Math
# Factoring Study Guide: What It Means, Why It Works, and How to Do It
This version explains the method, why it works, and what happens to the numbers and variables at each step—not just which formula to use.
Your difficulty scale stays the same:
- Level 1: Easier than your review sheet.
- Level 2: Equal to your review sheet.
- Level 3: WAY harder—hidden patterns, several methods combined, and unfamiliar reasoning.
The guide has four main parts:
1. Binomials
2. Trinomials
3. Grouping and repeated expressions
4. Applications
Each part contains methods → mental clues → worked examples.
---
# Part 0: What Is Factoring Actually Doing?
## 0A. What factoring means
Factoring rewrites addition or subtraction as multiplication without changing the value of the expression.
For example:
\[
6x+12=6(x+2).
\]
The left side is two terms being added. The right side is two factors being multiplied.
### Why are they equal?
Because multiplication distributes:
\[
6(x+2)=6\cdot x+6\cdot2=6x+12.
\]
Factoring runs the distributive property backward:
\[
\underbrace{6x+12}_{\text{expanded form}}
\quad\longrightarrow\quad
\underbrace{6(x+2)}_{\text{factored form}}.
\]
The expression’s appearance changes. Its value does not.
For instance, when \(x=3\):
\[
6(3)+12=30,
\]
and:
\[
6(3+2)=30.
\]
That numerical check illustrates the equality. The distributive property proves it for every value of \(x\).
---
## 0B. What factoring is used for
### 1. Solving equations
Suppose:
\[
x^2-5x+6=0.
\]
Factoring gives:
\[
(x-2)(x-3)=0.
\]
A product is zero only if at least one factor is zero:
\[
x-2=0
\quad\text{or}\quad
x-3=0.
\]
Therefore:
\[
\boxed{x=2\text{ or }x=3}.
\]
Why does factoring help? It replaces one quadratic equation with two simple linear equations.
Important: Factoring an expression does not automatically mean you should solve it. You set factors equal to zero only when you have an equation whose product equals zero.
---
### 2. Simplifying fractions with variables
\[
\frac{x^2-9}{x-3}
=
\frac{(x-3)(x+3)}{x-3}
=
x+3,
\qquad x\ne3.
\]
The cancellation works because you are dividing out a multiplicative factor:
\[
\frac{x-3}{x-3}=1.
\]
You cannot cancel pieces merely because they appear in sums.
The restriction \(x\ne3\) remains because the original denominator cannot be zero.
---
### 3. Finding possible dimensions
If a rectangle has area:
\[
x^2+7x+12=(x+3)(x+4),
\]
then \(x+3\) and \(x+4\) are possible side-length expressions because:
\[
\text{area}=\text{length}\times\text{width}.
\]
---
### 4. Revealing structure
Expanded form can hide useful information.
For example:
\[
x^2-2x+1=(x-1)^2.
\]
The factored form immediately shows that:
- the expression is a square;
- it is never negative for real \(x\);
- it equals zero when \(x=1\).
---
## 0C. How you prove a factorization is correct
Multiply the proposed factors. If you recover the original expression exactly, the factorization is correct.
Example:
\[
(x-4)(x+3)
\]
expands to:
\[
x^2+3x-4x-12=x^2-x-12.
\]
This proves:
\[
\boxed{x^2-x-12=(x-4)(x+3)}.
\]
Checking one numerical value can catch some mistakes, but it does not prove two expressions are always equal. Expanding and matching every term does.
---
## 0D. What “factor completely” means
Keep factoring until the remaining polynomial factors cannot be factored further in the allowed number system.
In this guide, we use integer or rational coefficients, unless stated otherwise.
For example:
\[
x^4-16=(x^2-4)(x^2+4)
\]
is not complete because:
\[
x^2-4=(x-2)(x+2).
\]
The complete answer is:
\[
\boxed{(x-2)(x+2)(x^2+4)}.
\]
---
# Part 1: Binomials
A binomial has two terms.
Examples:
\[
9x^2-25,\qquad x^3+8,\qquad 5x^7y^2-45x^3y^4.
\]
## 1A. Methods and Why They Work
### Method 1: Greatest common factor—GCF
If every term contains the same factor, move that factor outside parentheses.
\[
AB+AC=A(B+C).
\]
### What happens to the numbers?
You divide each original term by the factor you remove to find what stays inside.
Example:
\[
12x^3+18x^2.
\]
Remove \(6x^2\):
\[
\frac{12x^3}{6x^2}=2x,
\qquad
\frac{18x^2}{6x^2}=3.
\]
Therefore:
\[
12x^3+18x^2=6x^2(2x+3).
\]
Multiplying by \(6x^2\) puts back exactly what you removed.
---
### Method 2: Difference of squares
\[
\boxed{A^2-B^2=(A-B)(A+B)}.
\]
### Why does it work?
Expand:
\[
(A-B)(A+B)
=
A^2+AB-AB-B^2.
\]
The middle terms cancel:
\[
AB-AB=0.
\]
That leaves:
\[
A^2-B^2.
\]
The opposite signs are what cause the cancellation.
---
### Method 3: Sum and difference of cubes
\[
\boxed{A^3-B^3=(A-B)(A^2+AB+B^2)}
\]
\[
\boxed{A^3+B^3=(A+B)(A^2-AB+B^2)}.
\]
### Why does the difference-of-cubes formula work?
Distribute:
\[
(A-B)(A^2+AB+B^2)
\]
\[
=A^3+A^2B+AB^2-A^2B-AB^2-B^3.
\]
The mixed terms cancel in pairs, leaving:
\[
A^3-B^3.
\]
The sum-of-cubes formula works through the same kind of cancellation.
Sign shortcut: SOAP
- Same sign in the binomial.
- Opposite sign in the trinomial’s middle.
- Always Positive at the end.
---
## 1B. What to Look For in Your Head
- GCF first: a square or cube pattern may be hidden behind it.
- Square: numerical coefficient is a square and variable exponents are even.
- Cube: numerical coefficient is a cube and variable exponents are multiples of \(3\).
- Negative leading term: removing a negative GCF may make the next step easier.
- Large powers: expect to factor more than once.
Examples:
\[
49x^8=(7x^4)^2
\]
because squaring \(7x^4\) gives:
\[
7^2x^{4\cdot2}=49x^8.
\]
Similarly:
\[
216x^9=(6x^3)^3
\]
because:
\[
6^3x^{3\cdot3}=216x^9.
\]
---
## 1C. Worked Examples
### Level 1: Factor \(12x^2-27\)
#### Step 1 — Method: GCF
The greatest common numerical factor of \(12\) and \(27\) is \(3\).
Divide both terms by \(3\):
\[
12x^2\div3=4x^2,
\qquad
-27\div3=-9.
\]
So:
\[
12x^2-27=3(4x^2-9).
\]
#### Step 2 — Method: Difference of squares
Inside the parentheses:
\[
4x^2=(2x)^2,\qquad9=3^2.
\]
The expressions being squared are \(2x\) and \(3\). Put them into one subtraction factor and one addition factor:
\[
4x^2-9=(2x-3)(2x+3).
\]
#### Answer
\[
\boxed{3(2x-3)(2x+3)}.
\]
#### Why the answer works
\[
(2x-3)(2x+3)=4x^2+6x-6x-9=4x^2-9.
\]
Multiplying by \(3\) gives the original:
\[
12x^2-27.
\]
---
### Level 2: Factor \(5x^7y^2-45x^3y^4\)
#### Step 1 — Method: GCF
The shared factors are:
- \(5\), from the coefficients;
- \(x^3\), the smaller shared power of \(x\);
- \(y^2\), the smaller shared power of \(y\).
So the GCF is:
\[
5x^3y^2.
\]
#### Step 2 — Divide each term by the GCF
For the first term:
\[
\frac{5x^7y^2}{5x^3y^2}
=
x^{7-3}y^{2-2}
=
x^4.
\]
For the second term:
\[
\frac{-45x^3y^4}{5x^3y^2}
=
-9x^{3-3}y^{4-2}
=
-9y^2.
\]
Dividing powers with the same base subtracts the exponents.
Therefore:
\[
5x^7y^2-45x^3y^4
=
5x^3y^2(x^4-9y^2).
\]
#### Step 3 — Method: Difference of squares
\[
x^4=(x^2)^2,
\qquad
9y^2=(3y)^2.
\]
Thus:
\[
x^4-9y^2=(x^2-3y)(x^2+3y).
\]
#### Answer
\[
\boxed{5x^3y^2(x^2-3y)(x^2+3y)}.
\]
#### Why the answer works
The last two factors multiply to:
\[
x^4+3x^2y-3x^2y-9y^2=x^4-9y^2.
\]
Multiplying by the GCF restores both original terms.
---
### Level 3: Factor \(x^{12}-y^{12}\) completely
Challenge: The same expression fits several patterns. You must keep checking every factor.
#### Step 1 — Method: Difference of squares
Since:
\[
x^{12}=(x^6)^2,\qquad y^{12}=(y^6)^2,
\]
we get:
\[
x^{12}-y^{12}=(x^6-y^6)(x^6+y^6).
\]
The square roots here are \(x^6\) and \(y^6\), not \(x^3\) and \(y^3\).
#### Step 2 — Method: Difference of squares again
For the first factor:
\[
x^6-y^6=(x^3)^2-(y^3)^2.
\]
Therefore:
\[
x^6-y^6=(x^3-y^3)(x^3+y^3).
\]
#### Step 3 — Method: Cube formulas
\[
x^3-y^3=(x-y)(x^2+xy+y^2),
\]
\[
x^3+y^3=(x+y)(x^2-xy+y^2).
\]
The cube formulas remove the mixed terms through cancellation when multiplied.
#### Step 4 — Return to the factor \(x^6+y^6\)
This is a sum of cubes because:
\[
x^6=(x^2)^3,\qquad y^6=(y^2)^3.
\]
Use \(A=x^2\) and \(B=y^2\):
\[
x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4).
\]
Notice how the formula’s pieces are calculated:
\[
A^2=x^4,\qquad AB=x^2y^2,\qquad B^2=y^4.
\]
#### Answer
\[
\boxed{
\begin{aligned}
&(x-y)(x+y)(x^2+xy+y^2)\\
&\quad\cdot(x^2-xy+y^2)(x^2+y^2)(x^4-x^2y^2+y^4).
\end{aligned}
}
\]
#### Why the whole answer works
Reverse the steps:
1. The cube factors reconstruct \(x^3-y^3\) and \(x^3+y^3\).
2. Their product reconstructs \(x^6-y^6\).
3. The other two factors reconstruct \(x^6+y^6\).
4. Those multiply to \(x^{12}-y^{12}\).
---
# Part 2: Trinomials
A trinomial has three terms.
Examples:
\[
x^2+2x-15,
\qquad
7x^2-11x-6,
\qquad
9x^4-145x^2+16.
\]
## 2A. Methods and Why They Work
### Method 1: Sum-and-product factoring
For:
\[
x^2+bx+c,
\]
look for:
\[
(x+m)(x+n).
\]
### Why do the numbers need the correct sum and product?
Expand:
\[
(x+m)(x+n)=x^2+nx+mx+mn.
\]
Combine the middle terms:
\[
=x^2+(m+n)x+mn.
\]
Therefore:
\[
m+n=b,\qquad mn=c.
\]
The sum controls the middle coefficient. The product controls the constant.
---
### Method 2: The \(ac\) method
For:
\[
ax^2+bx+c,
\]
find two numbers with:
\[
\text{product}=ac,\qquad \text{sum}=b.
\]
Use them to split the middle term, then group.
### Why does splitting the middle term work?
If \(m+n=b\), then:
\[
mx+nx=(m+n)x=bx.
\]
So replacing \(bx\) with \(mx+nx\) does not change the expression.
The product condition \(mn=ac\) is what makes the split compatible with the cross-products of a factorization.
For example, if:
\[
(px+q)(rx+s),
\]
then the two middle terms are \(psx\) and \(qrx\). Their coefficients multiply to:
\[
(ps)(qr)=(pr)(qs)=ac.
\]
---
### Method 3: Substitution
If an expression looks quadratic in a repeated quantity, rename that quantity.
\[
x^4-5x^2+4
\]
becomes:
\[
u^2-5u+4
\]
when \(u=x^2\).
Substitution changes the notation, not the value. You must replace \(u\) with \(x^2\) again afterward.
---
## 2B. What to Look For in Your Head
- Positive constant, negative middle: both numbers are negative.
- Negative constant: the numbers have opposite signs.
- Fractional coefficients: factor out a fraction to leave integers.
- Exponents like \(8,4,0\): use \(u=x^4\).
- Repeated parentheses: treat them as one object instead of expanding.
---
## 2C. Worked Examples
### Level 1: Factor \(x^2+2x-15\)
#### Step 1 — Method: Sum and product
We need two numbers whose:
\[
\text{sum}=2,\qquad\text{product}=-15.
\]
Since the product is negative, one number must be positive and the other negative.
#### Step 2 — Choose \(5\) and \(-3\)
\[
5+(-3)=2,
\]
so their cross-products will produce \(2x\).
Also:
\[
5(-3)=-15,
\]
so their constant product will produce \(-15\).
#### Answer
\[
\boxed{(x+5)(x-3)}.
\]
#### Proof by multiplication
\[
(x+5)(x-3)=x^2-3x+5x-15=x^2+2x-15.
\]
---
### Level 2A: Factor \(\frac{14}{3}x^2-\frac{22}{3}x-4\)
#### Step 1 — Method: Remove a common fractional factor
Take out \(\frac23\).
Dividing by \(\frac23\) means multiplying by its reciprocal, \(\frac32\):
\[
\frac{14}{3}\div\frac23
=
\frac{14}{3}\cdot\frac32=7,
\]
\[
-\frac{22}{3}\div\frac23=-11,
\]
\[
-4\div\frac23=-6.
\]
Therefore:
\[
\frac23(7x^2-11x-6).
\]
#### Step 2 — Method: \(ac\)
\[
ac=7(-6)=-42.
\]
We need two numbers with product \(-42\) and sum \(-11\):
\[
-14,\quad3.
\]
#### Step 3 — Split the middle term
\[
-11x=-14x+3x.
\]
So:
\[
\frac23(7x^2-14x+3x-6).
\]
Nothing has changed: the two replacement terms still add to \(-11x\).
#### Step 4 — Method: Grouping
\[
\frac23\left[(7x^2-14x)+(3x-6)\right].
\]
The first group has GCF \(7x\):
\[
7x^2\div7x=x,\qquad -14x\div7x=-2.
\]
The second group has GCF \(3\):
\[
3x\div3=x,\qquad -6\div3=-2.
\]
Thus:
\[
\frac23\left[7x(x-2)+3(x-2)\right].
\]
#### Step 5 — Factor the repeated expression
Both terms contain \(x-2\), so:
\[
7x(x-2)+3(x-2)=(x-2)(7x+3).
\]
#### Answer
\[
\boxed{\frac23(x-2)(7x+3)}.
\]
---
### Level 2B: Factor \(-18x^5+290x^3-32x\)
#### Step 1 — Method: Negative GCF
Remove \(-2x\):
\[
-18x^5\div(-2x)=9x^4,
\]
\[
290x^3\div(-2x)=-145x^2,
\]
\[
-32x\div(-2x)=16.
\]
Therefore:
\[
-2x(9x^4-145x^2+16).
\]
#### Step 2 — Method: Substitution
Let:
\[
u=x^2.
\]
Then \(x^4=u^2\), so the inside becomes:
\[
9u^2-145u+16.
\]
#### Step 3 — Method: \(ac\)
\[
ac=9(16)=144.
\]
The numbers \(-144\) and \(-1\) work:
\[
(-144)(-1)=144,
\qquad
-144-1=-145.
\]
#### Step 4 — Split and group
\[
9u^2-144u-u+16
\]
\[
=9u(u-16)-1(u-16).
\]
The negative \(1\) matters:
\[
-1(u-16)=-u+16.
\]
Factor the shared expression:
\[
(u-16)(9u-1).
\]
#### Step 5 — Undo the substitution
\[
-2x(x^2-16)(9x^2-1).
\]
#### Step 6 — Method: Difference of squares twice
\[
x^2-16=(x-4)(x+4),
\]
\[
9x^2-1=(3x-1)(3x+1).
\]
#### Answer
\[
\boxed{-2x(x-4)(x+4)(3x-1)(3x+1)}.
\]
---
### Level 3: Factor \(\frac16[16(x-2)^{12}-97(x-2)^6+81]\)
#### Step 1 — Method: Hidden quadratic substitution
Let:
\[
u=(x-2)^6.
\]
Then:
\[
u^2=(x-2)^{12}.
\]
The expression becomes:
\[
\frac16(16u^2-97u+81).
\]
We keep \(\frac16\) outside because it multiplies the entire expression.
#### Step 2 — Method: \(ac\)
\[
16(81)=1296.
\]
We need a product of \(1296\) and a sum of \(-97\):
\[
-81,\quad-16.
\]
Indeed:
\[
(-81)(-16)=1296,\qquad -81-16=-97.
\]
#### Step 3 — Split and group
\[
16u^2-81u-16u+81
\]
\[
=u(16u-81)-1(16u-81).
\]
Therefore:
\[
\frac16(u-1)(16u-81).
\]
#### Step 4 — Use a simpler name for \(x-2\)
Let:
\[
t=x-2.
\]
Since \(u=t^6\), we now have:
\[
\frac16(t^6-1)(16t^6-81).
\]
#### Step 5 — Factor \(t^6-1\)
Use squares first:
\[
t^6-1=(t^3-1)(t^3+1).
\]
Then cubes:
\[
t^3-1=(t-1)(t^2+t+1),
\]
\[
t^3+1=(t+1)(t^2-t+1).
\]
Thus:
\[
t^6-1=(t-1)(t+1)(t^2+t+1)(t^2-t+1).
\]
#### Step 6 — Factor \(16t^6-81\)
The square roots are:
\[
4t^3,\qquad9,
\]
because:
\[
(4t^3)^2=16t^6,\qquad9^2=81.
\]
Therefore:
\[
16t^6-81=(4t^3-9)(4t^3+9).
\]
#### Step 7 — Replace \(t\) with \(x-2\)
\[
t-1=x-3,\qquad t+1=x-1.
\]
For the quadratic factors:
\[
t^2+t+1=(x-2)^2+(x-2)+1=x^2-3x+3,
\]
\[
t^2-t+1=(x-2)^2-(x-2)+1=x^2-5x+7.
\]
#### Answer
\[
\boxed{
\begin{aligned}
\frac16&(x-3)(x-1)(x^2-3x+3)(x^2-5x+7)\\
&\cdot[4(x-2)^3-9][4(x-2)^3+9].
\end{aligned}
}
\]
The remaining factors do not factor further over the rational numbers.
Big idea: The exponents were intimidating, but the first task was still ordinary trinomial factoring.
---
# Part 3: Grouping and Repeated Expressions
## 3A. Methods and Why They Work
### Method 1: Group four terms into pairs
Start with:
\[
AB+AC+DB+DC.
\]
Factor each pair:
\[
A(B+C)+D(B+C).
\]
Now \(B+C\) is a shared factor:
\[
\boxed{(A+D)(B+C)}.
\]
Grouping uses the distributive property backward twice.
### Method 2: Factor repeated parentheses directly
\[
4x^2(2x-3)+x(2x-3)-18(2x-3)
\]
already has a shared factor.
You can treat \(2x-3\) like one object. There is no need to expand it.
### Method 3: Rearrange or remove a negative
If your parentheses do not match:
- try a different pairing;
- check whether one pair is the negative of the other.
---
## 3B. What to Look For in Your Head
Your main question is:
> “What can I remove from each group so the leftovers match?”
For example:
\[
6x^2-9x-4x+6
\]
becomes:
\[
3x(2x-3)-2(2x-3).
\]
The outside factors are different. That is fine. The inside factors match.
---
## 3C. Worked Examples
### Level 1: Factor \(x^3+2x^2+5x+10\)
#### Step 1 — Method: Pair the terms
\[
(x^3+2x^2)+(5x+10).
\]
#### Step 2 — Remove each group’s GCF
From the first group, remove \(x^2\):
\[
x^3\div x^2=x,\qquad2x^2\div x^2=2.
\]
From the second group, remove \(5\):
\[
5x\div5=x,\qquad10\div5=2.
\]
So:
\[
x^2(x+2)+5(x+2).
\]
#### Step 3 — Remove the common binomial
Both terms contain \(x+2\). The remaining outside pieces are \(x^2\) and \(5\):
\[
\boxed{(x+2)(x^2+5)}.
\]
#### Why it works
Distributing \(x+2\) restores:
\[
x^2(x+2)+5(x+2),
\]
which expands to the original four terms.
---
### Level 2A: Factor \(x^3+3x^2-121x-363\)
#### Step 1 — Method: Grouping
\[
(x^3+3x^2)+(-121x-363).
\]
#### Step 2 — Factor each group
The first group gives:
\[
x^2(x+3).
\]
The second group needs a negative GCF:
\[
-121x\div(-121)=x,
\]
\[
-363\div(-121)=3.
\]
So:
\[
x^2(x+3)-121(x+3).
\]
#### Step 3 — Remove \(x+3\)
\[
(x+3)(x^2-121).
\]
The remaining pieces are \(x^2\) and \(-121\).
#### Step 4 — Method: Difference of squares
\[
121=11^2.
\]
Thus:
\[
x^2-121=(x-11)(x+11).
\]
#### Answer
\[
\boxed{(x+3)(x-11)(x+11)}.
\]
---
### Level 2B: Factor \(4x^2(2x-3)+x(2x-3)-18(2x-3)\)
#### Step 1 — Method: Common expression
Every term contains \(2x-3\).
Factoring it out leaves the multiplier from each term:
\[
(2x-3)(4x^2+x-18).
\]
This works exactly like:
\[
4x^2A+xA-18A=A(4x^2+x-18).
\]
#### Step 2 — Method: \(ac\)
For \(4x^2+x-18\):
\[
ac=4(-18)=-72.
\]
Choose \(9\) and \(-8\):
\[
9(-8)=-72,\qquad9-8=1.
\]
#### Step 3 — Split and group
\[
4x^2+9x-8x-18
\]
\[
=x(4x+9)-2(4x+9).
\]
The second group works because:
\[
-8x\div(-2)=4x,\qquad-18\div(-2)=9.
\]
Therefore:
\[
4x^2+x-18=(x-2)(4x+9).
\]
#### Answer
\[
\boxed{(2x-3)(x-2)(4x+9)}.
\]
---
### Level 3: Factor \(3x^5-6x^4y+12xy^4-24y^5\)
#### Step 1 — Method: Overall GCF
Remove \(3\):
\[
3(x^5-2x^4y+4xy^4-8y^5).
\]
Each coefficient is divided by \(3\); the variables remain unchanged.
#### Step 2 — Method: Grouping
\[
3[(x^5-2x^4y)+(4xy^4-8y^5)].
\]
From the first group, remove \(x^4\):
\[
x^4(x-2y).
\]
From the second group, remove \(4y^4\):
\[
4y^4(x-2y).
\]
Thus:
\[
3[x^4(x-2y)+4y^4(x-2y)].
\]
#### Step 3 — Remove the shared binomial
\[
3(x-2y)(x^4+4y^4).
\]
This looks finished—but it is not.
#### Step 4 — Method: Create a difference of squares
Add and subtract the same term:
\[
x^4+4y^4
=
x^4+4x^2y^2+4y^4-4x^2y^2.
\]
Why is this allowed?
Because:
\[
4x^2y^2-4x^2y^2=0.
\]
We changed the arrangement, not the value.
#### Step 5 — Recognize the square we created
\[
x^4+4x^2y^2+4y^4=(x^2+2y^2)^2.
\]
The middle term comes from:
\[
2(x^2)(2y^2)=4x^2y^2.
\]
Also:
\[
4x^2y^2=(2xy)^2.
\]
Therefore:
\[
x^4+4y^4=(x^2+2y^2)^2-(2xy)^2.
\]
#### Step 6 — Method: Difference of squares
\[
=(x^2+2y^2-2xy)(x^2+2y^2+2xy).
\]
#### Answer
\[
\boxed{
3(x-2y)(x^2-2xy+2y^2)(x^2+2xy+2y^2)
}.
\]
Big idea: Sometimes the pattern is not visible until you create it by adding zero in a useful form.
---
# Part 4: Applications
## 4A. Methods and Why They Work
### Method 1: Compare coefficients
Two polynomials are identical only when their corresponding coefficients match.
If:
\[
ax^2+bx+c=px^2+qx+r
\]
for every \(x\), then:
\[
a=p,\qquad b=q,\qquad c=r.
\]
This lets you find missing coefficients by expanding factors.
### Method 2: Work backward from a pattern
If you recognize a factor from a square or cube formula, reconstruct the original expression.
### Method 3: Factor before substituting
A complicated expression may simplify into exactly the quantities you are given.
### Method 4: Interpret factors as dimensions
For a rectangle:
\[
A=LW.
\]
For a rectangular prism:
\[
V=LWH.
\]
The factors must represent positive lengths in a physical situation.
---
## 4B. What to Look For in Your Head
- Known factor + missing coefficient: compare products.
- Given \(m+n\) and \(m-n\): look for \(m^2-n^2\).
- Volume: try to produce three factors.
- Area to perimeter: factor first, then add side lengths.
- Reverse cube question: identify \(A^2\), \(AB\), and \(B^2\).
- Large fraction: factor numerator and denominator before doing arithmetic.
---
## 4C. Worked Examples
### Level 1: Find a missing coefficient
If \(x+2\) is a factor of:
\[
x^2+kx+6,
\]
find \(k\).
#### Step 1 — Method: Determine the other factor
Write:
\[
(x+2)(x+b).
\]
The other leading term is \(x\) because:
\[
x\cdot x=x^2.
\]
#### Step 2 — Match the constant
The constant comes from:
\[
2b=6.
\]
Divide by \(2\):
\[
b=3.
\]
#### Step 3 — Find the middle coefficient
\[
(x+2)(x+3)=x^2+3x+2x+6.
\]
The two cross-products add:
\[
3x+2x=5x.
\]
Therefore:
\[
\boxed{k=5}.
\]
---
### Level 2A: Find \(k\) when \(5x+8\) is a factor
Given:
\[
15x^2+kx-32.
\]
#### Step 1 — Method: Match the leading product
Write:
\[
(5x+8)(3x+b).
\]
Why \(3x\)?
Because:
\[
5x\cdot3x=15x^2.
\]
#### Step 2 — Match the constant product
\[
8b=-32.
\]
Divide by \(8\):
\[
b=-4.
\]
#### Step 3 — Find the cross-products
\[
5x(-4)=-20x,
\]
\[
8(3x)=24x.
\]
Together:
\[
-20x+24x=4x.
\]
Therefore:
\[
\boxed{k=4}.
\]
The full expression is:
\[
(5x+8)(3x-4)=15x^2+4x-32.
\]
---
### Level 2B: Reconstruct the original cube expression
The standard trinomial factor is:
\[
49x^2+28x+16.
\]
Find the original sum or difference of cubes.
#### Step 1 — Method: Recognize the formula pieces
\[
49x^2=(7x)^2,\qquad16=4^2.
\]
So:
\[
A=7x,\qquad B=4.
\]
#### Step 2 — Check the middle product
\[
AB=(7x)(4)=28x.
\]
It matches the middle term exactly.
#### Step 3 — Use the sign to choose the formula
The trinomial has \(+AB\), so it belongs to:
\[
A^3-B^3=(A-B)(A^2+AB+B^2).
\]
#### Step 4 — Calculate the cubes
\[
A^3=(7x)^3=343x^3,
\]
\[
B^3=4^3=64.
\]
#### Answer
\[
\boxed{343x^3-64}.
\]
---
### Level 2C: Find possible tank dimensions
The volume is:
\[
27x^4+54x^3-8x-16.
\]
#### Step 1 — Method: Grouping
\[
(27x^4+54x^3)+(-8x-16).
\]
Remove \(27x^3\) from the first group:
\[
27x^3(x+2).
\]
Remove \(-8\) from the second group:
\[
-8(x+2).
\]
So:
\[
27x^3(x+2)-8(x+2).
\]
#### Step 2 — Remove \(x+2\)
\[
(x+2)(27x^3-8).
\]
#### Step 3 — Method: Difference of cubes
\[
27x^3=(3x)^3,\qquad8=2^3.
\]
Thus:
\[
27x^3-8=(3x-2)(9x^2+6x+4).
\]
The trinomial comes from:
\[
(3x)^2=9x^2,\qquad(3x)(2)=6x,\qquad2^2=4.
\]
#### Step 4 — Interpret the factors
One possible set of dimensions is:
\[
\boxed{L=x+2,\quad W=3x-2,\quad H=9x^2+6x+4}.
\]
Their product equals the volume expression.
For these dimensions to be positive, use:
\[
\boxed{x>\frac23}.
\]
---
### Level 2D: Evaluate using a sum and difference
Given:
\[
m+n=4,\qquad m-n=-5,
\]
evaluate:
\[
\frac18m^2(m+n)-\frac18n^2(m+n).
\]
#### Step 1 — Method: Common-expression factoring
Both terms contain:
\[
\frac18(m+n).
\]
Removing that factor leaves \(m^2\) and \(-n^2\):
\[
\frac18(m+n)(m^2-n^2).
\]
#### Step 2 — Method: Difference of squares
\[
m^2-n^2=(m+n)(m-n).
\]
Therefore:
\[
\frac18(m+n)^2(m-n).
\]
The square appears because there are now two copies of \(m+n\).
#### Step 3 — Substitute
\[
\frac18(4)^2(-5).
\]
Calculate in order:
\[
4^2=16,\qquad16(-5)=-80,\qquad-80\div8=-10.
\]
#### Answer
\[
\boxed{-10}.
\]
---
### Level 2E: Area to perimeter—and the answer-key correction
Using the intended factor dimensions, the rectangle’s area is:
\[
5x^2-29x-88.
\]
#### Step 1 — Method: \(ac\)
\[
ac=5(-88)=-440.
\]
Choose \(-40\) and \(11\):
\[
(-40)(11)=-440,\qquad-40+11=-29.
\]
#### Step 2 — Split and group
\[
5x^2-40x+11x-88
\]
\[
=5x(x-8)+11(x-8).
\]
Notice:
\[
-88\div11=-8.
\]
That is why the second outside factor is positive \(11\).
#### Step 3 — Factor
\[
(x-8)(5x+11).
\]
Possible dimensions are:
\[
L=x-8,\qquad W=5x+11.
\]
#### Step 4 — Method: Perimeter formula
A rectangle has two copies of each side:
\[
P=2L+2W.
\]
So:
\[
P=2(x-8)+2(5x+11).
\]
Distribute each \(2\):
\[
P=2x-16+10x+22.
\]
Combine like terms:
\[
\boxed{P=12x+6}.
\]
Your key’s \(12x-38\) results from using \(5x-11\), which is not the correct factor.
---
### Level 3A: Two unknown coefficients and two known factors
Both \(2x-1\) and \(3x+2\) are factors of:
\[
P(x)=12x^3+kx^2+\ell x-4.
\]
Find \(k,\ell\), and factor completely.
#### Step 1 — Method: Multiply the known factors
\[
(2x-1)(3x+2)
\]
produces:
\[
6x^2+4x-3x-2=6x^2+x-2.
\]
#### Step 2 — Determine the remaining factor’s degree
The original polynomial has degree \(3\). The known product has degree \(2\).
Degrees add when nonzero polynomials multiply, so the remaining factor must have degree \(1\):
\[
ax+b.
\]
Thus:
\[
P(x)=(6x^2+x-2)(ax+b).
\]
#### Step 3 — Method: Compare leading coefficients
The only product producing \(x^3\) is:
\[
6x^2(ax)=6ax^3.
\]
Match it to \(12x^3\):
\[
6a=12\implies a=2.
\]
#### Step 4 — Compare constants
The constant product is:
\[
(-2)b.
\]
Match it to \(-4\):
\[
-2b=-4\implies b=2.
\]
#### Step 5 — Expand to identify \(k\) and \(\ell\)
\[
(6x^2+x-2)(2x+2)
\]
\[
=12x^3+12x^2+2x^2+2x-4x-4.
\]
Combine:
\[
12x^3+14x^2-2x-4.
\]
Therefore:
\[
\boxed{k=14,\qquad\ell=-2}.
\]
#### Step 6 — Factor completely
\[
2x+2=2(x+1).
\]
So:
\[
\boxed{P(x)=2(2x-1)(3x+2)(x+1)}.
\]
---
### Level 3B: High-power evaluation with hidden cancellation
Given:
\[
m+n=5,\qquad m-n=3,
\]
evaluate:
\[
E=\frac{m^{12}-n^{12}}{m^4+m^2n^2+n^4}.
\]
#### Step 1 — Method: Factor to match the denominator
First use difference of squares:
\[
m^{12}-n^{12}=(m^6-n^6)(m^6+n^6).
\]
Then factor:
\[
m^6-n^6=(m^2)^3-(n^2)^3.
\]
Using difference of cubes:
\[
m^6-n^6=(m^2-n^2)(m^4+m^2n^2+n^4).
\]
The denominator now appears as a factor of the numerator.
#### Step 2 — Cancel the common factor
\[
E=
\frac{(m^2-n^2)(m^4+m^2n^2+n^4)(m^6+n^6)}
{m^4+m^2n^2+n^4}.
\]
The given conditions imply \(m=4,n=1\), so the denominator is nonzero.
A nonzero expression divided by itself equals \(1\), leaving:
\[
E=(m^2-n^2)(m^6+n^6).
\]
#### Step 3 — Find \(m^2-n^2\)
Use difference of squares:
\[
m^2-n^2=(m+n)(m-n)=5(3)=15.
\]
#### Step 4 — Find \(m^2+n^2\)
Expand the two squares:
\[
(m+n)^2=m^2+2mn+n^2,
\]
\[
(m-n)^2=m^2-2mn+n^2.
\]
Adding cancels the \(mn\) terms:
\[
(m+n)^2+(m-n)^2=2(m^2+n^2).
\]
Therefore:
\[
m^2+n^2=\frac{5^2+3^2}{2}=\frac{34}{2}=17.
\]
#### Step 5 — Find \(m^2n^2\)
Subtract the same expanded squares:
\[
(m+n)^2-(m-n)^2=4mn.
\]
Thus:
\[
mn=\frac{25-9}{4}=4,
\]
so:
\[
m^2n^2=(mn)^2=16.
\]
#### Step 6 — Find \(m^6+n^6\) without taking separate sixth powers
The identity:
\[
A^3+B^3=(A+B)^3-3AB(A+B)
\]
works because expanding \((A+B)^3\) creates the extra mixed terms:
\[
3A^2B+3AB^2=3AB(A+B).
\]
Subtracting them leaves \(A^3+B^3\).
Use \(A=m^2\), \(B=n^2\):
\[
m^6+n^6=(m^2+n^2)^3-3m^2n^2(m^2+n^2).
\]
Substitute:
\[
17^3-3(16)(17)
=
4913-816
=
4097.
\]
#### Step 7 — Finish
\[
E=15(4097).
\]
#### Answer
\[
\boxed{61,\!455}.
\]
Big idea: Factoring reduced a fraction with twelfth powers to quantities controlled by the given sum and difference.
---
# Final Summary: What Every Step Should Mean to You
| Step you perform | What is actually happening |
|---|---|
| Remove a GCF | Divide each term by the shared factor, then place that factor outside |
| Split a middle term | Replace one term with two terms whose sum is exactly the original |
| Group | Arrange terms so the distributive property can be reversed |
| Use difference of squares | Choose opposite signs so the cross-products cancel |
| Use a cube formula | Choose factors whose mixed terms cancel |
| Substitute | Rename a repeated expression without changing its value |
| Cancel factors | Divide numerator and denominator by the same nonzero multiplier |
| Compare coefficients | Match the products that generate each power of the variable |
| Check your answer | Multiply back to prove the original expression is recovered |
## Your five-question checklist
1. What factor or pattern is shared?
2. Which method am I using?
3. What does that method do to each number and variable?
4. Why does this step preserve the original expression?
5. Can I multiply backward and recover what I started with?
The goal is not just to recognize a formula. It is to understand why the factors multiply back into the original expression.