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2. $(x^{4}+4x^{3}-28x^{2}-45x + 26)div(x + 7)$

Question

  1. $(x^{4}+4x^{3}-28x^{2}-45x + 26)div(x + 7)$

Explanation:

Step1: Divide the leading terms

Divide the leading term of the dividend \(x^4\) by the leading term of the divisor \(x\) to get \(x^3\). Multiply the divisor \((x + 7)\) by \(x^3\) to get \(x^4+7x^3\). Subtract this from the dividend:
\[

$$\begin{align*} &(x^4 + 4x^3 - 28x^2 - 45x + 26)-(x^4+7x^3)\\ =&x^4 + 4x^3 - 28x^2 - 45x + 26 - x^4 - 7x^3\\ =& - 3x^3 - 28x^2 - 45x + 26 \end{align*}$$

\]

Step2: Divide the new leading terms

Divide the leading term of the new dividend \(-3x^3\) by the leading term of the divisor \(x\) to get \(-3x^2\). Multiply the divisor \((x + 7)\) by \(-3x^2\) to get \(-3x^3-21x^2\). Subtract this from the new dividend:
\[

$$\begin{align*} &(-3x^3 - 28x^2 - 45x + 26)-(-3x^3-21x^2)\\ =&-3x^3 - 28x^2 - 45x + 26 + 3x^3 + 21x^2\\ =& - 7x^2 - 45x + 26 \end{align*}$$

\]

Step3: Divide the new leading terms

Divide the leading term of the new dividend \(-7x^2\) by the leading term of the divisor \(x\) to get \(-7x\). Multiply the divisor \((x + 7)\) by \(-7x\) to get \(-7x^2-49x\). Subtract this from the new dividend:
\[

$$\begin{align*} &(-7x^2 - 45x + 26)-(-7x^2-49x)\\ =&-7x^2 - 45x + 26 + 7x^2 + 49x\\ =&4x + 26 \end{align*}$$

\]

Step4: Divide the new leading terms

Divide the leading term of the new dividend \(4x\) by the leading term of the divisor \(x\) to get \(4\). Multiply the divisor \((x + 7)\) by \(4\) to get \(4x + 28\). Subtract this from the new dividend:
\[

$$\begin{align*} &(4x + 26)-(4x + 28)\\ =&4x + 26 - 4x - 28\\ =& - 2 \end{align*}$$

\]

Step5: Write the result

The quotient is \(x^3-3x^2 - 7x + 4\) and the remainder is \(-2\). So, \(\frac{x^4 + 4x^3 - 28x^2 - 45x + 26}{x + 7}=x^3-3x^2 - 7x + 4-\frac{2}{x + 7}\)

Answer:

The quotient is \(x^3 - 3x^2 - 7x + 4\) and the remainder is \(-2\), so \((x^4 + 4x^3 - 28x^2 - 45x + 26)\div(x + 7)=x^3 - 3x^2 - 7x + 4-\frac{2}{x + 7}\)