Question

question: use the long - division method to find the result when - 2x^4 + 6x^3 - x^2 + 13x + 18 is divided by - x^2 - 3. if there is a remainder, express the result in the form q(x)+\frac{r(x)}{b(x)}. answer. submit answer score: 1/2 penalty: 1 off polynomial long division (level 5) watch video show examples

Explanation:

Step1: Set up long - division

We are dividing the polynomial \(f(x)=-2x^{4}+6x^{3}-x^{2}+13x + 18\) by \(d(x)=-x^{2}-3\). We write it in the long - division format: \(\frac{-2x^{4}+6x^{3}-x^{2}+13x + 18}{-x^{2}-3}\).

Step2: Divide the leading terms

The leading term of the dividend is \(-2x^{4}\) and the leading term of the divisor is \(-x^{2}\). So, \(\frac{-2x^{4}}{-x^{2}} = 2x^{2}\). This is the first term of the quotient \(q(x)\).

Step3: Multiply and subtract

Multiply the divisor \(-x^{2}-3\) by \(2x^{2}\) to get \(-2x^{4}-6x^{2}\). Subtract this from the dividend: \((-2x^{4}+6x^{3}-x^{2}+13x + 18)-(-2x^{4}-6x^{2})=6x^{3}+5x^{2}+13x + 18\).

Step4: Repeat the process

Now, divide the leading term of the new dividend \(6x^{3}\) by the leading term of the divisor \(-x^{2}\), we get \(\frac{6x^{3}}{-x^{2}}=-6x\). This is the next term of the quotient. Multiply the divisor \(-x^{2}-3\) by \(-6x\) to get \(6x^{3}+18x\). Subtract this from the new dividend: \((6x^{3}+5x^{2}+13x + 18)-(6x^{3}+18x)=5x^{2}-5x + 18\).

Step5: Divide the leading term again

Divide the leading term of the new dividend \(5x^{2}\) by the leading term of the divisor \(-x^{2}\), we get \(\frac{5x^{2}}{-x^{2}}=-5\). Multiply the divisor \(-x^{2}-3\) by \(-5\) to get \(5x^{2}+15\). Subtract this from the new dividend: \((5x^{2}-5x + 18)-(5x^{2}+15)=-5x + 3\).

Step6: Write the result

The quotient \(q(x)=2x^{2}-6x - 5\) and the remainder \(r(x)=-5x + 3\). So the result in the form \(q(x)+\frac{r(x)}{d(x)}\) is \(2x^{2}-6x - 5+\frac{-5x + 3}{-x^{2}-3}\).

Answer:

\(2x^{2}-6x - 5+\frac{-5x + 3}{-x^{2}-3}\)