Factoring Polynomials Using Synthetic Division

[kenny1999]

Homework Statement

Let H(x) = x^4 + 5x^3 + 10x^2 - x - 5
(a) Find the quotient and remainder when H(x) is divided by x^2 + 3x + 3
(b) If H(x) + Ax + B is divisible by x^2 + 3x + 3, find the values of A and B
(c) If H(x) + Cx^2 + Dx is divisible by x^2 + 3x + 3, find the values of C and D

Homework Equations

The Attempt at a Solution

I am able to work out part (a)
Quotient should be x^2 + 2x + 1 and Remainder should be -10x - 8

For part (b),
my attempt is
from part (a), it is deduced that
H(x) = Quotient x Divisor + Remainder
H(x) = (x^2 + 3x + 3) (x^2 + 2x + 1) + (-10x - 8)
then
H(x) + 10x + 8 = (x^2 + 3x + 3) (x^2 + 2x + 1)

then it is seen that A = 10 and B = 8

but I really doubt it is correct
because I think H(x) = (x^2 + 3x + 3) (x^2 + 2x + 1) + (-10x - 8)
is in fact an identity, since it is true for all x, it is definitely not an equation.

But why -10x - 8 could be 'thrown' to the left hand side

 

[SteamKing]

It's simple algebra.

If H(x) = Q*D + R, then it also stands to reason that
H(x) - R = Q*D

Nothing is being 'thrown' anywhere.

 

[SammyS]

Homework Statement

Let H(x) = x^4 + 5x^3 + 10x^2 - x - 5
(a) Find the quotient and remainder when H(x) is divided by x^2 + 3x + 3
(b) If H(x) + Ax + B is divisible by x^2 + 3x + 3, find the values of A and B
(c) If H(x) + Cx^2 + Dx is divisible by x^2 + 3x + 3, find the values of C and D

Homework Equations

The Attempt at a Solution

I am able to work out part (a)
Quotient should be x^2 + 2x + 1 and Remainder should be -10x - 8

For part (b),
my attempt is
from part (a), it is deduced that
H(x) = Quotient x Divisor + Remainder
H(x) = (x^2 + 3x + 3) (x^2 + 2x + 1) + (-10x - 8)
then
H(x) + 10x + 8 = (x^2 + 3x + 3) (x^2 + 2x + 1)

then it is seen that A = 10 and B = 8

but I really doubt it is correct
because I think H(x) = (x^2 + 3x + 3) (x^2 + 2x + 1) + (-10x - 8)
is in fact an identity, since it is true for all x, it is definitely not an equation.

But why -10x - 8 could be 'thrown' to the left hand side

If $\displaystyle \ H(x) = (x^2 + 3x + 3) (x^2 + 2x + 1) + (-10x - 8)\,,\$ for all x, then doesn't it follow that

$\displaystyle H(x)+(10x+8) = (x^2 + 3x + 3) (x^2 + 2x + 1) + (-10x - 8)+(10x+8)\ ?$​