Question

what is the quotient of $(x^4 - 3x^3 + 6x^2 - 12x + 8) \div (x - 1)$?
a $x^3 + 2x^2 + 4x + 8$
b $x^3 + 2x^2 - 4x - 8$
c $x^3 - 2x^2 - 4x + 8$
d $x^3 - 2x^2 + 4x - 8$

Explanation:

Step1: Use polynomial long division

Divide the leading term of the dividend \(x^4 - 3x^3 + 6x^2 - 12x + 8\) by the leading term of the divisor \(x - 1\), which is \(x\). So, \(\frac{x^4}{x}=x^3\).
Multiply the divisor \(x - 1\) by \(x^3\) to get \(x^4 - x^3\).
Subtract this from the dividend: \((x^4 - 3x^3 + 6x^2 - 12x + 8)-(x^4 - x^3)= - 2x^3 + 6x^2 - 12x + 8\).

Step2: Repeat the process

Divide the leading term of the new dividend \(-2x^3\) by \(x\) to get \(-2x^2\).
Multiply the divisor \(x - 1\) by \(-2x^2\) to get \(-2x^3 + 2x^2\).
Subtract this from the new dividend: \((-2x^3 + 6x^2 - 12x + 8)-(-2x^3 + 2x^2)=4x^2 - 12x + 8\).

Step3: Repeat again

Divide the leading term of the new dividend \(4x^2\) by \(x\) to get \(4x\).
Multiply the divisor \(x - 1\) by \(4x\) to get \(4x^2 - 4x\).
Subtract this from the new dividend: \((4x^2 - 12x + 8)-(4x^2 - 4x)= - 8x + 8\).

Step4: Repeat once more

Divide the leading term of the new dividend \(-8x\) by \(x\) to get \(-8\).
Multiply the divisor \(x - 1\) by \(-8\) to get \(-8x + 8\).
Subtract this from the new dividend: \((-8x + 8)-(-8x + 8)=0\).

Putting together the terms we got: \(x^3 - 2x^2 + 4x - 8\)? Wait, no, wait, let's check again. Wait, in step 1, when we did the first subtraction: \(x^4 - 3x^3 - (x^4 - x^3)=x^4 - 3x^3 - x^4 + x^3=-2x^3\), correct. Then step 2: divide \(-2x^3\) by \(x\) is \(-2x^2\), multiply divisor by \(-2x^2\) is \(-2x^3 + 2x^2\), subtract from \(-2x^3 + 6x^2\) gives \(4x^2\), correct. Then step 3: divide \(4x^2\) by \(x\) is \(4x\), multiply divisor by \(4x\) is \(4x^2 - 4x\), subtract from \(4x^2 - 12x\) gives \(-8x\), correct. Then step 4: divide \(-8x\) by \(x\) is \(-8\), multiply divisor by \(-8\) is \(-8x + 8\), subtract from \(-8x + 8\) gives 0. Wait, but the quotient terms are \(x^3 - 2x^2 + 4x - 8\)? But let's check with synthetic division. The root for \(x - 1\) is \(x = 1\). Using synthetic division:

Coefficients of dividend: 1 (x⁴), -3 (x³), 6 (x²), -12 (x), 8 (constant)

Bring down the 1.

Multiply by 1: 1. Add to -3: -2.

Multiply by 1: -2. Add to 6: 4.

Multiply by 1: 4. Add to -12: -8.

Multiply by 1: -8. Add to 8: 0.

So the quotient polynomial is \(x^3 - 2x^2 + 4x - 8\), which is option D. Wait, but earlier when I did long division, I think I made a mistake in the sign. Wait, let's re - do the long division:

Dividend: \(x^4 - 3x^3 + 6x^2 - 12x + 8\)

Divisor: \(x - 1\)

First term: \(x^4\div x=x^3\). Multiply divisor by \(x^3\): \(x^4 - x^3\). Subtract from dividend:

\((x^4 - 3x^3 + 6x^2 - 12x + 8)-(x^4 - x^3)=x^4 - 3x^3 + 6x^2 - 12x + 8 - x^4 + x^3=-2x^3 + 6x^2 - 12x + 8\)

Second term: \(-2x^3\div x=-2x^2\). Multiply divisor by \(-2x^2\): \(-2x^3 + 2x^2\). Subtract:

\((-2x^3 + 6x^2 - 12x + 8)-(-2x^3 + 2x^2)=-2x^3 + 6x^2 - 12x + 8 + 2x^3 - 2x^2 = 4x^2 - 12x + 8\)

Third term: \(4x^2\div x = 4x\). Multiply divisor by \(4x\): \(4x^2 - 4x\). Subtract:

\((4x^2 - 12x + 8)-(4x^2 - 4x)=4x^2 - 12x + 8 - 4x^2 + 4x=-8x + 8\)

Fourth term: \(-8x\div x=-8\). Multiply divisor by \(-8\): \(-8x + 8\). Subtract:

\((-8x + 8)-(-8x + 8)=-8x + 8 + 8x - 8 = 0\)

So the quotient is \(x^3 - 2x^2 + 4x - 8\), which is option D.

Answer:

D. \(x^3 - 2x^2 + 4x - 8\)