Factoring
Cheat sheet.

Key method

A practical order is: common factor, familiar identity, grouping or trinomial, then substitution. Verify every proposed factorization by multiplication. Over the reals, some polynomials stop at irreducible quadratic factors.

Example

Factor $2x^3-8x$. Every term contains $2x$, giving $2x(x^2-4)$. The remaining expression is a difference of squares, so

To solve $2x^3-8x=0$, set each factor equal to zero: $x=0,2,-2$. Expanding the answer recovers both original terms.

Avoid this mistake

Factoring is an identity, not division. Dividing an equation by $x$ without checking $x=0$ can lose a solution.

Key method

The distributive law works in reverse:

A polynomial factor can also be common. Keep its parentheses intact while treating it as one object.

Example

In $18x^3y-12x^2y^2$, the coefficient factor is $6$, and the shared variable part is $x^2y$. Thus

For $x(x+1)+4(x+1)$, the repeated object is $x+1$, giving $(x+1)(x+4)$.

Avoid this mistake

Do not take a variable outside if it is absent from one term. For example, $x^2+3$ has no common factor $x$ as a polynomial.

Key method

For $ax^2+bx+c$, seek numbers whose product is $ac$ and sum is $b$. If none exist among integers, use the discriminant or quadratic formula rather than forcing integer factors.

Example

Factor $6x^2+11x+3$. The product $ac=18$ and sum $b=11$ suggest $9$ and $2$:

Hence $(3x+1)(2x+3)$. The two zeros are $-1/3$ and $-3/2$.

Avoid this mistake

Matching only the constant term is insufficient. Check the middle coefficient as well as the leading coefficient.

Key method

If the two parentheses differ by a sign, factor a negative from one group. If they genuinely differ, try another grouping.

Example

Factor $x^3+2x^2-3x-6$:

Over the reals this can continue to $(x+2)(x-\sqrt3)(x+\sqrt3)$. Over integer coefficients, the first product is the natural stopping point.

Avoid this mistake

Factoring $-3x-6$ as $-3(x-2)$ changes the constant term. Distribute the negative to check both signs.

Key method

A repeated linear factor gives a repeated zero. Its graph touches the horizontal axis rather than crossing there.

Example

In $4x^2-12x+9$, the outer terms are $(2x)^2$ and $3^2$, and $-12x=-2(2x)(3)$. Therefore

Its only zero is $x=3/2$, with multiplicity two; its value cannot be negative.

Avoid this mistake

$A^2+B^2$ is missing the mixed term and is not $(A+B)^2$.

Key method

Identify both squares explicitly before substituting. This identity is valid for real or complex values.

Example

For $9x^2-25$, use $A=3x$ and $B=5$ to obtain $(3x-5)(3x+5)$. For $(x+2)^2-9$, use $A=x+2$ and $B=3$:

This avoids an unnecessary expansion.

Avoid this mistake

The sum $A^2+B^2$ does not equal $(A+B)(A-B)$. Over the reals, $x^2+1$ has no linear factors.

Key method

The sign in the first factor matches the original; the middle sign in the quadratic is opposite.

Example

Since $8x^3+27=(2x)^3+3^3$, its factors are

Multiplying gives $8x^3-12x^2+18x+12x^2-18x+27$: the middle terms cancel. This is why changing the quadratic's middle sign breaks the identity.

Avoid this mistake

A sum of cubes is not a cube of a sum: $(A+B)^3$ also contains two mixed terms.

Key method

For subtraction, the signs alternate because odd powers of $-B$ are negative.

Example

For $8x^3-36x^2+54x-27$, try $A=2x$, $B=3$. The mixed terms are $-3(2x)^2(3)=-36x^2$ and $3(2x)(3^2)=54x$. Thus the expression is $(2x-3)^3$.

Avoid this mistake

A repeated cubic factor crosses the axis at its zero; it does not behave like a repeated square. Multiplicity matters.

Key method

For integer $n\ge1$,

For odd $n$, the corresponding sum factors with alternating signs in the second factor.

Example

For $n=5$,

On multiplication, every mixed term cancels. For $x^6-y^6$, first use the difference of squares $(x^3-y^3)(x^3+y^3)$, then apply both cube identities.

Avoid this mistake

The same sum formula does not work for even exponents: substituting $x=-y$ into $x^2+y^2$ usually does not give zero.

Key method

For $x^4+bx^2+c$, put $u=x^2$. The expression becomes $u^2+bu+c$. When solving over the reals, remember that $u=x^2\ge0$.

Example

Factor $x^4-5x^2+4$. With $u=x^2$,

Substitute back to get $(x^2-1)(x^2-4)$, then factor both differences of squares:

Avoid this mistake

Do not stop a solution at $u=1,4$ when the question asks for $x$. Each positive square value gives two real roots.

Key method

Complete a square by supplying the missing middle term, then treat the remainder separately. For a monic quadratic:

This also identifies the vertex of its graph.

Example

To factor $x^4+4$, add and subtract $4x^2$:

A difference of squares now gives

Both factors are positive for real $x$, since they equal $(x\mp1)^2+1$.

Avoid this mistake

Adding a term alone changes the problem. Write the compensating subtraction on the same line.