Question

find the quotient and remainder using long division.

\\\frac{x^4 - 4x^3 + 6x - 7}{x - 1}\\

the quotient is

the remainder is

Explanation:

Step 1: Set up the division

Write the dividend with a placeholder for the missing \(x^2\) term.

$$x^4 - 4x^3 + 0x^2 + 6x - 7$$

Step 2: Divide the first term

Divide \(x^4\) by \(x\) to get the first term.

$$\frac{x^4}{x} = x^3$$

Step 3: Multiply and subtract

Multiply \(x^3\) by \((x - 1)\) and subtract from the dividend.

$$(x^4 - 4x^3) - (x^4 - x^3) = -3x^3$$

Step 4: Divide the next term

Divide \(-3x^3\) by \(x\) to get the second term.

$$\frac{-3x^3}{x} = -3x^2$$

Step 5: Multiply and subtract

Multiply \(-3x^2\) by \((x - 1)\) and subtract from the remainder.

$$(-3x^3 + 0x^2) - (-3x^3 + 3x^2) = -3x^2$$

Step 6: Divide the third term

Divide \(-3x^2\) by \(x\) to get the third term.

$$\frac{-3x^2}{x} = -3x$$

Step 7: Multiply and subtract

Multiply \(-3x\) by \((x - 1)\) and subtract from the remainder.

$$(-3x^2 + 6x) - (-3x^2 + 3x) = 3x$$

Step 8: Divide the final term

Divide \(3x\) by \(x\) to get the constant term.

$$\frac{3x}{x} = 3$$

Step 9: Find the final remainder

Multiply \(3\) by \((x - 1)\) and subtract to find the remainder.

$$(3x - 7) - (3x - 3) = -4$$

Answer:

The quotient is \(x^3 - 3x^2 - 3x + 3\)
The remainder is \(-4\)