Question

(-8x⁴ - 14x³ + 21x - 11) ÷ (-4x + 1) =

Explanation:

Step1: Use polynomial long division

Divide the leading term of the dividend \(-8x^4\) by the leading term of the divisor \(-4x\): \(\frac{-8x^4}{-4x} = 2x^3\).
Multiply the divisor \(-4x + 1\) by \(2x^3\): \(2x^3(-4x + 1) = -8x^4 + 2x^3\).
Subtract this from the dividend: \((-8x^4 - 14x^3 + 21x - 11) - (-8x^4 + 2x^3) = -16x^3 + 21x - 11\).

Step2: Divide the new leading term

Divide \(-16x^3\) by \(-4x\): \(\frac{-16x^3}{-4x} = 4x^2\).
Multiply the divisor by \(4x^2\): \(4x^2(-4x + 1) = -16x^3 + 4x^2\).
Subtract: \((-16x^3 + 21x - 11) - (-16x^3 + 4x^2) = -4x^2 + 21x - 11\).

Step3: Divide the new leading term

Divide \(-4x^2\) by \(-4x\): \(\frac{-4x^2}{-4x} = x\).
Multiply the divisor by \(x\): \(x(-4x + 1) = -4x^2 + x\).
Subtract: \((-4x^2 + 21x - 11) - (-4x^2 + x) = 20x - 11\).

Step4: Divide the new leading term

Divide \(20x\) by \(-4x\): \(\frac{20x}{-4x} = -5\).
Multiply the divisor by \(-5\): \(-5(-4x + 1) = 20x - 5\).
Subtract: \((20x - 11) - (20x - 5) = -6\).

So the quotient is \(2x^3 + 4x^2 + x - 5\) with a remainder of \(-6\). We can write the result as \(2x^3 + 4x^2 + x - 5 - \frac{6}{-4x + 1}\) or simplify the fraction: \(2x^3 + 4x^2 + x - 5 + \frac{6}{4x - 1}\). But usually, we present it as the quotient plus the remainder over the divisor: \(2x^3 + 4x^2 + x - 5 + \frac{-6}{-4x + 1} = 2x^3 + 4x^2 + x - 5 + \frac{6}{4x - 1}\) (or keep the remainder as is: \(2x^3 + 4x^2 + x - 5 + \frac{-6}{-4x + 1}\)). However, if we consider the division without simplifying the remainder fraction further, the main quotient part is \(2x^3 + 4x^2 + x - 5\) and the remainder is \(-6\), so the division result is \(2x^3 + 4x^2 + x - 5 + \frac{-6}{-4x + 1}\) which can be written as \(2x^3 + 4x^2 + x - 5 - \frac{6}{-4x + 1}\) or simplified to \(2x^3 + 4x^2 + x - 5 + \frac{6}{4x - 1}\). But let's check the long division again for accuracy. Wait, maybe I made a mistake in the subtraction steps. Let's redo the long division carefully:

Dividend: \(-8x^4 -14x^3 + 0x^2 + 21x -11\) (added the missing \(x^2\) term)
Divisor: \(-4x + 1\)

1. First term: \(\frac{-8x^4}{-4x} = 2x^3\). Multiply divisor by \(2x^3\): \(-8x^4 + 2x^3\). Subtract from dividend: \((-8x^4 -14x^3 + 0x^2 + 21x -11) - (-8x^4 + 2x^3) = -16x^3 + 0x^2 + 21x -11\).

2. Second term: \(\frac{-16x^3}{-4x} = 4x^2\). Multiply divisor by \(4x^2\): \(-16x^3 + 4x^2\). Subtract: \((-16x^3 + 0x^2 + 21x -11) - (-16x^3 + 4x^2) = 0x^3 -4x^2 + 21x -11\).

3. Third term: \(\frac{-4x^2}{-4x} = x\). Multiply divisor by \(x\): \(-4x^2 + x\). Subtract: \((0x^3 -4x^2 + 21x -11) - (-4x^2 + x) = 0x^3 + 0x^2 + 20x -11\).

4. Fourth term: \(\frac{20x}{-4x} = -5\). Multiply divisor by \(-5\): \(20x -5\). Subtract: \((20x -11) - (20x -5) = -6\).

Ah, there we go, the missing \(x^2\) term was important. So the division gives us \(2x^3 + 4x^2 + x -5\) with a remainder of \(-6\). So the result is \(2x^3 + 4x^2 + x -5 + \frac{-6}{-4x + 1}\), which simplifies to \(2x^3 + 4x^2 + x -5 + \frac{6}{4x - 1}\) (since \(\frac{-6}{-4x + 1} = \frac{6}{4x - 1}\)). Alternatively, we can write it as \(2x^3 + 4x^2 + x - 5 - \frac{6}{-4x + 1}\), but the standard form is quotient + remainder/divisor. So the final result is \(2x^3 + 4x^2 + x - 5 + \frac{-6}{-4x + 1}\) or simplified \(2x^3 + 4x^2 + x - 5 + \frac{6}{4x - 1}\). But let's check if the remainder is correct. Let's multiply the quotient by the divisor and add the remainder:

\((2x^3 + 4x^2 + x -5)(-4x + 1) + (-6)\)

First, multiply \(2x^3(-4x + 1) = -8x^4 + 2x^3\)

\(4x^2(-4x + 1) = -16x^3 + 4x^2\)

\(x(-4x + 1) = -4x^2 + x\)

\(-5(-4x + 1) = 20x -5\)

Now add all these together:

\(-8x^4 + 2x^3 -16x^3 + 4x^2 -4x^2 + x + 20x -5 -6\)

Combine like terms:

\(-8x^4 + (2x^3 -16x^3) + (4x^2 -4x^2) + (x + 20x) + (-5 -6)\)

\(-8x^4 -14x^3 + 0x^2 + 21x -11\), which matches the original dividend. So the division is correct.

Answer:

\(2x^3 + 4x^2 + x - 5 + \frac{-6}{-4x + 1}\) (or simplified as \(2x^3 + 4x^2 + x - 5 + \frac{6}{4x - 1}\))