If ax^3+bx^2+x-6 has (x+2) as a factor and leaves a remainder 4 when divided by (x-2), find the value of a and b

a = 0

b = 2

Step-by-step explanation:

Synthetic division gives rise to two equations:

b -2a = 2

b +2a = 2

The solution to these equations is (a, b) = (0, 2).

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The remainder from the division is the lower-right sum shown in the attached tables. For the factor x+2, the remainder from dividing your cubic by x+2 is (4b-8a-8). Setting this to zero and putting it into standard form gives the first equation shown above.

The remainder from division by (x-2) is (4b+8a-4). Setting this to 4 and putting the result into standard form gives the second equation above.

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Pen skips and chocolate smudges make the first attachment less "professional" than it might otherwise be. Please forgive. The graph in the second attachment verifies this result.

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

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step-by-step explanation: