Factoring Polynomials

Algebra is arithmetic with a blank space in it. A letter such as x stands for a number you have not found yet, and solving means finding the number that makes the sentence true.

Factoring Polynomials: the central idea

Factoring reverses multiplication. It rewrites a polynomial as a product whose expansion returns the original expression exactly.

Words you need

Factor

A factor is one of the expressions being multiplied, so in $(x-3)(x+3)$ both $(x-3)$ and $(x+3)$ are factors.

Greatest common factor

The greatest common factor is the largest expression that divides evenly into every term, such as $3x$ inside $12x^3-27x$.

Difference of squares

A difference of squares is a binomial like $a^2-b^2$ in which one perfect square is subtracted from another, and it always factors as $(a-b)(a+b)$.

Factoring by grouping

Factoring by grouping is the method of splitting four terms into two pairs, pulling a factor out of each pair, and then pulling out the binomial the pairs share.

Prime polynomial

A prime polynomial is one that cannot be broken into smaller factors with integer coefficients, such as $x^2+9$.

Zero-product property

The zero-product property says that if a product equals zero then at least one of its factors equals zero, which is what turns a factored equation into a list of easy equations.

What to know before this lesson

Know the distributive property, greatest common factors, exponent rules, and multiplication of binomials.

If one of those prerequisites is uncertain, use the Algebra subject guide to locate the earlier concept before memorizing a procedure.

Factoring polynomials: a worked example

Every step, with the arithmetic

1. Step 1 - Write the polynomial$12x^3-27x$
2. Step 2 - Find the greatest common factor of the numbers$12=3\cdot4$ and $27=3\cdot9$, so 3 is the largest shared number
3. Step 3 - Find the greatest common factor of the variables$x^3$ and $x$ both contain $x$, so the full GCF is $3x$
4. Step 4 - Divide each term by $3x$$12x^3\div3x=4x^2$ and $27x\div3x=9$, leaving $3x(4x^2-9)$
5. Step 5 - Recognize the pattern inside$4x^2-9=(2x)^2-3^2$, a difference of squares
6. Step 6 - Apply the difference-of-squares rule$3x(2x-3)(2x+3)$
7. Step 7 - Multiply back to check$(2x-3)(2x+3)=4x^2-9$, and $3x(4x^2-9)=12x^3-27x$

$x^2-9$ is $x^2-3^2$, so the difference-of-squares pattern gives $(x-3)(x+3)$. The middle terms cancel when expanded.

The factoring patterns worth memorizing

The step-by-step method for factoring polynomials

1. Remove the greatest common factor before looking for a special pattern.
2. Count terms and test structures such as a difference of squares, trinomial, grouping, or sum and difference of cubes.
3. Multiply the factors back and compare every coefficient with the original polynomial.

How to check your answer

Expand the proposed factors. For $(x-3)(x+3)$ the result is $x^2+3x-3x-9=x^2-9$.

Substitute a proposed value into the original statement—not only the last simplified line. For functions, also inspect domain, intercepts, and whether the graph agrees.

A mistake that changes the mathematics

A sum of squares such as $x^2+9$ does not factor into real linear factors using the difference-of-squares rule.

Where you will actually use this

Finding when a ball hits the ground

A ball kicked upward has height $h=-16t^2+64t$. Factoring gives $h=-16t(t-4)$, so the height is zero at $t=0$ and $t=4$. The ball is in the air for 4 seconds, and you never needed the quadratic formula.

Recovering the dimensions of a rectangle

If a rectangle's area is $x^2+7x+12$ square units, factoring to $(x+3)(x+4)$ shows the sides are $x+3$ and $x+4$. When $x=5$ that is an 8 by 9 rectangle with an area of 72.

Simplifying an algebraic fraction

In $\dfrac{x^2-9}{x^2+7x+12}$, factoring both parts gives $\dfrac{(x-3)(x+3)}{(x+3)(x+4)}$. The shared factor $(x+3)$ cancels and leaves $\dfrac{x-3}{x+4}$ for every $x$ except $-3$.

How factoring polynomials connects to the rest of algebra

• The quadratic formula — The formula solves the quadratics that will not factor, and a discriminant that is a perfect square is the promise that factoring would have worked.
• Solving linear equations — After factoring, the zero-product property turns each factor into a small linear equation such as $2x-3=0$, giving $x=\tfrac32$.
• Fractions — Cancelling a shared factor in an algebraic fraction is the same move as simplifying $\tfrac69$ to $\tfrac23$: you divide the top and bottom by a common factor.

Try a transfer problem

Factor $6x^2-x-2$ completely, then identify which pair of terms creates the required middle coefficient.

Show the worked answer

There is no common factor in $6x^2-x-2$, so use the ac method. Multiply $a\cdot c=6\cdot(-2)=-12$ and look for two numbers with product $-12$ and sum $-1$. Those numbers are $+3$ and $-4$. Split the middle term with them: $6x^2+3x-4x-2$. Group the pairs: $3x(2x+1)-2(2x+1)$. Both groups share $(2x+1)$, so the factorization is $(2x+1)(3x-2)$. Check by expanding: $6x^2-4x+3x-2=6x^2-x-2$. The pair of terms that creates the required middle coefficient is $+3x$ and $-4x$, since $3x-4x=-x$. Set each factor to zero for the roots: $x=-\tfrac12$ and $x=\tfrac23$.

Work without copying the example. When finished, use the relevant focused calculator or formula reference to check the setup and result, then correct the first line where your reasoning changed. When the method feels reliable, move to algebra practice questions.

Questions about factoring polynomials

How do I know a polynomial is completely factored?

It is complete when no factor still has a common factor and no factor still matches a pattern. Stopping at $3x(4x^2-9)$ is not done, because $4x^2-9$ is a difference of squares and becomes $(2x-3)(2x+3)$. The finished answer is $3x(2x-3)(2x+3)$.

Why does $x^2+9$ not factor?

Because it is a sum of squares, not a difference. Factoring it would need two numbers whose product is 9 and whose sum is 0, and no pair of real numbers does that. Over the real numbers $x^2+9$ is prime, though $x^2-9$ factors immediately as $(x-3)(x+3)$.

What exactly is the ac method?

It is the way to factor $ax^2+bx+c$ when $a$ is not 1. Multiply $a$ by $c$, find two numbers with that product and with sum $b$, use them to split the middle term into two terms, then factor by grouping. For $2x^2+7x+3$, $ac=6$ and the pair is 1 and 6, giving $2x^2+x+6x+3=(2x+1)(x+3)$.

Can every quadratic be factored?

Not with nice numbers. Factoring with integers only works when the discriminant $b^2-4ac$ is a perfect square. For $x^2-4x+1$ the discriminant is 12, so the roots are $2\pm\sqrt3$ and no integer factoring exists. Use the quadratic formula there.