Question

find each quotient using long division.

1. $(k^3 - 10k^2 + k + 1) \div (k - 1)$
2. $(x^4 + 4x^3 - 28x^2 - 45x + 26) \div (x + 7)$
3. $(20c^3 + 22c^2 - 7c + 7) \div (5c - 2)$
4. $(3n^4 + 6n^3 - 15n^2 + 32n - 25) \div (n + 4)$

Explanation:

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Problem 1: $(k^3 - 10k^2 + k + 1) \div (k - 1)$

Step1: Divide leading terms

$\frac{k^3}{k} = k^2$
Multiply divisor by $k^2$: $k^2(k-1)=k^3 - k^2$
Subtract from dividend:
$(k^3 -10k^2 +k +1)-(k^3 -k^2) = -9k^2 +k +1$

Step2: Divide new leading term

$\frac{-9k^2}{k} = -9k$
Multiply divisor by $-9k$: $-9k(k-1)=-9k^2 +9k$
Subtract:
$(-9k^2 +k +1)-(-9k^2 +9k) = -8k +1$

Step3: Divide new leading term

$\frac{-8k}{k} = -8$
Multiply divisor by $-8$: $-8(k-1)=-8k +8$
Subtract:
$(-8k +1)-(-8k +8) = -7$

Answer:

for 1:
$k^2 -9k -8 - \frac{7}{k-1}$
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Problem 2: $(x^4 + 4x^3 -28x^2 -45x +26) \div (x+7)$

Step1: Divide leading terms

$\frac{x^4}{x} = x^3$
Multiply divisor by $x^3$: $x^3(x+7)=x^4 +7x^3$
Subtract from dividend:
$(x^4 +4x^3 -28x^2 -45x +26)-(x^4 +7x^3) = -3x^3 -28x^2 -45x +26$

Step2: Divide new leading term

$\frac{-3x^3}{x} = -3x^2$
Multiply divisor by $-3x^2$: $-3x^2(x+7)=-3x^3 -21x^2$
Subtract:
$(-3x^3 -28x^2 -45x +26)-(-3x^3 -21x^2) = -7x^2 -45x +26$

Step3: Divide new leading term

$\frac{-7x^2}{x} = -7x$
Multiply divisor by $-7x$: $-7x(x+7)=-7x^2 -49x$
Subtract:
$(-7x^2 -45x +26)-(-7x^2 -49x) = 4x +26$

Step4: Divide new leading term

$\frac{4x}{x} = 4$
Multiply divisor by $4$: $4(x+7)=4x +28$
Subtract:
$(4x +26)-(4x +28) = -2$