The Central Idea
Factoring reverses multiplication.
If
\[ (x+2)(x+5)=x^2+7x+10, \]then reading the relationship backward gives
\[ \boxed{x^2+7x+10=(x+2)(x+5)}. \]Factoring a trinomial therefore means identifying which binomials could have produced it.
Always Check for a GCF First
Before using any trinomial or special-form method, inspect the entire polynomial for a greatest common factor.
Consider
\[ 6x^2+18x+12. \]The GCF is \(6\):
\[ 6(x^2+3x+2). \]Now factor the trinomial:
\[ x^2+3x+2=(x+1)(x+2). \]Therefore,
\[ \boxed{6x^2+18x+12=6(x+1)(x+2)}. \]Factoring out the GCF first makes the remaining structure easier to see.
Quadratic Trinomials
A quadratic trinomial has the form
\[ ax^2+bx+c, \]where \(a\neq0\).
When \(a=1\), the trinomial has the simpler form
\[ x^2+bx+c. \]We will begin with this monic case.
Factoring \(x^2+bx+c\)
Suppose
\[ x^2+bx+c=(x+m)(x+n). \]Expanding the right side gives
\[ x^2+(m+n)x+mn. \]Therefore, to factor
\[ x^2+bx+c, \]we need two numbers \(m\) and \(n\) such that
\[ \boxed{m+n=b} \]and
\[ \boxed{mn=c}. \]Example: Positive Constant Term
Factor
\[ x^2+7x+12. \]We need two numbers that multiply to \(12\) and add to \(7\).
The numbers are
\[ 3 \]and
\[ 4. \]Therefore,
\[ \boxed{x^2+7x+12=(x+3)(x+4)}. \]Sign Patterns for Monic Trinomials
The signs of \(b\) and \(c\) give useful information.
If
\[ x^2+bx+c \]has \(c>0\), then the two factor numbers have the same sign.
- If \(b>0\), both are positive.
- If \(b<0\), both are negative.
If \(c<0\), the two factor numbers have opposite signs.
Example: Both Signs Negative
Factor
\[ x^2-9x+20. \]We need two numbers whose product is \(20\) and whose sum is \(-9\).
The numbers are
\[ -4 \]and
\[ -5. \]Therefore,
\[ \boxed{x^2-9x+20=(x-4)(x-5)}. \]Example: Opposite Signs
Factor
\[ x^2+x-12. \]We need two numbers whose product is \(-12\) and whose sum is \(1\).
The numbers are
\[ 4 \]and
\[ -3. \]Therefore,
\[ \boxed{x^2+x-12=(x+4)(x-3)}. \]Use Factor Pairs Systematically
When the correct pair is not immediately obvious, list factor pairs.
To factor
\[ x^2+11x+24, \]consider the positive factor pairs of \(24\):
\[ 1\cdot24, \] \[ 2\cdot12, \] \[ 3\cdot8, \] \[ 4\cdot6. \]The pair whose sum is \(11\) is
\[ 3+8=11. \]Therefore,
\[ \boxed{x^2+11x+24=(x+3)(x+8)}. \]Prime Trinomials over the Integers
Not every quadratic trinomial factors into binomials with integer coefficients.
Consider
\[ x^2+x+1. \]We would need two integers whose product is \(1\) and whose sum is \(1\).
The possible integer pairs are
\[ 1,\ 1 \]and
\[ -1,\ -1. \]Their sums are \(2\) and \(-2\), not \(1\).
Thus this polynomial is not factorable into linear factors with integer coefficients.
At this stage we may call it prime over the integers.
When the Leading Coefficient Is Not 1
Now consider a trinomial of the form
\[ ax^2+bx+c \]with \(a\neq1\).
For example,
\[ 2x^2+7x+3. \]We cannot simply search for two numbers whose product is \(3\) and sum is \(7\), because the leading coefficient changes the multiplication structure.
The AC Method
A reliable method for factoring
\[ ax^2+bx+c \]is to multiply
\[ a\cdot c. \]Then find two numbers whose product is
\[ ac \]and whose sum is
\[ b. \]These numbers are used to split the middle term, after which factoring by grouping can be applied.
Example: AC Method
Factor
\[ 2x^2+7x+3. \]Here,
\[ a=2,\qquad b=7,\qquad c=3. \]Compute
\[ ac=2\cdot3=6. \]We need two numbers whose product is \(6\) and sum is \(7\):
\[ 6 \]and
\[ 1. \]Split the Middle Term
Rewrite
\[ 7x \]as
\[ 6x+x. \]Then
\[ 2x^2+7x+3 = 2x^2+6x+x+3. \]Group:
\[ (2x^2+6x)+(x+3). \]Factor each group:
\[ 2x(x+3)+1(x+3). \]Therefore,
\[ \boxed{2x^2+7x+3=(2x+1)(x+3)}. \]Example: Non-Monic Trinomial with a Negative Term
Factor
\[ 3x^2+x-2. \]Here,
\[ ac=3(-2)=-6. \]We need two numbers whose product is \(-6\) and sum is \(1\).
The numbers are
\[ 3 \]and
\[ -2. \]Split the middle term:
\[ 3x^2+3x-2x-2. \]Group:
\[ 3x(x+1)-2(x+1). \]Therefore,
\[ \boxed{3x^2+x-2=(3x-2)(x+1)}. \]Example: Larger Coefficients
Factor
\[ 6x^2+11x+3. \]Compute
\[ ac=6\cdot3=18. \]We need two numbers whose product is \(18\) and sum is \(11\).
The numbers are
\[ 9 \]and
\[ 2. \]Split:
\[ 6x^2+9x+2x+3. \]Group:
\[ 3x(2x+3)+1(2x+3). \]Therefore,
\[ \boxed{6x^2+11x+3=(3x+1)(2x+3)}. \]Example: Negative Constant Term
Factor
\[ 8x^2-2x-3. \]Compute
\[ ac=8(-3)=-24. \]We need two numbers whose product is \(-24\) and sum is \(-2\).
The numbers are
\[ 4 \]and
\[ -6. \]Split:
\[ 8x^2+4x-6x-3. \]Group:
\[ 4x(2x+1)-3(2x+1). \]Therefore,
\[ \boxed{8x^2-2x-3=(4x-3)(2x+1)}. \]Why the AC Method Works
Suppose
\[ ax^2+bx+c \]factors as
\[ (px+q)(rx+s). \]Expanding gives
\[ prx^2+(ps+qr)x+qs. \]Therefore,
\[ a=pr, \] \[ b=ps+qr, \]and
\[ c=qs. \]The two middle contributions \(ps\) and \(qr\) have product
\[ (ps)(qr)=prqs=ac. \]That is why the AC method searches for two numbers whose product is \(ac\) and whose sum is \(b\).
Factoring by Inspection
Once the multiplication structure becomes familiar, some trinomials can be factored directly.
For
\[ 2x^2+5x+2, \]we may recognize
\[ (2x+1)(x+2) \]because
\[ (2x)(x)=2x^2 \]and
\[ 1\cdot2=2, \]while the cross-products produce
\[ 4x+x=5x. \]Inspection can be faster, but the AC method remains a dependable fallback.
Perfect-Square Trinomials
Recall the special products
\[ (a+b)^2=a^2+2ab+b^2 \]and
\[ (a-b)^2=a^2-2ab+b^2. \]Reading these backward gives the factoring identities
\[ \boxed{a^2+2ab+b^2=(a+b)^2} \]and
\[ \boxed{a^2-2ab+b^2=(a-b)^2}. \]Recognizing a Perfect-Square Trinomial
A trinomial is a perfect square when:
- the first term is a square;
- the last term is a square;
- the middle term is twice the product of the square roots of those terms.
The sign of the middle term determines whether the binomial contains addition or subtraction.
Example: Perfect Square with Addition
Factor
\[ x^2+10x+25. \]The first term is
\[ x^2=(x)^2. \]The last term is
\[ 25=5^2. \]Twice their product is
\[ 2(x)(5)=10x. \]Therefore,
\[ \boxed{x^2+10x+25=(x+5)^2}. \]Example: Perfect Square with Subtraction
Factor
\[ 4x^2-12x+9. \]The first term is
\[ (2x)^2. \]The last term is
\[ 3^2. \]The middle term is
\[ -2(2x)(3)=-12x. \]Therefore,
\[ \boxed{4x^2-12x+9=(2x-3)^2}. \]Difference of Squares
Recall the conjugate identity
\[ (a+b)(a-b)=a^2-b^2. \]Reading it backward gives
\[ \boxed{a^2-b^2=(a+b)(a-b)}. \]This is one of the most useful factoring patterns in algebra.
Example: Difference of Squares
Factor
\[ x^2-49. \]Recognize
\[ x^2=(x)^2 \]and
\[ 49=7^2. \]Therefore,
\[ \boxed{x^2-49=(x+7)(x-7)}. \]Example: Difference of Squares with Coefficients
Factor
\[ 9x^2-25. \]Write
\[ 9x^2=(3x)^2 \]and
\[ 25=5^2. \]Therefore,
\[ \boxed{9x^2-25=(3x+5)(3x-5)}. \]Difference of Higher Powers That Are Squares
The square pattern can involve powers larger than \(2\).
Consider
\[ x^6-16. \]Since
\[ x^6=(x^3)^2 \]and
\[ 16=4^2, \]we have
\[ \boxed{x^6-16=(x^3+4)(x^3-4)}. \]Factor Repeated Differences of Squares Completely
Sometimes factoring once reveals another difference of squares.
Consider
\[ x^4-16. \]First,
\[ x^4-16=(x^2+4)(x^2-4). \]Then
\[ x^2-4=(x+2)(x-2). \]Therefore,
\[ \boxed{x^4-16=(x^2+4)(x+2)(x-2)}. \]Over the real numbers, \(x^2+4\) does not factor into real linear factors.
Sum of Squares Is Different
The difference-of-squares identity does not apply to
\[ a^2+b^2. \]For example,
\[ x^2+9 \]does not factor as
\[ (x+3)(x-3), \]because that product equals
\[ x^2-9. \]The sign is essential.
Factoring After Removing a GCF
Special forms often appear only after a common factor is removed.
Factor
\[ 3x^3-27x. \]First factor the GCF:
\[ 3x(x^2-9). \]Now factor the difference of squares:
\[ x^2-9=(x+3)(x-3). \]Therefore,
\[ \boxed{3x^3-27x=3x(x+3)(x-3)}. \]Factoring a Perfect Square After a GCF
Factor
\[ 2x^3+12x^2+18x. \]First factor the GCF \(2x\):
\[ 2x(x^2+6x+9). \]The trinomial is a perfect square:
\[ x^2+6x+9=(x+3)^2. \]Therefore,
\[ \boxed{2x^3+12x^2+18x=2x(x+3)^2}. \]Factoring by Grouping Can Lead to a Special Form
Consider
\[ x^3-2x^2-4x+8. \]Group:
\[ x^2(x-2)-4(x-2). \]Factor the common binomial:
\[ (x-2)(x^2-4). \]Then factor the difference of squares:
\[ x^2-4=(x+2)(x-2). \]Therefore,
\[ \boxed{x^3-2x^2-4x+8=(x-2)^2(x+2)}. \]Complete Factoring Is a Sequence of Decisions
Factoring is rarely just one isolated technique.
A polynomial may require several stages:
- remove a GCF;
- factor by grouping;
- factor a trinomial;
- recognize a perfect-square form;
- factor a difference of squares;
- repeat until no further known factorization applies.
A Practical Factoring Decision Tree
When asked to factor a polynomial, use this order.
- GCF: Is there a factor common to every term?
- Number of terms: Are there two, three, or four terms?
- Two terms: Does the expression match a special form such as a difference of squares?
- Three terms: Is it a perfect-square trinomial or a factorable quadratic trinomial?
- Four terms: Can grouping create a repeated factor?
- Repeat: Can any resulting factor be factored further?
Factoring \(ax^2+bx+c\): A Reliable Procedure
- Factor out any GCF.
- Identify \(a\), \(b\), and \(c\).
- Compute \(ac\).
- Find two integers whose product is \(ac\) and whose sum is \(b\).
- Split the middle term using those two numbers.
- Factor by grouping.
- Check whether any factor can be factored further.
- Expand to verify the result.
Checking by Multiplication
Factoring and multiplication are inverse processes, so expansion provides a direct check.
Suppose
\[ 6x^2+11x+3=(3x+1)(2x+3). \]Expand the right side:
\[ 6x^2+9x+2x+3. \]Combine:
\[ 6x^2+11x+3. \]The original polynomial is recovered.
Checking a Perfect-Square Factorization
Suppose
\[ x^2-14x+49=(x-7)^2. \]Expand:
\[ (x-7)(x-7) = x^2-7x-7x+49. \]Therefore,
\[ x^2-14x+49. \]The factorization is correct.
Checking a Difference of Squares
Suppose
\[ 16x^2-81=(4x+9)(4x-9). \]The factors are conjugates, so the middle terms cancel:
\[ (4x)^2-9^2 = 16x^2-81. \]The original expression returns.
Common Error: Forgetting the GCF
Consider
\[ 4x^2+12x+8. \]A student may notice
\[ x^2+3x+2=(x+1)(x+2), \]but the original polynomial has a factor of \(4\).
The complete factorization is
\[ \boxed{4(x+1)(x+2)}. \]Common Error: Finding Numbers with the Right Product but Wrong Sum
To factor
\[ x^2+8x+15, \]the factor pair \(1\) and \(15\) has the correct product but the wrong sum.
The correct pair is
\[ 3,\ 5 \]because
\[ 3\cdot5=15 \]and
\[ 3+5=8. \]Therefore,
\[ \boxed{x^2+8x+15=(x+3)(x+5)}. \]Common Error: Ignoring Signs
To factor
\[ x^2-5x-14, \]the numbers must multiply to \(-14\), so they must have opposite signs.
The pair
\[ -7,\ 2 \]gives
\[ -7+2=-5. \]Therefore,
\[ \boxed{x^2-5x-14=(x-7)(x+2)}. \]Common Error: Using the Monic Rule When \(a\neq1\)
For
\[ 2x^2+7x+3, \]looking only for two numbers whose product is \(3\) and sum is \(7\) will not reveal the correct factors.
The leading coefficient \(2\) must be included in the structure.
Using \(ac=6\) leads to the correct factorization:
\[ \boxed{(2x+1)(x+3)}. \]Common Error: Misidentifying a Perfect Square
Consider
\[ x^2+8x+16. \]The first and last terms are squares:
\[ x^2 \]and
\[ 4^2. \]The middle term must equal
\[ 2(x)(4)=8x, \]which it does.
Therefore,
\[ \boxed{x^2+8x+16=(x+4)^2}. \]Checking the middle term is essential.
Common Error: Factoring a Sum of Squares as Conjugates
The expression
\[ x^2+25 \]does not factor as
\[ (x+5)(x-5). \]That product is
\[ x^2-25. \]The difference-of-squares identity requires subtraction.
Common Error: Stopping Before the Factorization Is Complete
Consider
\[ x^4-81. \]Factoring once gives
\[ (x^2+9)(x^2-9). \]But
\[ x^2-9 \]is another difference of squares:
\[ (x+3)(x-3). \]Therefore,
\[ \boxed{x^4-81=(x^2+9)(x+3)(x-3)}. \]Example: Complete Factoring Workflow
Factor completely:
\[ 12x^3-48x. \]First factor the GCF:
\[ 12x(x^2-4). \]Recognize the difference of squares:
\[ x^2-4=(x+2)(x-2). \]Therefore,
\[ \boxed{12x^3-48x=12x(x+2)(x-2)}. \]Example: Complete Trinomial Workflow
Factor completely:
\[ 10x^2+35x+30. \]First factor the GCF:
\[ 5(2x^2+7x+6). \]For the trinomial,
\[ ac=2\cdot6=12. \]The numbers \(3\) and \(4\) multiply to \(12\) and add to \(7\).
Split:
\[ 2x^2+3x+4x+6. \]Group:
\[ x(2x+3)+2(2x+3). \]Therefore,
\[ \boxed{10x^2+35x+30=5(x+2)(2x+3)}. \]Example: Complete Factoring with Grouping and a Special Form
Factor completely:
\[ 2x^3+4x^2-8x-16. \]First factor the GCF:
\[ 2(x^3+2x^2-4x-8). \]Group:
\[ 2[x^2(x+2)-4(x+2)]. \]Factor the repeated binomial:
\[ 2(x+2)(x^2-4). \]Factor the difference of squares:
\[ x^2-4=(x+2)(x-2). \]Therefore,
\[ \boxed{2(x+2)^2(x-2)}. \]Practice 1: Monic Trinomial
Factor:
\[ x^2+9x+20. \]Practice 2: Monic Trinomial with Negative Signs
Factor:
\[ x^2-11x+24. \]Practice 3: Opposite Signs
Factor:
\[ x^2-2x-15. \]Practice 4: AC Method
Factor:
\[ 2x^2+9x+4. \]Practice 5: AC Method with a Negative Constant
Factor:
\[ 6x^2+x-2. \]Practice 6: Perfect-Square Trinomial
Factor:
\[ x^2+12x+36. \]Practice 7: Perfect-Square Trinomial with Coefficients
Factor:
\[ 9x^2-24x+16. \]Practice 8: Difference of Squares
Factor completely:
\[ 25x^2-64. \]Practice 9: GCF and Difference of Squares
Factor completely:
\[ 5x^3-45x. \]Practice 10: GCF and Trinomial
Factor completely:
\[ 8x^2+20x+12. \]Practice 11: Grouping and Special Form
Factor completely:
\[ x^3+3x^2-4x-12. \]Practice 12: Prime over the Integers?
Determine whether
\[ x^2+4x+7 \]factors into binomials with integer coefficients.
Practice Answers
Practice 1
The numbers \(4\) and \(5\) multiply to \(20\) and add to \(9\):
\[ \boxed{x^2+9x+20=(x+4)(x+5)}. \]Practice 2
The numbers \(-3\) and \(-8\) multiply to \(24\) and add to \(-11\):
\[ \boxed{x^2-11x+24=(x-3)(x-8)}. \]Practice 3
The numbers \(3\) and \(-5\) multiply to \(-15\) and add to \(-2\):
\[ \boxed{x^2-2x-15=(x+3)(x-5)}. \]Practice 4
Here,
\[ ac=2\cdot4=8. \]The numbers \(1\) and \(8\) add to \(9\):
\[ 2x^2+x+8x+4. \]Group:
\[ x(2x+1)+4(2x+1). \]Therefore,
\[ \boxed{2x^2+9x+4=(x+4)(2x+1)}. \]Practice 5
Compute
\[ ac=6(-2)=-12. \]The numbers \(4\) and \(-3\) add to \(1\):
\[ 6x^2+4x-3x-2. \]Group:
\[ 2x(3x+2)-1(3x+2). \]Therefore,
\[ \boxed{6x^2+x-2=(2x-1)(3x+2)}. \]Practice 6
\[ x^2+12x+36 = x^2+2(x)(6)+6^2. \]Therefore,
\[ \boxed{(x+6)^2}. \]Practice 7
\[ 9x^2-24x+16 = (3x)^2-2(3x)(4)+4^2. \]Therefore,
\[ \boxed{(3x-4)^2}. \]Practice 8
\[ 25x^2-64 = (5x)^2-8^2. \]Therefore,
\[ \boxed{(5x+8)(5x-8)}. \]Practice 9
Factor the GCF \(5x\):
\[ 5x(x^2-9). \]Then factor the difference of squares:
\[ \boxed{5x(x+3)(x-3)}. \]Practice 10
Factor the GCF \(4\):
\[ 4(2x^2+5x+3). \]The trinomial factors as
\[ (2x+3)(x+1). \]Therefore,
\[ \boxed{4(2x+3)(x+1)}. \]Practice 11
Group:
\[ x^2(x+3)-4(x+3). \]Factor:
\[ (x+3)(x^2-4). \]Then
\[ x^2-4=(x+2)(x-2). \]Therefore,
\[ \boxed{(x+3)(x+2)(x-2)}. \]Practice 12
We need two integers whose product is \(7\) and whose sum is \(4\).
The integer factor pairs are
\[ 1,\ 7 \]and
\[ -1,\ -7. \]Their sums are \(8\) and \(-8\).
Therefore, the trinomial does not factor into binomials with integer coefficients.
What to Remember
Factoring trinomials means reversing binomial multiplication.
For
\[ x^2+bx+c, \]find numbers \(m\) and \(n\) satisfying
\[ \boxed{m+n=b} \]and
\[ \boxed{mn=c}. \]For
\[ ax^2+bx+c, \]the AC method searches for two numbers whose product is
\[ ac \]and whose sum is
\[ b. \]The important special forms are
\[ \boxed{a^2+2ab+b^2=(a+b)^2}, \] \[ \boxed{a^2-2ab+b^2=(a-b)^2}, \]and
\[ \boxed{a^2-b^2=(a+b)(a-b)}. \]Always remove the GCF first, and always inspect the resulting factors to see whether factoring can continue.
Where We Go Next
Factoring rewrites a polynomial as a product. Division asks the reverse operational question: given a polynomial and a known divisor, what quotient and remainder result?
The next lesson, Polynomial and Synthetic Division, develops long division for polynomials and then introduces synthetic division as a compact coefficient-based method for linear divisors.