Question

divide the expression.
\\(\frac{6x^4 - 4x^2 + 6}{x^2 + 2}\\)
\\(\frac{6x^4 - 4x^2 + 6}{x^2 + 2} = \square\\) (simplify your answer.)

Explanation:

Step1: Perform polynomial long division or rewrite the numerator

We can rewrite the numerator \(6x^{4}-4x^{2}+6\) to factor out \(x^{2}+2\). First, note that \(6x^{4}=6x^{2}\cdot x^{2}\), and we can express the numerator as:
\[ \begin{align*} 6x^{4}-4x^{2}+6&=6x^{2}(x^{2}+2)-12x^{2}-4x^{2}+6\\ &=6x^{2}(x^{2}+2)-16x^{2}+6 \end{align*} \]
Now, rewrite \(-16x^{2}\) as \(-16(x^{2}+2)+32\):
\[ \begin{align*} 6x^{2}(x^{2}+2)-16x^{2}+6&=6x^{2}(x^{2}+2)-16(x^{2}+2)+32 + 6\\ &=6x^{2}(x^{2}+2)-16(x^{2}+2)+38 \end{align*} \]
Wait, that's not the right way. Let's do polynomial long division. Divide \(6x^{4}-4x^{2}+6\) by \(x^{2}+2\).

The first term: \(\frac{6x^{4}}{x^{2}} = 6x^{2}\). Multiply \(x^{2}+2\) by \(6x^{2}\) to get \(6x^{4}+12x^{2}\). Subtract this from the numerator:

\((6x^{4}-4x^{2}+6)-(6x^{4}+12x^{2})=-16x^{2}+6\)

Next term: \(\frac{-16x^{2}}{x^{2}}=-16\). Multiply \(x^{2}+2\) by \(-16\) to get \(-16x^{2}-32\). Subtract this from \(-16x^{2}+6\):

\((-16x^{2}+6)-(-16x^{2}-32)=38\)

So, \(\frac{6x^{4}-4x^{2}+6}{x^{2}+2}=6x^{2}-16+\frac{38}{x^{2}+2}\)? Wait, that can't be. Wait, maybe I made a mistake in the sign. Wait, let's check again.

Wait, the numerator: \(6x^{4}-4x^{2}+6\). Let's try to factor differently. Let \(y = x^{2}\), then the numerator is \(6y^{2}-4y + 6\), and the denominator is \(y + 2\).

Using polynomial long division for \(6y^{2}-4y + 6\) divided by \(y + 2\):

First term: \(\frac{6y^{2}}{y}=6y\). Multiply \(y + 2\) by \(6y\) to get \(6y^{2}+12y\). Subtract from \(6y^{2}-4y + 6\):

\((6y^{2}-4y + 6)-(6y^{2}+12y)=-16y + 6\)

Next term: \(\frac{-16y}{y}=-16\). Multiply \(y + 2\) by \(-16\) to get \(-16y-32\). Subtract from \(-16y + 6\):

\((-16y + 6)-(-16y-32)=38\)

So, \(\frac{6y^{2}-4y + 6}{y + 2}=6y-16+\frac{38}{y + 2}\). Substituting back \(y = x^{2}\), we get \(\frac{6x^{4}-4x^{2}+6}{x^{2}+2}=6x^{2}-16+\frac{38}{x^{2}+2}\). But that doesn't seem right. Wait, maybe the original problem has a typo? Wait, no, maybe I made a mistake. Wait, let's check the numerator again. If the numerator was \(6x^{4}-16x^{2}+6\), then it would work. Wait, the original numerator is \(6x^{4}-4x^{2}+6\). Wait, maybe the user made a typo, but assuming the problem is correct, let's re-express.

Wait, another approach: Let's assume that the division results in a polynomial (no remainder). So, let \(6x^{4}-4x^{2}+6=(x^{2}+2)(ax^{2}+bx + c)\). Expanding the right side: \(ax^{4}+bx^{3}+(c + 2a)x^{2}+2bx + 2c\). Comparing coefficients:

- \(x^{4}\): \(a = 6\)
- \(x^{3}\): \(b = 0\)
- \(x^{2}\): \(c + 2a=-4\). Since \(a = 6\), \(c + 12=-4\), so \(c=-16\)
- \(x\): \(2b = 0\) (which matches \(b = 0\))
- constant term: \(2c = 6\). But \(2c=2\times(-16)=-32 eq6\). So, there is a remainder. But the problem says "Simplify your answer", maybe it's a mistake, or maybe I misread the problem. Wait, the original problem: numerator is \(6x^{4}-4x^{2}+6\), denominator \(x^{2}+2\). Wait, maybe the numerator is \(6x^{4}-16x^{2}+38\)? No, the user wrote \(6x^{4}-4x^{2}+6\). Wait, maybe I made a mistake in long division. Let's do it again.

Divide \(6x^{4}+0x^{3}-4x^{2}+0x + 6\) by \(x^{2}+0x + 2\).

First term: \(6x^{4}\div x^{2}=6x^{2}\). Multiply divisor by \(6x^{2}\): \(6x^{4}+0x^{3}+12x^{2}\). Subtract from dividend:

\((6x^{4}+0x^{3}-4x^{2}+0x + 6)-(6x^{4}+0x^{3}+12x^{2})=0x^{3}-16x^{2}+0x + 6\)

Next term: \(-16x^{2}\div x^{2}=-16\). Multiply divisor by \(-16\): \(0x^{3}+0x^{2}-32\). Subtract from the previous result:

\((0x^{3}-16x^{2}+0x + 6)-(-16x^{2}+0x - 32)=38\)

So, the division gives \(6x^{2}-16+\frac{38}{x^{2}+2}\). But the problem says "Simplify your answer", maybe it's a mistake, or maybe the numerator was supposed to be \(6x^{4}-16x^{2}+38\), but as per the given problem, this is the result. Wait, but maybe I misread the problem. Let me check again. The problem is to divide \(\frac{6x^{4}-4x^{2}+6}{x^{2}+2}\). So, the answer is \(6x^{2}-16+\frac{38}{x^{2}+2}\). But that seems odd. Wait, maybe the original numerator was \(6x^{4}-16x^{2}+38\), then the remainder would be zero. Alternatively, maybe the user made a typo. But assuming the problem is correct, this is the result.

Wait, no, maybe I made a mistake in the sign when subtracting. Let's do the subtraction again:

Dividend: \(6x^{4}+0x^{3}-4x^{2}+0x + 6\)

Divisor: \(x^{2}+0x + 2\)

First step: Multiply divisor by \(6x^{2}\): \(6x^{4}+0x^{3}+12x^{2}\)

Subtract from dividend: \((6x^{4}+0x^{3}-4x^{2}+0x + 6)-(6x^{4}+0x^{3}+12x^{2})=0x^{3}-16x^{2}+0x + 6\) (correct)

Second step: Multiply divisor by \(-16\): \(0x^{3}+0x^{2}-32\)

Subtract from the above result: \((0x^{3}-16x^{2}+0x + 6)-(-16x^{2}+0x - 32)=0x^{3}+0x^{2}+0x + 38\) (correct). So, the remainder is 38. So, the result is \(6x^{2}-16+\frac{38}{x^{2}+2}\).

But the problem says "Simplify your answer", maybe it's a mistake, but this is the result.

Wait, maybe the original problem has a different numerator. For example, if the numerator is \(6x^{4}-16x^{2}+38\), then \(\frac{6x^{4}-16x^{2}+38}{x^{2}+2}=6x^{2}-16+\frac{38}{x^{2}+2}\)? No, wait, \(6x^{2}(x^{2}+2)=6x^{4}+12x^{2}\), then \(6x^{4}-16x^{2}+38 - (6x^{4}+12x^{2})=-28x^{2}+38\), then \(-28x^{2}+38=-28(x^{2}+2)+56 + 38=-28(x^{2}+2)+94\). No, that's not right. I think I'm confused. Maybe the problem is correct, and the answer is \(6x^{2}-16+\frac{38}{x^{2}+2}\).

Answer:

\(6x^{2}-16+\frac{38}{x^{2}+2}\)