Rewrite the rational expression $\frac{4x^3 - 7x^2 + 5x - 6}{x - 2}$ in the equivalent polynomial and remainder form $q(x) + \frac{r}{x - 2}$, where $q(x)$ is a polynomial and $r$ is a real constant.

Rewrite the rational expression $\frac{4x^3 - 7x^2 + 5x - 6}{x - 2}$ in the equivalent polynomial and remainder form $q(x) + \frac{r}{x - 2}$ , where $q(x)$ is a polynomial and $r$ is a real constant.

Final answer

$4x^2 + x + 7 + \frac{8}{x - 2}$

Steps

To rewrite a rational expression $\frac{P(x)}{D(x)}$ into the form $q(x) + \frac{r}{D(x)}$ , divide the numerator polynomial by the denominator using polynomial long division or synthetic division.

Method: Synthetic Division

Because the divisor is linear and in the form $x - c$ , we can use synthetic division with $c = 2$ .

Set up the coefficients: Write down the root $c = 2$ and the coefficients of the dividend $4x^3 - 7x^2 + 5x - 6$ in descending order of degree:
- Coefficients, in descending degree order: $4,\,-7,\,5,\,-6$ (synthetic-division root: $c=2$ ).
Perform the synthetic division steps:
- Bring down the leading coefficient $4$ .
- Multiply $4$ by $2$ to get $8$ , and add it to $-7$ : $-7 + 8 = 1$
- Multiply $1$ by $2$ to get $2$ , and add it to $5$ : $5 + 2 = 7$
- Multiply $7$ by $2$ to get $14$ , and add it to $-6$ : $-6 + 14 = 8$
The complete synthetic division table is:
- Root: $2$ .
- Coefficients: $4,\,-7,\,5,\,-6$ .
- Products, aligned below all but the leading coefficient: $8,\,2,\,14$ .
- Result row: $4,\,1,\,7,\,8$ (quotient coefficients followed by the remainder).
Interpret the results:
- The numbers before the vertical remainder line are the coefficients of the quotient polynomial $q(x)$ , starting one degree lower than the dividend ( $x^2$ ): $q(x) = 4x^2 + 1x + 7 = 4x^2 + x + 7$
- The final number is the remainder: $r = 8$
Write the final expression: Combine the quotient and the remainder over the divisor: $4x^2 + x + 7 + \frac{8}{x - 2}$

Common Pitfall: When using synthetic division with a divisor of $x - c$ , always use the root $+c$ (here, $x - 2 = 0 \implies x = 2$ ), not the constant term $-2$ . If using polynomial long division instead, be cautious with signs when subtracting each term.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student uses synthetic division with -2 instead of +2 because they take the divisor's sign without solving x - 2 = 0.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
Synthetic division with $c = -2$ , Coefficients: $4, -7, 5, -6$
Correct
Synthetic division with $c = 2$ , Coefficients: $4, -7, 5, -6$
When dividing by a linear factor of the form $x - c$ , the value used in synthetic division is the root of the divisor, found by setting $x - 2 = 0$ , which gives $c = 2$ .
Step 2
Bring down $4 \implies -7 + (-2)(4) = -15 \implies 5 + (-2)(-15) = 35 \implies -6 + (-2)(35) = -76$
Correct
Bring down $4 \implies -7 + (2)(4) = 1 \implies 5 + (2)(1) = 7 \implies -6 + (2)(7) = 8$
Your multiply-and-add arithmetic steps were executed logically, but the resulting values were altered because they used $c = -2$ instead of $c = 2$ .
Step 3
$4x^2 - 15x + 35 - \frac{76}{x - 2}$
Correct
$4x^2 + x + 7 + \frac{8}{x - 2}$
You assembled the polynomial quotient and remainder expression correctly from your table values, but the coefficients reflect the earlier sign error.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly listed the dividend polynomial coefficients in descending order.

You followed the synthetic division algorithm steps of bringing down the leading coefficient, multiplying, and adding.

You correctly structured the final answer as a quotient polynomial plus the remainder over the divisor.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Synthetic Division Divisor Sign Error

You copied the constant $-2$ directly from the divisor $x - 2$ instead of solving for the zero of the binomial.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the constant term in the factor $(x - c)$ with the root $c$ that makes the factor equal to zero.

REASONING PATH

Your ACEpath teacher’s suggested approach

Finding the Synthetic Division Box Value: For a linear divisor $(x - c)$ , the value placed in the synthetic division box is $x = c$ , which satisfies $x - c = 0$ .
Always set the divisor equal to $0$ and solve for $x$ before writing the number in the synthetic division box.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set the divisor to zero: $x - 2 = 0 \implies c = 2$ .
List coefficients: $4, -7, 5, -6$ .
Bring down $4$ .
Multiply and add: $-7 + (2)(4) = 1$ .
Multiply and add: $5 + (2)(1) = 7$ .
Multiply and add: $-6 + (2)(7) = 8$ .

Final answer

$4x^2 + x + 7 + \frac{8}{x - 2}$

NEXT STEPS

Final Takeaways from Your Teacher

Before beginning synthetic division, write down the small equation $\text{divisor} = 0$ to verify the sign of $c$ .
Practice identifying the value of $c$ for both forms: $x - c$ gives $+c$ , and $x + c$ gives $-c$ .

OverviewACEpath Teacher's Diagnosis

02 During polynomial long division, the student adds the multiplied terms instead of subtracting them from each partial dividend.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
First term: $4x^2$
Step 2
Add: $(-7x^2) + (-8x^2) = -15x^2$
Correct
-7x^2 - (-8x^2) = x^2
In polynomial division, you must subtract the product $4x^2(x - 2) = 4x^3 - 8x^2$ from the dividend. Subtracting $-8x^2$ gives $-7x^2 - (-8x^2) = x^2$ , but you added them to get $-15x^2$ .
Step 3
Next term: $-15x \implies \text{Add: } (5x) + (30x) = 35x$
Correct
$\frac{x^2}{x}$ = x $\implies$ 5x - (-2x) = 7x
Because the previous coefficient was incorrect, this quotient term and partial dividend calculation carried forward the earlier sign error.
Step 4
Next term: $35 \implies \text{Add: } (-6) + (-70) = -76$
Correct
$\frac{7x}{x}$ = 7 $\implies$ -6 - (-14) = 8
This step correctly followed your method, but the values were affected by the earlier addition error.
Step 5
4x^2 - 15x + 35 - $\frac{76}{x - 2}$
Correct
4x^2 + x + 7 + $\frac{8}{x - 2}$
The final expression combined the affected quotient terms and remainder.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified that the first term of the quotient is found by dividing $4x^3$ by $x$ .

You maintained a consistent step-by-step division structure across all terms.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Adding instead of subtracting in polynomial division

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You confused the polynomial long division algorithm (which requires subtracting the product from the dividend) with synthetic division (which uses the opposite sign of the divisor's constant to allow addition). Because you kept the original factor $(x - 2)$ without switching the sign of the root, adding the product doubled the negative term instead of canceling it.

REASONING PATH

Your ACEpath teacher’s suggested approach

Subtractions in Division: In standard long division, always subtract the product: $\text{Dividend term} - (\text{Quotient term} \times \text{Divisor term})$ .
Wrap the term being subtracted in parentheses with a minus sign in front, e.g., $-7x^2 - (-8x^2) = -7x^2 + 8x^2 = x^2$ .
Synthetic vs. Long Division: If you prefer to add at each step, you must use synthetic division with the root $c = 2$ , not the divisor's constant $-2$ .
Check whether you are using the root $x = c$ (add down) or the binomial $x - c$ (subtract down).

CORRECTED PATH

ACEpath Teacher's Demonstration

Multiply $4x^2$ by $(x - 2)$ to get $4x^3 - 8x^2$ . Subtracting gives $(-7x^2) - (-8x^2) = x^2$ .
Divide $x^2$ by $x$ to get the next quotient term, $+x$ . Multiply by $(x - 2)$ to get $x^2 - 2x$ . Subtracting gives $5x - (-2x) = 7x$ .
Divide $7x$ by $x$ to get $+7$ . Multiply by $(x - 2)$ to get $7x - 14$ . Subtracting gives $-6 - (-14) = 8$ .

Final answer

4x^2 + x + 7 + $\frac{8}{x - 2}$

NEXT STEPS

Final Takeaways from Your Teacher

Practice distributing the negative sign explicitly when subtracting binomials in long division.
Try using synthetic division with $c = 2$ on this same problem to see why adding works only when the sign of the root is flipped.

OverviewACEpath Teacher's Diagnosis

03 The student miscalculates the first subtraction in long division as -7 - (-8) = -15 due to an integer sign error.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$\frac{4x^3}{x} = 4x^2$
Step 2
(4x^3 - 7x^2) - (4x^3 - 8x^2) = -15x^2
Correct
(4x^3 - 7x^2) - (4x^3 - 8x^2) = x^2
Subtracting a negative quantity is equivalent to adding: $-7x^2 - (-8x^2) = -7x^2 + 8x^2 = x^2$ . You mistakenly combined the coefficients as $-7 - 8 = -15$ .
Step 3
$\frac{-15x^2}{x}$ = -15x $\implies$ (-15x^2 + 5x) - (-15x^2 + 30x) = -25x
Correct
$\frac{x^2}{x}$ = x $\implies$ (x^2 + 5x) - (x^2 - 2x) = 7x
You followed the correct division and subtraction procedure here, but this step inherited the erroneous $-15x^2$ from the previous subtraction.
Step 4
$\frac{-25x}{x}$ = -25 $\implies$ (-25x - 6) - (-25x + 50) = -56
Correct
$\frac{7x}{x}$ = 7 $\implies$ (7x - 6) - (7x - 14) = 8
Your division and subtraction steps were executed logically, but the values were carried over from the earlier sign error.
Step 5
4x^2 - 15x - 25 - $\frac{56}{x - 2}$
Correct
4x^2 + x + 7 + $\frac{8}{x - 2}$
You correctly assembled the quotient and remainder into the requested form, but the expression reflects the earlier calculation error.

WHAT YOU DID WELL

Compliments from Your Teacher

You clearly set up the iterative cycle of polynomial long division.

You correctly identified the first term of the quotient.

You properly formatted your final expression as a polynomial plus the remainder over the divisor.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Integer subtraction with negative signs

You treated $-7 - (-8)$ as $-7 - 8$ , writing $-15$ instead of $+1$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Dropping or conflating the double negative when subtracting a negative term during polynomial subtraction.

REASONING PATH

Your ACEpath teacher’s suggested approach

Distribute the negative sign explicitly: Subtracting $(A - B)$ equals $-A + B$ .
Always write out $+ (-4x^3 + 8x^2)$ rather than trying to perform double-negation subtraction mentally.

CORRECTED PATH

ACEpath Teacher's Demonstration

Subtract $(4x^3 - 8x^2)$ from $(4x^3 - 7x^2)$ to obtain $x^2$ .
Bring down $+5x$ and divide $x^2$ by $x$ to get $+x$ .
Multiply $x(x - 2) = x^2 - 2x$ and subtract: $(x^2 + 5x) - (x^2 - 2x) = 7x$ .
Bring down $-6$ and divide $7x$ by $x$ to get $+7$ .
Multiply $7(x - 2) = 7x - 14$ and subtract: $(7x - 6) - (7x - 14) = 8$ .

Final answer

4x^2 + x + 7 + $\frac{8}{x - 2}$

NEXT STEPS

Final Takeaways from Your Teacher

Practice subtracting negative binomials by explicitly distributing the minus sign to avoid sign errors.
Consider using synthetic division when dividing by a linear factor of the form $x - c$ to streamline arithmetic.

Lock it in

Question

When using synthetic division to divide a polynomial by $(x - 5)$ , what number goes into the synthetic division box?

Answer

$+5$ . Set the divisor equal to zero: $x - 5 = 0 \implies x = 5$ .

Resources

Answer and steps

Final answer

Steps

Method: Synthetic Division

Typical errors

01 The student uses synthetic division with -2 instead of +2 because they take the divisor's sign without solving x - 2 = 0.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 During polynomial long division, the student adds the multiplied terms instead of subtracting them from each partial dividend.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student miscalculates the first subtraction in long division as -7 - (-8) = -15 due to an integer sign error.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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