Lesson 1.5: Factoring Polynomials

New Concept

Factoring breaks down complex polynomials into simpler products, like reverse multiplication. You'll master several key methods, from finding the GCF to using special formulas, to factor various types of expressions and solve problems more easily.

What’s next

You'll now dive into these techniques with worked examples and a series of practice cards to build your factoring skills.

Property

The greatest common factor (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials. To factor out the GCF:

1. Identify the GCF of the coefficients.
2. Identify the GCF of the variables (using the lowest power for each variable).
3. Combine to find the expression's GCF.
4. Determine what the GCF must be multiplied by to get each term.
5. Write the factored expression as the GCF multiplied by the remaining terms.

Examples

• To factor $8a^2b^3 + 12ab^2 + 20a^3b$, the GCF is $4ab$. Factoring this out gives $4ab(2ab^2 + 3b + 5a^2)$.
• To factor $15m^4n - 25m^2n^2 + 5mn$, the GCF is $5mn$. The factored form is $5mn(3m^3 - 5mn + 1)$.
• To factor $12x^2(y-1) + 8x(y-1)$, the GCF is $4x(y-1)$. This results in $4x(y-1)(3x + 2)$.

Explanation

Think of this as reverse distribution. You are finding the largest number and variable part that every single term shares and pulling it out to the front. This simplifies the expression inside the parentheses, making it easier to work with.

Property

A trinomial of the form $x^2 + bx + c$ can be written in factored form as $(x + p)(x + q)$ where $pq = c$ and $p + q = b$.
To factor, list the factors of $c$ and find the pair, $p$ and $q$, that has a sum of $b$.
Then write the factored expression as $(x+p)(x+q)$. Some polynomials, called prime, cannot be factored.

Examples

• To factor $x^2 + 7x + 12$, we need two numbers that multiply to $12$ and add to $7$. These are $3$ and $4$, so the factors are $(x+3)(x+4)$.
• To factor $y^2 - 8y + 15$, we need numbers that multiply to $15$ and add to $-8$. These are $-3$ and $-5$, giving the factors $(y-3)(y-5)$.
• To factor $z^2 + 3z - 10$, we need numbers that multiply to $-10$ and add to $3$. These are $5$ and $-2$, so the factors are $(z+5)(z-2)$.

Explanation

You're looking for two magic numbers that multiply to the last term ($c$) and add up to the coefficient of the middle term ($b$). This process reverses the FOIL method you use to multiply two binomials together.

Property

To factor a trinomial in the form $ax^2 + bx + c$ by grouping, find two numbers, $p$ and $q$, that have a product of $ac$ and a sum of $b$.
Rewrite the expression as $ax^2 + px + qx + c$.
Then, factor the GCF from the first pair of terms and the GCF from the second pair, and finally factor out the common binomial factor.

Examples

• To factor $3x^2 + 10x + 8$, find two numbers that multiply to $ac=24$ and add to $b=10$. The numbers are $6$ and $4$. Rewrite as $3x^2+6x+4x+8$, which factors to $3x(x+2)+4(x+2)$, giving $(3x+4)(x+2)$.
• To factor $2y^2 - 5y - 12$, find numbers that multiply to $ac=-24$ and add to $b=-5$. The numbers are $-8$ and $3$. Rewrite as $2y^2-8y+3y-12$, which factors to $2y(y-4)+3(y-4)$, giving $(2y+3)(y-4)$.
• To factor $4z^2 + 13z - 12$, find numbers that multiply to $ac=-48$ and add to $b=13$. The numbers are $16$ and $-3$. Rewrite as $4z^2+16z-3z-12$, which factors to $4z(z+4)-3(z+4)$, giving $(4z-3)(z+4)$.

Explanation

When the leading coefficient isn't 1, we split the middle term into two new terms. This allows us to factor the polynomial in two halves, revealing a common binomial that can then be factored out for the final answer.

Property

A perfect square trinomial can be written as the square of a binomial:

To use this, confirm the first and last terms are perfect squares ($a^2$ and $b^2$) and the middle term is twice their product ($2ab$).

Examples

• To factor $9x^2 + 12x + 4$, notice $a=3x$ and $b=2$. The middle term is $2(3x)(2) = 12x$. So, the factored form is $(3x+2)^2$.
• To factor $16y^2 - 40y + 25$, notice $a=4y$ and $b=5$. The middle term is $2(4y)(5)=40y$. So, the factored form is $(4y-5)^2$.
• To factor $x^2 + 18x + 81$, notice $a=x$ and $b=9$. The middle term is $2(x)(9)=18x$. So, the factored form is $(x+9)^2$.

Explanation

This is a special shortcut. If you spot that the first and last terms are perfect squares, check the middle term. If it's $2$ times the product of the square roots of the outer terms, you can factor it instantly.

Property

A difference of squares is a perfect square subtracted from a perfect square. It can be rewritten as two factors containing the same terms but opposite signs.

To factor, confirm both terms are perfect squares and write the factored form. A sum of squares cannot be factored.

Examples

• To factor $49x^2 - 16$, recognize this as $(7x)^2 - 4^2$. The factors are $(7x+4)(7x-4)$.
• To factor $100y^4 - 81z^2$, recognize this as $(10y^2)^2 - (9z)^2$. The factors are $(10y^2+9z)(10y^2-9z)$.
• To factor $a^2 - 1$, recognize this as $a^2 - 1^2$. The factors are $(a+1)(a-1)$.

Explanation

When you see a perfect square minus another perfect square, it factors into two identical binomials, one with a plus and one with a minus. This causes the middle terms from FOIL to cancel out, leaving just the first and last terms.

Property

We can factor the sum of two cubes as

We can factor the difference of two cubes as

Use the acronym SOAP (Same, Opposite, Always Positive) to remember the signs.

Examples

• To factor $y^3 + 27$, recognize this as $y^3 + 3^3$. Using the sum of cubes formula, we get $(y+3)(y^2 - 3y + 9)$.
• To factor $8z^3 - 1$, recognize this as $(2z)^3 - 1^3$. Using the difference of cubes formula, we get $(2z-1)(4z^2 + 2z + 1)$.
• To factor $2x^3 + 128$, first factor out the GCF of $2$ to get $2(x^3+64)$. This is $2(x^3+4^3)$, which factors to $2(x+4)(x^2-4x+16)$.

Explanation

This is a special formula for factoring two perfect cubes. The SOAP acronym (Same, Opposite, Always Positive) helps you remember the signs for the binomial and trinomial factors. The resulting trinomial part cannot be factored further.

Property

Expressions with fractional or negative exponents can be factored by pulling out a GCF.
Look for the variable or exponent that is common to each term of the expression and pull out that variable or exponent raised to the lowest power.
These expressions follow the same factoring rules as those with integer exponents.

Examples

• To factor $x^{\frac{3}{2}} + x^{\frac{1}{2}}$, the lowest power is $x^{\frac{1}{2}}$. Factoring this out gives $x^{\frac{1}{2}}(x+1)$.
• To factor $5x(x - 1)^{-\frac{1}{2}} + 2(x - 1)^{\frac{1}{2}}$, the lowest exponent is $-\frac{1}{2}$. Factoring out $(x-1)^{-\frac{1}{2}}$ leaves $(x-1)^{-\frac{1}{2}}(5x + 2(x-1))$, which simplifies to $(x-1)^{-\frac{1}{2}}(7x-2)$.
• To factor $4(y+5)^{-\frac{1}{4}} + 8y(y+5)^{\frac{3}{4}}$, factor out $4(y+5)^{-\frac{1}{4}}$ to get $4(y+5)^{-\frac{1}{4}}(1+2y(y+5))$, which simplifies to $4(y+5)^{-\frac{1}{4}}(2y^2+10y+1)$.

Explanation

This method is just like finding a GCF. Identify the common base and factor out the one with the smallest exponent, even if it's negative or a fraction. Remember that when you divide powers, you subtract the exponents.