Help..4 remainder theorems...........plz...solve?

Question

Q4: The expressions 8x^3 + ax^2 + bx - 9 leaves remainders -95 and 3 when divided by x + 2 and 2x - 3 respectively. Calculate the value of a and b.

Answer 1

f(x)=8x^3+ax^2+bx-9
f(-2)=-64+4a-2b-9=-95
so4a-2b=-22
2a-b=-11
f(3/2)=27+9/4a+3/2b-9=3
9/4a+3/2b=-15
multiplying by 4
9a+6b=-60
dividing by 3
3a+2b=-20
4a-2b=-22
7a=-42
a=-6
substituting
-12-b=-11
b=-1

Answer 2

Let p(x) = 8x^3 + ax^2 + bx - 9

Since we get a remainder of -95 when we divide by (x + 2), it follows that

p(-2) = -95.
However, we can also solve for p(-2) by directly plugging it into our function.

p(-2) = 8(-2)^3 + a(-2)^2 + b(-2) - 9
p(-2) = 8(-8) + a(4) - 2b - 9
p(-2) = -64 + 4a - 2b - 9
p(-2) = -73 + 4a - 2b

Now, we equate this to -95.
-95 = -73 + 4a - 2b
-22 = 4a - 2b

We also know that when we divide by (2x - 3), we get 3. To get our "root" (which isn't really a root because there's a remainder), we equate 2x - 3 = 0 to get x = 3/2. Therefore

p(3/2) = 3. But,
p(3/2) = 8(3/2)^3 + a(3/2)^2 + b(3/2) - 9
p(3/2) = 8(27/8) + a(9/4) + b(3/2) - 9
p(3/2) = 27 + a(9/4) + b(3/2) - 9
p(3/2) = 18 + a(9/4) + b(3/2)

Now, we equate this to 3.

3 = 18 + a(9/4) + b(3/2)
-15 = a(9/4) + b(3/2)

To get rid of all fractions, let's multiply everything by 4.
-60 = 9a + 6b, and to simplify, let's divide everything by 3.
-20 = 3a + 2b

Therefore, we have two equations and two unknowns:

-22 = 4a - 2b
-20 = 3a + 2b

This is best solved by elimination; let's add the two equations, giving us

-42 = 7a; therefore, a = -6
Since we now have a, we can get b.
-20 = 3(-6) + 2b
-20 = -18 + 2b
-2 = 2b, giving us b = -1

Therefore, a = -6, b = -1.

Answer 3

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Answer 4

what about (x+2)(2x-3)-89