A rectangular solid has a square base with side length $s(x) = 2x - 3$ and height $h(x) = 3x + 1$. A cylindrical hole of radius $r(x) = x - 1$ is drilled completely through the center from the top face to the bottom face. Express the remaining volume $V(x)$ of the solid as a simplified polynomial in terms of $x$ and the constant $\pi$.

A rectangular solid has a square base with side length $s(x) = 2x - 3$ and height $h(x) = 3x + 1$ . A cylindrical hole of radius $r(x) = x - 1$ is drilled completely through the center from the top face to the bottom face. Express the remaining volume $V(x)$ of the solid as a simplified polynomial in terms of $x$ and the constant $\pi$ .

Final answer

$V(x) = (12 - 3\pi)x^3 - (32 - 5\pi)x^2 + (21 - \pi)x + (9 - \pi)$

Steps

Find the volume of the rectangular solid

To find the remaining volume $V(x)$ , subtract the volume of the cylindrical hole from the total volume of the rectangular solid. Both shapes share the same height, $h(x) = 3x + 1$ .

The solid has a square base with side length $s(x) = 2x - 3$ and height $h(x) = 3x + 1$ . Its volume is: $V_{\text{solid}} = [s(x)]^2 \cdot h(x) = (2x - 3)^2(3x + 1)$

First, expand the square of the binomial: $(2x - 3)^2 = 4x^2 - 12x + 9$

Next, multiply by the height $(3x + 1)$ : $V_{\text{solid}} = (4x^2 - 12x + 9)(3x + 1)$ Distributing and combining the terms according to degree: $V_{\text{solid}} = 12x^3 - 32x^2 + 21x + 9$

Find the volume of the cylindrical hole

The cylinder has radius $r(x) = x - 1$ and height $h(x) = 3x + 1$ . Its volume is: $V_{\text{hole}} = \pi [r(x)]^2 \cdot h(x) = \pi (x - 1)^2(3x + 1)$

First, expand the squared radius: $(x - 1)^2 = x^2 - 2x + 1$

Next, multiply by the height: $(x^2 - 2x + 1)(3x + 1) = 3x(x^2 - 2x + 1) + 1(x^2 - 2x + 1)$ $= (3x^3 - 6x^2 + 3x) + (x^2 - 2x + 1)$ $= 3x^3 - 5x^2 + x + 1$

Multiplying by $\pi$ : $V_{\text{hole}} = \pi(3x^3 - 5x^2 + x + 1)$

Subtract the hole's volume from the solid's volume

The remaining volume is: $V(x) = V_{\text{solid}} - V_{\text{hole}}$ $V(x) = (12x^3 - 32x^2 + 21x + 9) - \pi(3x^3 - 5x^2 + x + 1)$

Group like powers of $x$ :

For $x^3$ : $(12 - 3\pi)x^3$
For $x^2$ : $-32x^2 - (-5\pi x^2) = -(32 - 5\pi)x^2$
For $x$ : $21x - \pi x = (21 - \pi)x$
Constant term: $9 - \pi = (9 - \pi)$

Combining these gives the simplified polynomial: $V(x) = (12 - 3\pi)x^3 - (32 - 5\pi)x^2 + (21 - \pi)x + (9 - \pi)$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student uses diameter instead of radius when computing the volume of the cylinder, squaring 2(x - 1) or omitting the division, or using π·r·h instead of π·r²·h by forgetting to square the radius.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
$V_{\text{solid}} = (2x - 3)^2 (3x + 1) \implies (4x^2 - 12x + 9)(3x + 1) \implies 12x^3 + 4x^2 - 36x^2 - 12x + 27x + 9 \implies 12x^3 - 32x^2 + 15x + 9$
Step 2
$V_{\text{cylinder}} = \pi \cdot r(x) \cdot h(x) \implies \pi (x - 1)(3x + 1) \implies \pi (3x^2 - 2x - 1)$
Correct
$V_{\text{cylinder}} = \pi [r(x)]^2 h(x) = \pi (x - 1)^2(3x + 1) = \pi(3x^3 - 5x^2 + x + 1)$
The volume of a cylinder is base area times height, $V = \pi r^2 h$ . You omitted the exponent $2$ on the radius $r(x) = x - 1$ , multiplying only $\pi r h$ .
Step 3
V(x) = 12x^3 - 32x^2 + 15x + 9 - $\pi$ (3x^2 - 2x - 1)
Correct
$V(x) = V_{\text{solid}} - V_{\text{cylinder}} = (12 - 3\pi)x^3 - (32 - 5\pi)x^2 + (15 - \pi)x + (9 - \pi)$
Because the hole volume was missing the squared radius factor, your subtraction carried that error forward. The remaining terms should also be grouped by powers of $x$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly expanded the square of the binomial $(2x - 3)^2 = 4x^2 - 12x + 9$ .

You accurately distributed the trinomial with $(3x + 1)$ and combined like terms to find the solid's volume.

You correctly identified that the remaining volume is found by subtracting the cylinder's volume from the solid's volume.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Formula Error: Missing Exponent in Cylinder Volume

You wrote $\pi \cdot r(x) \cdot h(x)$ instead of $\pi [r(x)]^2 h(x)$ , forgetting to square the radius before multiplying by the height.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Conflating linear radius with circular cross-sectional area: a cylinder's volume depends on the area of its circular base, which requires $r^2$ , not $r$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Cylinder Volume Formula: The volume of any right cylinder is base area times height: $V = A_{\text{base}} \cdot h = (\pi r^2)h$ .
Always write down $\pi r^2 h$ with an explicit exponent $2$ before substituting algebraic expressions for $r$ .
Polynomial Simplification: When a problem asks for a simplified polynomial in $x$ and $\pi$ , group like powers of $x$ together: $(a + b\pi)x^k$ .
Collect all $x^3$ , $x^2$ , $x$ , and constant terms so that each power of $x$ appears exactly once.

CORRECTED PATH

ACEpath Teacher's Demonstration

Square the radius binomial: $[r(x)]^2 = (x - 1)^2 = x^2 - 2x + 1$ .
Multiply by the height: $(x^2 - 2x + 1)(3x + 1) = 3x^3 - 5x^2 + x + 1$ .
Multiply by $\pi$ : $V_{\text{hole}} = \pi(3x^3 - 5x^2 + x + 1)$ .
Subtract from $V_{\text{solid}}$ : $V(x) = (12x^3 - 32x^2 + 15x + 9) - \pi(3x^3 - 5x^2 + x + 1)$ .
Group coefficients by powers of $x$ : $(12 - 3\pi)x^3 - (32 - 5\pi)x^2 + (15 - \pi)x + (9 - \pi)$ .

Final answer

$V(x) = (12 - 3\pi)x^3 - (32 - 5\pi)x^2 + (15 - \pi)x + (9 - \pi)$

NEXT STEPS

Final Takeaways from Your Teacher

Memorize cylinder volume as $(\text{circle area}) \times \text{height} = \pi r^2 h$ to avoid dropping the square.
Practice squaring binomial radii like $(x - 1)^2$ before multiplying by polynomial heights.
Group like powers of $x$ into combined coefficients such as $(a - b\pi)x^n$ when writing final polynomial forms.

OverviewACEpath Teacher's Diagnosis

02 The student squares the binomial (2x - 3) as 4x² + 9 by forgetting the cross term -12x when expanding the base area of the solid.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
$V_{\text{solid}} = (2x - 3)^2(3x + 1) \implies (4x^2 + 9)(3x + 1) \implies 12x^3 + 4x^2 + 27x + 9$
Correct
$V_{\text{solid}} = (2x - 3)^2(3x + 1) = (4x^2 - 12x + 9)(3x + 1) = 12x^3 - 32x^2 + 21x + 9$
In your second line, you expanded $(2x - 3)^2$ as $4x^2 + 9$ . Remember that squaring a binomial follows $(a - b)^2 = a^2 - 2ab + b^2$ , which includes the middle cross term $-2(2x)(3) = -12x$ . As a result, the correct expansion is $4x^2 - 12x + 9$ , leading to $12x^3 - 32x^2 + 21x + 9$ after distributing $(3x + 1)$ .
Step 2
$V_{\text{cylinder}} = \pi(x - 1)^2(3x + 1) \implies \pi(x^2 - 2x + 1)(3x + 1) \implies \pi(3x^3 + x^2 - 6x^2 - 2x + 3x + 1) \implies \pi(3x^3 - 5x^2 + x + 1)$
Step 3
V(x) = 12x^3 + 4x^2 + 27x + 9 - $\pi$ (3x^3 - 5x^2 + x + 1)
Correct
V(x) = (12 - 3 $\pi$ )x^3 - (32 - 5 $\pi$ )x^2 + (21 - $\pi$ )x + (9 - $\pi$ )
Your subtraction setup is logically correct, but the expression carries over the earlier mistake from $V_{\text{solid}}$ . Furthermore, expressing the final answer as a simplified polynomial in terms of $x$ requires grouping matching powers of $x$ together with their $\pi$ coefficients.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified the geometric formulas for both the rectangular solid and the cylinder.

You accurately expanded $(x - 1)^2 = x^2 - 2x + 1$ and multiplied it by $(3x + 1)$ to find the volume of the cylindrical hole.

You correctly set up the subtraction $V(x) = V_{\text{solid}} - V_{\text{cylinder}}$ to determine the remaining volume.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Missing middle term in binomial square

You squared the individual terms in $(2x - 3)$ to get $4x^2 + 9$ , forgetting the cross term $-2ab = -12x$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating exponentiation as distributive over subtraction by writing $(a - b)^2 = a^2 - b^2$ or $a^2 + b^2$ , rather than multiplying the binomial by itself using FOIL or the identity $(a - b)^2 = a^2 - 2ab + b^2$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Binomial Square Identity: For any terms $a$ and $b$ , $(a \pm b)^2 = a^2 \pm 2ab + b^2$ .
Always write out both the squared terms and the doubled cross term, or write $(a \pm b)(a \pm b)$ and distribute fully.
Grouping Polynomial Coefficients: When polynomials involve a constant like $\pi$ , collect like powers of $x$ so that each power has a single combined coefficient.
Group terms by degree: $ax^n - \pi bx^n = (a - b\pi)x^n$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Expand the square of the base: $(2x - 3)^2 = 4x^2 - 12x + 9$ .
Multiply by the height: $(4x^2 - 12x + 9)(3x + 1) = 12x^3 - 32x^2 + 21x + 9$ .
Subtract $V_{\text{cylinder}}$ : $V(x) = (12x^3 - 32x^2 + 21x + 9) - \pi(3x^3 - 5x^2 + x + 1)$ .
Group coefficients of like powers of $x$ : $(12 - 3\pi)x^3 - (32 - 5\pi)x^2 + (21 - \pi)x + (9 - \pi)$ .

Final answer

V(x) = (12 - 3 $\pi$ )x^3 - (32 - 5 $\pi$ )x^2 + (21 - $\pi$ )x + (9 - $\pi$ )

NEXT STEPS

Final Takeaways from Your Teacher

Practice expanding binomial squares with coefficients greater than $1$ , such as $(3x - 5)^2$ , paying close attention to the $-2ab$ term.
Check your expansions by testing a simple number (for example, setting $x = 2$ ) in both the factored and expanded forms.
Group like powers of $x$ when subtracting algebraic expressions involving constants like $\pi$ to ensure the polynomial is fully simplified.

OverviewACEpath Teacher's Diagnosis

03 The student adds the cylinder volume to the solid volume instead of subtracting it when finding the remaining volume.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
$V_{\text{solid}} = (2x - 3)^2(3x + 1) \implies (4x^2 - 12x + 9)(3x + 1) \implies 12x^3 - 32x^2 + 15x + 9$
Step 2
$V_{\text{cylinder}} = \pi(x - 1)^2(3x + 1) \implies \pi(x^2 - 2x + 1)(3x + 1) \implies \pi(3x^3 - 5x^2 + x + 1)$
Step 3
$V(x) = V_{\text{solid}} + V_{\text{cylinder}} \implies 12x^3 - 32x^2 + 15x + 9 + \pi(3x^3 - 5x^2 + x + 1)$
Correct
$V(x) = V_{\text{solid}} - V_{\text{cylinder}} = (12 - 3\pi)x^3 - (32 - 5\pi)x^2 + (15 - \pi)x + (9 - \pi)$
Because the hole is removed from the solid, its volume must be subtracted from the total volume rather than added. Furthermore, the terms should be grouped by powers of $x$ to form a simplified polynomial.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly applied the formula for the volume of a rectangular prism with a square base.

You accurately expanded the binomial squares and multiplied the polynomials.

You correctly determined the volume formula for the cylindrical hole.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Operation Sign Error (Addition Instead of Subtraction)

You wrote an addition sign between the solid volume and the cylinder volume instead of subtracting the removed space.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing composite volume addition with remaining volume subtraction when material is removed/drilled out, along with stopping before combining like powers of $x$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Material Removal Operation: When a hole or void is drilled or cut out from a solid, the remaining volume is $V_{\text{remaining}} = V_{\text{total}} - V_{\text{removed}}$ .
Always subtract the cutout volume from the outer solid's volume.
Polynomial Grouping by Degree: A simplified polynomial in $x$ requires combining all like powers: $(a - b\pi)x^k$ .
Distribute the negative sign through the subtracted cylinder polynomial and group coefficients for each power of $x$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Set up the subtraction for remaining volume: $V(x) = V_{\text{solid}} - V_{\text{cylinder}}$ .
Substitute your expanded expressions: $V(x) = (12x^3 - 32x^2 + 15x + 9) - \pi(3x^3 - 5x^2 + x + 1)$ .
Distribute $\pi$ and group terms by degree of $x$ : $(12 - 3\pi)x^3 + (-32 + 5\pi)x^2 + (15 - \pi)x + (9 - \pi)$ .

Final answer

$V(x) = (12 - 3\pi)x^3 - (32 - 5\pi)x^2 + (15 - \pi)x + (9 - \pi)$

NEXT STEPS

Final Takeaways from Your Teacher

Read physical setup keywords carefully: words like 'drilled', 'hollowed', or 'cut out' signal subtraction of volumes.
Practice combining terms with constant coefficients like $\pi$ by factoring out $x^k$ for each degree.

Lock it in

Question

What is the formula for the volume of a right circular cylinder with radius $r$ and height $h$ ?

Answer

$V = \pi r^2 h$ . Memory cue: Volume requires three dimensions of length, so radius must be squared ( $r^2$ ) to make area before multiplying by height ( $h$ ).

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student uses diameter instead of radius when computing the volume of the cylinder, squaring 2(x - 1) or omitting the division, or using π·r·h instead of π·r²·h by forgetting to square the radius.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student squares the binomial (2x - 3) as 4x² + 9 by forgetting the cross term -12x when expanding the base area of the solid.

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 adds the cylinder volume to the solid volume instead of subtracting it when finding the remaining volume.

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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