# Find values of the constants in the following identity: x^4+Ax^3 + 5x^2 + x + 3 = (x^2+4)(x^2-x+B)+Cx+ D

Another question on identities:

$$x^4+Ax^3 + 5x^2 + x + 3 = (x^2+4)(x^2-x+B)+Cx+ D$$

How can I find the coefficients for this?

I've got as far as multiplying out the brackets to get:

$$x^4+Ax^3 + 5x^2 + x + 3 = (x^4-x^3+Bx^2+4x^2-4x+4B)+Cx+ D$$

It would be useful to get a hint at least on where to go next.

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You're almost to the final solution. Just combine all coefficients of various powers or $x$ and compare each of them individually:

$$x^4+Ax^3 + 5x^2 + x + 3 = x^4-x^3+ (B+4)x^2+(C-4)x+4B+ D$$ You'll get

\begin{align} A &= -1 \\ B &= +5 - 4 \\ &= 1 \\ C &= +1 +4 \\ &= 5 \\ D &= 3 - 4B \\ &= -1 \end{align}

### Edit

To explain a little as OP mentioned:

The process is as simple as taking everything onto one side of $=$:

$$(x^4 - x^4) + (A x^3 + x^3) + (5 x^2 - (B + 4) x^2 ) + (x - (C - 4)x ) + \left(3 - (4B + D)\right) = 0 \\ \implies (A + 1) x^3 + (1 - B) x^2 + (4 - C) x + (3 - 4B - D) = 0 \\ \therefore \pmatrix{ A + 1 \\ 1 - B \\ 4 - C \\ 3 - 4B - D } = \pmatrix{ 0 \\ 0 \\ 0 \\ 0 }$$ which gives you the result.

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@AbhraAbirKundu Thanks. I used the exponent :/ I should sleep now. – hjpotter92 Jun 11 '13 at 17:38
You are welcome... – Abhra Abir Kundu Jun 11 '13 at 17:40
does this mean you should simplify further by removing x^4 from the rhs and lhs?, divide both sides by x ? – peter_gent Jun 11 '13 at 22:59
I'm missing some steps or intuition on how exactly you compare. can you please explain – peter_gent Jun 11 '13 at 23:30
@peter_gent Okay, check the edit to my reply. – hjpotter92 Jun 12 '13 at 4:53

Now you can equate the coefficients of each power of $x$ from both sides.

Like you equate the coefficient of $x^3$ to get $A=-1$

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