Cn anyone solve these...?

Question

the polynomial x^3 + ax^2 + bx - 3 leaves remainders of 27 when divided by x - 2 and a remainder of 3 when divided by x + 1. Calculate the remainder when the polynomial is divided by x - 1.

Answer 1

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Answer 2

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

Since we get a remainder when dividing p(x) by (x - 2), it follows that p(2) = 27. However, p(2) is also equal to what is below:

p(2) = 2^3 + a(2)^2 + b(2) - 3
p(2) = 8 + 4a + 2b - 3
p(2) = 5 + 4a + 2b

So now we can equate this to 27, getting
27 = 5 + 4a + 2b
22 = 4a + 2b, and reducing this, we get
11 = 2a + b

We also know we get a remainder of 3 when dividing p(x) by (x + 1), so we know that p(-1) = 3. But, p(-1) is also equal to:

p(-1) = (-1)^3 + a(-1)^2 + b(-1) - 3
p(-1) = -1 + a - b - 3
p(-1) = -4 + a - b

Equating this to 3, we get
3 = -4 + a - b
7 = a - b

Two equations, two unknowns:

11 = 2a + b
7 = a - b

To solve this, we can use elimination and add the two equations.

18 = 3a
Therefore, a = 6
Plugging this into the second equation,
7 = 6 - b, therefore,
1 = -b, so b = -1

So a = 6 and b = -1

Answer 3

By remainder theorem,

2^3+a2^2+2b - 3 = 27
which can be reduced to
2a+b = 11......(1)

(-1)^3+a-b - 3 = 3
which can be reduced to
a-b = 7......(2)

Solving the system of (1) and (2) gives,
a = 6, b = -1

Therefore, the remainder when the polynomial is divided by x-1 is: 1+6-1-3 = 3

Answer 4

using the remainder theorem
f(2)=2^3+a(2)^2+b(2)-3=27
so 4a+2b=22 (1)
f(-1)=(-1)^3+a(-1)^2+b(-1)-3=3
so a-b=7 (2)
(2)*2
2a-2b=14
4a+2b=22
adding
6a=36
dividing by 6
a=6
aub in (2)
6-b=7
-b=1
b=-1
so the poly is
x^3+6x^2-b-3
f(1) =1+6-1-3
=3
so the remainderwhen divided by x-1 is 3