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- PinkgreenLv 71 month agoFavorite Answer
Let f(x)=2x^3+ax^2+bx-3, then

f(1)=2+a+b-3=8

=>

a+b=9-------(1)

f(-2)=-16+4a-2b-3=-7

=>

2a-b+6------(2)

according to the remainder theorem.

Solving the system (1) & (2) for a, b, get

a=5

b=4

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- David1217Lv 51 month ago
You have two unknowns, so you need to set up a system of two equations. You can make this system of equations by dividing the polynomials by (x-1) and setting the remainder equal to 8, and then dividing by (x+2) and setting that remainder equal to -7.

You can do long division of polynomials to get remainders. Look up how to do long division of polynomials if you do not know how to do it.

Dividing 2x^3 + ax^2 + bx - 3 by (x-1), you get a remainder of -3-b+a-2 = -5-b+a. So you have:

-5 -b + a = 8 or a + b = 13

Dividing 2x^3 + ax^2 + bx - 3 by (x+2), you get a remainder of -2b+4a-19. So you have:

-2b+4a-19 = -7 or -2b +4a = 12

So your system of equations is:

a + b = 13

-2b +4a = 12

-2b +4(13-b) = 12

-2b -4b +52 = 12

-6b = -40

b = 40/6 = 20/3

a = 13 - 20/3 = 19/3

Note: This is the process you use to find a and b, but there is a possibility that I made a mistake somewhere, so you should do the problem out yourself to check.

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- PopeLv 71 month ago
Let p(x) = 2x³ + ax² + bx - 3.

By the remainder theorem of polynomials, p(1) = 8 and p(-2) = -7.

p(1) = 8

2 + a + b - 3 = 8

a + b = 9

p(-2) = -7

-16 + 4a - 2b - 3 = -7

4a - 2b = 12

2a - b = 6

It is now a system of two linear equations in a and b:

a + b = 9

2a - b = 6

a = 5

b = 4

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