Question

section 2 - dividing polynomials
for each problem, divide the polynomials, show all work and circle the final answer.
3 points each

1. $x^3 - 5x^2 + 10x + 1 \div (x - 6)$

Explanation:

Step1: Use Polynomial Long Division

Divide the leading term of the dividend \(x^3 - 5x^2 + 10x + 1\) by the leading term of the divisor \(x - 6\). The leading term of the dividend is \(x^3\) and of the divisor is \(x\), so \(\frac{x^3}{x}=x^2\). Multiply the divisor \(x - 6\) by \(x^2\) to get \(x^3 - 6x^2\). Subtract this from the dividend: \((x^3 - 5x^2 + 10x + 1)-(x^3 - 6x^2)=x^2 + 10x + 1\).

Step2: Divide Again

Now, divide the leading term of the new dividend \(x^2 + 10x + 1\) (which is \(x^2\)) by the leading term of the divisor \(x\), giving \(\frac{x^2}{x}=x\). Multiply the divisor \(x - 6\) by \(x\) to get \(x^2 - 6x\). Subtract this from \(x^2 + 10x + 1\): \((x^2 + 10x + 1)-(x^2 - 6x)=16x + 1\).

Step3: Divide Once More

Divide the leading term of \(16x + 1\) (which is \(16x\)) by \(x\), giving \(16\). Multiply the divisor \(x - 6\) by \(16\) to get \(16x - 96\). Subtract this from \(16x + 1\): \((16x + 1)-(16x - 96)=97\).

Answer:

The result of the division is \(x^2 + x + 16+\frac{97}{x - 6}\) (or, in quotient - remainder form, the quotient is \(x^2 + x + 16\) and the remainder is \(97\)). If we write it as \(x^2 + x + 16+\frac{97}{x - 6}\), we can also express the division as \((x^3 - 5x^2 + 10x + 1)=(x - 6)(x^2 + x + 16)+97\). The final answer (quotient with remainder) is \(x^2 + x + 16+\frac{97}{x - 6}\) (or quotient \(x^2 + x + 16\), remainder \(97\)).