Breaking polynomials apart into the factors that built them.
A store sells boxes of pencils. One box holds 6 pencils, another holds 9, and a third holds 15. A teacher wants to split each box into equal groups without mixing types. The largest group size that works for all three is 3 — because 3 divides evenly into 6, 9, and 15. That number is the greatest common factor. Factoring a polynomial works the same way: you find the largest expression that divides evenly into every term.
The greatest common factor, or GCF, is the largest monomial that divides every term in a polynomial. Factoring it out is always the first step before anything else.
Factoring out the GCF from 12x³ + 8x²
12x3+8x2
Look at the coefficients first: GCF of 12 and 8 is 4. Then look at the variables: both have at least x², so x² is the variable part of the GCF.
4x2(3x+2)
Divide each term by 4x². The 12x³ becomes 3x, the 8x² becomes 2.
Factoring out the GCF from 6x³ − 9x² + 3x
6x3−9x2+3x
GCF of 6, 9, and 3 is 3. Every term has at least x¹, so the GCF is 3x.
3x(2x2−3x+1)
Divide each term by 3x. Check by distributing back: 3x·2x² = 6x³, 3x·3x = 9x², 3x·1 = 3x. Signs match.
Once you have factored out the GCF, you often have a trinomial left. Factoring a trinomial means rewriting it as a product of two binomials. When the leading coefficient is 1, you look for two numbers that multiply to the constant term and add to the middle coefficient.
For x2+bx+c, find two numbers p and q such that p⋅q=c and p+q=b. Then:
x2+bx+c=(x+p)(x+q)
Factoring x² + 7x + 12
p⋅q=12,p+q=7
You need two numbers whose product is 12 and whose sum is 7. List factor pairs of 12: (1,12), (2,6), (3,4). The pair 3 and 4 adds to 7.
(x+3)(x+4)
Place those two numbers as the constants in the binomials.
(x+3)(x+4)=x2+4x+3x+12=x2+7x+12✓
Distribute to verify. FOIL gives back the original.
Factoring x² − 5x + 6
p⋅q=6,p+q=−5
The product is positive and the sum is negative, so both numbers must be negative. Pairs of negatives that multiply to 6: (-1,-6), (-2,-3). The pair -2 and -3 adds to -5.
(x−2)(x−3)✓
Both signs are negative. Distribute to check: x² - 3x - 2x + 6 = x² - 5x + 6.
When the leading coefficient is not 1, the strategy changes. You cannot just hunt for a simple pair — you need to split the middle term. This method is called factoring by grouping, and it works for any trinomial ax2+bx+c with a=1.
The process: multiply a and c together. Find two numbers that multiply to a⋅c and add to b. Use those numbers to split the middle term. Then factor by grouping.
Factoring 2x² + 7x + 3
a⋅c=2⋅3=6
Multiply the leading coefficient by the constant. You need two numbers that multiply to 6 and add to 7.
6=1⋅6,1+6=7
The numbers are 1 and 6. They multiply to 6 and add to 7.
2x2+1x+6x+3
Replace 7x with 1x + 6x. The polynomial now has four terms.
x(2x+1)+3(2x+1)
Group the first two terms and factor out x, then group the last two and factor out 3. Both groups reveal the same binomial: (2x + 1).
(x+3)(2x+1)
The shared binomial (2x+1) becomes one factor. The outsides, x and 3, form the other.
(x+3)(2x+1)=2x2+x+6x+3=2x2+7x+3✓
FOIL confirms the result.
Factoring 6x² − 11x + 4
a⋅c=6⋅4=24
Start by multiplying the leading coefficient and the constant.
−8⋅(−3)=24,−8+(−3)=−11
You need a product of 24 and a sum of -11. Both numbers must be negative because the product is positive and the sum is negative. -8 and -3 work.
6x2−8x−3x+4
Split -11x into -8x and -3x.
2x(3x−4)−1(3x−4)
Factor 2x from the first group and -1 from the second. The -1 is important — without it, the signs in the second group would be wrong.
(2x−1)(3x−4)
The shared binomial (3x - 4) factors out. The remaining pieces are 2x and -1.
(2x−1)(3x−4)=6x2−8x−3x+4=6x2−11x+4✓
Verified by FOIL.
Some problems combine both steps. Always factor out the GCF first, then factor the remaining trinomial.
Factoring 4x³ + 10x² − 6x completely
2x(2x2+5x−3)
Factor out the GCF first. GCF of 4, 10, and 6 is 2. Every term has at least x, so the GCF is 2x.
a⋅c=2⋅(−3)=−6,6+(−1)=5
Now factor the trinomial inside. Multiply 2 and -3 to get -6. Find two numbers that multiply to -6 and add to 5: that's 6 and -1.
2x(2x2+6x−x−3)
Split 5x into 6x and -x.
2x[2x(x+3)−1(x+3)]
Factor 2x from the first group and -1 from the second. The shared binomial is (x+3).
2x(2x−1)(x+3)
Complete factorization: the GCF out front stays, and the trinomial becomes two binomials.
On Parts II, III, and IV of the Algebra II Regents, factoring appears constantly — not as an isolated skill but as the first step in solving equations, simplifying rational expressions, and identifying zeros of polynomial functions. A factoring error early in a multi-part problem will cascade. The graders follow through scoring, so a correct process built on a wrong factorization can still earn partial credit, but only if every subsequent step is logically consistent with your incorrect factor.
When using the grouping method on a problem like 6x² − 11x + 4, students split the middle term correctly but then factor out a positive 1 from the second group instead of a negative 1, writing 2x(3x − 4) + 1(−3x + 4) and getting stuck because the binomials don't match. The binomials must be identical before you can pull them out. If they differ by a sign, you factored the wrong sign from the second group — go back and factor out the negative.