c

The 4a and  c  cancel out. Then:-

8 + 2b = -8 - 2b

4b = -16

b = -4    answer

Consider this example: say we want to convert x = 11 0101 1001₂ to base 3.

The binary expansion of x is

x = 1×2⁹ + 1×2⁸ + 1×2⁶ + 1×2⁴ + 1×2³ + 1×2⁰

which in base 10 reduces to

x = 512 + 256 + 64 + 16 + 8 + 1 = 857

Recall the first several powers of 3:

3⁰ = 1; 3¹ = 3; 3² = 9; 3³ = 27; 3⁴ = 81; 3⁵ = 243; 3⁶ = 729

Now it's a matter of computing quotients and remainders. Pick the largest power of 3 that fits into the target number, then check the remainder and repeat.

• How many times does the 729 go into 857? 1 time, and it leaves a remainder of 128, or

857 = 1×729 + 128 = 1×3⁶ + 128

• How many times does 81 go into 128? Also 1 time, so

128 = 1×81 + 47 = 1×3⁴ + 47

• How many times does 27 go into 47? Also 1 time,

47 = 1×27 + 20 = 1×3³ + 20

• How many times does 9 go into 20? 2 times,

20 = 18 + 2 = 2×9 + 2 = 2×3² + 2

• How many times does 1 go into 2? Also 2 times,

2 = 2 + 1 = 2×1 + 0 = 2×3⁰

So we found

857 = 1×3⁶ + 1×3⁴ + 1×3³ + 2×3² + 2×3⁰

which is the ternary expansion of x,

x = 1 011 202₃