Polynomials?

Question

When the polynomial P(x) = ax^4 + bx^3 + 3x^2 - 2x + 3 is divided by (x-1)(x-2) the quotient is Q(x) and the remainder is (x+1).
Find the vaules of 'a' and 'b'.

Answer 1

From the problem, you see that

P(x) = Q(x)*(x-1)(x-2) + (x+1)
ax^4 + bx^3 + 3x^2 - 2x + 3 = Q(x)*(x-1)(x-2) + (x+1) .

To solve for a and b, we substitute a value on x such that the expression Q(x)*(x-1)(x-2) will become 0. Immediately, we see that these values are x=1 and x=2.

For x=1, we have
a(1)^4 + b(1)^3 + 3(1)^2 - 2(1) + 3 = (0) + (1+1) .
a + b + 3 - 2 + 3 = 2
a + b = -2. [our equation #1]

For x=2, we have
a(2)^4 + b(2)^3 + 3(2)^2 - 2(2) + 3 = (0) + (2+1) .
16a + 8b + 12 - 4 + 3 = 3
16a + 8b = -8 [our equation #2]

Thus we have a system of two simultaneous linear equation:
a + b = -2
16a + 8b = -8

There are many ways to solve this; for now lets use the method of substitution. From the first equation, we have
a = -2 - b.

We substitute this to the variable `a` on the 2nd equation. We have:
16(-2 - b) + 8b = -8
-32 -16b +8b = -8
-8b = 24
b = -3.
Then a = -2 - b = -2 - (-3) = 1.

:)

Answer 2

ax^4 + bx^3 + 3x^2 - 2x + 3 =(x-1)(x-2)Q(x)+(x+1).

ax^4 + bx^3 + 3x^2 - 3x +2=(x-1)(x-2)Q(x)

x-1=0
For x=1 a+b+3-3+2=0

a+b=-2

x-2=0
For x=2
16a+8b+12-6+2=0

16a+8b+8=0

2a+b=-1

a=1
b=-3

Answer 3

(x-1)(x-2)(Q(x)) + (x+1) = ax^4 + bx^3 + 3x^2 -2x + 3

LHS = (x^2-3x+2)(Q(x)) + (x+1)
Highest power on the RHS is x^4
Q(x) is of the O(x^2)
Q(x) = (dx^2 + ex + f)

(x^2-3x+2)(dx^2+ex+f) + (x+1) = ax^4 + bx^3 + 3x^2 -2x + 3

Equating coefficients of x^0
2f+1=3
f=1

Equating coefficients of x
-3f+2e+1=-2
-3+2e+1=-2
e=0

Equating coefficients of x^2
2d-3e+f=3
2d+1=3
d=1

(x^2-3x+2)(dx^2+f) + (x+1) = ax^4 + bx^3 + 3x^2 -2x + 3
(x^2-3x+2)(x^2+1) + (x+1) = ax^4 + bx^3 + 3x^2 -2x + 3
Expanding this out
x^4-3x^3+3x^2-3x+2+x+1=ax^4 + bx^3 + 3x^2 -2x + 3

Equating coefficients
a=1
b=-3