Question

use the long division method to find the result when $2x^4 - 3x^3 - 6x + 3$ is divided by $x^2 - 3x + 3$. if there is a remainder, express the result in the form $q(x) + \frac{r(x)}{b(x)}$.

Explanation:

Step1: Divide the leading terms

Divide the leading term of the dividend \(2x^{4}-3x^{3}-6x + 3\) (which is \(2x^{4}\)) by the leading term of the divisor \(x^{2}-3x + 3\) (which is \(x^{2}\)). So, \(\frac{2x^{4}}{x^{2}}=2x^{2}\). This is the first term of the quotient \(q(x)\).

Step2: Multiply the divisor by \(2x^{2}\)

Multiply the entire divisor \(x^{2}-3x + 3\) by \(2x^{2}\): \(2x^{2}(x^{2}-3x + 3)=2x^{4}-6x^{3}+6x^{2}\).

Step3: Subtract this from the dividend

Subtract \(2x^{4}-6x^{3}+6x^{2}\) from the dividend \(2x^{4}-3x^{3}-6x + 3\):
\[ \begin{align*} &(2x^{4}-3x^{3}-6x + 3)-(2x^{4}-6x^{3}+6x^{2})\\ =&2x^{4}-3x^{3}-6x + 3 - 2x^{4}+6x^{3}-6x^{2}\\ =&(2x^{4}-2x^{4})+(-3x^{3}+6x^{3})-6x^{2}-6x + 3\\ =&3x^{3}-6x^{2}-6x + 3 \end{align*} \]

Step4: Divide the new leading term

Now, take the new dividend \(3x^{3}-6x^{2}-6x + 3\) and divide its leading term \(3x^{3}\) by the leading term of the divisor \(x^{2}\). So, \(\frac{3x^{3}}{x^{2}} = 3x\). This is the next term of the quotient.

Step5: Multiply the divisor by \(3x\)

Multiply the divisor \(x^{2}-3x + 3\) by \(3x\): \(3x(x^{2}-3x + 3)=3x^{3}-9x^{2}+9x\).

Step6: Subtract this from the new dividend

Subtract \(3x^{3}-9x^{2}+9x\) from \(3x^{3}-6x^{2}-6x + 3\):
\[ \begin{align*} &(3x^{3}-6x^{2}-6x + 3)-(3x^{3}-9x^{2}+9x)\\ =&3x^{3}-6x^{2}-6x + 3 - 3x^{3}+9x^{2}-9x\\ =&(3x^{3}-3x^{3})+(-6x^{2}+9x^{2})+(-6x - 9x)+3\\ =&3x^{2}-15x + 3 \end{align*} \]

Step7: Divide the new leading term

Take the new dividend \(3x^{2}-15x + 3\) and divide its leading term \(3x^{2}\) by the leading term of the divisor \(x^{2}\). So, \(\frac{3x^{2}}{x^{2}}=3\). This is the next term of the quotient.

Step8: Multiply the divisor by \(3\)

Multiply the divisor \(x^{2}-3x + 3\) by \(3\): \(3(x^{2}-3x + 3)=3x^{2}-9x + 9\).

Step9: Subtract this from the new dividend

Subtract \(3x^{2}-9x + 9\) from \(3x^{2}-15x + 3\):
\[ \begin{align*} &(3x^{2}-15x + 3)-(3x^{2}-9x + 9)\\ =&3x^{2}-15x + 3 - 3x^{2}+9x - 9\\ =&(3x^{2}-3x^{2})+(-15x + 9x)+(3 - 9)\\ =&-6x - 6 \end{align*} \]

Now, the degree of the remainder \(-6x - 6\) is less than the degree of the divisor \(x^{2}-3x + 3\) (degree 2), so we stop.

The quotient \(q(x)=2x^{2}+3x\) (wait, no, in step 4 we had \(3x\) and step 1 was \(2x^{2}\), then after step 9, the remainder is \(-6x - 6\)? Wait, no, let's re - check:

Wait, after step 3, we had \(3x^{3}-6x^{2}-6x + 3\)

Step 4: divide \(3x^{3}\) by \(x^{2}\) to get \(3x\)

Step 5: multiply divisor by \(3x\) gives \(3x^{3}-9x^{2}+9x\)

Subtract from \(3x^{3}-6x^{2}-6x + 3\):

\((3x^{3}-6x^{2}-6x + 3)-(3x^{3}-9x^{2}+9x)=3x^{3}-6x^{2}-6x + 3 - 3x^{3}+9x^{2}-9x = 3x^{2}-15x + 3\)

Step 7: divide \(3x^{2}\) by \(x^{2}\) to get \(3\)

Step 8: multiply divisor by \(3\) gives \(3x^{2}-9x + 9\)

Subtract from \(3x^{2}-15x + 3\):

\((3x^{2}-15x + 3)-(3x^{2}-9x + 9)=3x^{2}-15x + 3 - 3x^{2}+9x - 9=-6x - 6\)

So the quotient \(q(x)=2x^{2}+3x + 3\)? Wait, no:

Wait, first term: \(2x^{2}\) (from \(2x^{4}\div x^{2}\))

Second term: \(3x\) (from \(3x^{3}\div x^{2}\))

Third term: \(3\) (from \(3x^{2}\div x^{2}\))

So \(q(x)=2x^{2}+3x + 3\)

And the remainder \(r(x)=-6x - 6\)

Wait, but let's do the long division properly:

Divide \(2x^{4}-3x^{3}+0x^{2}-6x + 3\) (we added the missing \(x^{2}\) term) by \(x^{2}-3x + 3\)

1. Divide \(2x^{4}\) by \(x^{2}\) to get \(2x^{2}\). Multiply divisor by \(2x^{2}\): \(2x^{2}(x^{2}-3x + 3)=2x^{4}-6x^{3}+6x^{2}\)

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

2. Divide \(3x^{3}\) by \(x^{2}\) to get \(3x\). Multiply divisor by \(3x\): \(3x(x^{2}-3x + 3)=3x^{3}-9x^{2}+9x\)

Subtract: \((3x^{3}-6x^{2}-6x + 3)-(3x^{3}-9x^{2}+9x)=3x^{2}-15x + 3\)

3. Divide \(3x^{2}\) by \(x^{2}\) to get \(3\). Multiply divisor by \(3\): \(3(x^{2}-3x + 3)=3x^{2}-9x + 9\)

Subtract: \((3x^{2}-15x + 3)-(3x^{2}-9x + 9)=-6x - 6\)

So the quotient \(q(x)=2x^{2}+3x + 3\) and the remainder \(r(x)=-6x - 6\), and the divisor \(b(x)=x^{2}-3x + 3\)

So the result is \(q(x)+\frac{r(x)}{b(x)}=(2x^{2}+3x + 3)+\frac{-6x - 6}{x^{2}-3x + 3}\) or we can factor out \(-6\) from the remainder: \((2x^{2}+3x + 3)+\frac{-6(x + 1)}{x^{2}-3x + 3}\)

Answer:

\(2x^{2}+3x + 3+\frac{-6x - 6}{x^{2}-3x + 3}\) (or \(2x^{2}+3x + 3-\frac{6x + 6}{x^{2}-3x + 3}\))