1.2.7: Factoring Trinomials (2024)

Factoring Trinomials of the Form \(x^2+bx+c\)

You have already learned how to multiply binomials using FOIL. Now you’ll need to “undo” this multiplication. To factor the trinomial means to start with the product, and end with the factors.

To figure out how we would factor a trinomial of the form \(x^2+bx+c\), such as \(x^2+5x+6\) and factor it to \((x+2)(x+3)\), let’s start with two general binomials of the form \((x+m)\) and \((x+n)\).

This tells us that to factor a trinomial of the form \(x^2+bx+c\), we need two factors \((x+m)\) and \((x+n)\) where the two numbers \(m\) and \(n\) multiply to \(c\) and add to \(b\).

Let’s summarize the steps we used to find the factors.

In the first example, all terms in the trinomial were positive. What happens when there are negative terms? Well, it depends which term is negative. Let’s look first at trinomials with only the middle term negative.

How do you get a positive product and a negative sum? We use two negative numbers.

Now, what if the last term in the trinomial is negative? Think about FOIL. The last term is the product of the last terms in the two binomials. A negative product results from multiplying two numbers with opposite signs. You have to be very careful to choose factors to make sure you get the correct sign for the middle term, too.

How do you get a negative product and a positive sum? We use one positive and one negative number.

When we factor trinomials, we must have the terms written in descending order—in order from highest degree to lowest degree.

Sometimes you’ll need to factor trinomials of the form \(x^2+bxy+cy^2\) with two variables, such as \(x^2+12xy+36y^2\). The first term, \(x^2\), is the product of the first terms of the binomial factors, \(xx\). The \(y^2\) in the last term means that the second terms of the binomial factors must each contain \(y\).

Some trinomials are prime. The only way to be certain a trinomial is prime is to list all the possibilities and show that none of them work.

Let’s summarize the method we just developed to factor trinomials of the form \(x^2+bx+c\).

Factoring Trinomials of the Form ax² + bx + cUsing Trial and Error

Our next step is to factor trinomials whose leading coefficient is not 1, trinomials of the form \(ax^2+bx+c\).

Remember to always check for a GCF first! Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes \(1\) and you can factor it by the methods we’ve used so far. Let’s do an example to see how this works.

What happens when the leading coefficient is not \(1\) and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method.

Let’s factor the trinomial \(3x^2+5x+2\).

From our earlier work, we expect this will factor into two binomials.

\[3x^2+5x+2\nonumber\]\[(\quad)(\quad)\nonumber\]

We know the first terms of the binomial factors will multiply to give us \(3x^2\). The only factors of \(3x^2\) are \(1x,\space 3x\). We can place them in the binomials.

\[(x+ \quad)(3x+\quad)\nonumber\]

Check: Does \(1x·3x=3x^2\)?

We know the last terms of the binomials will multiply to \(2\). Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of \(2\) are \(1\) and \(2\). But we now have two cases to consider as it will make a difference if we write \(1\), \(2\) or \(2\), \(1\) (the opposites are also a possibility, but since the middle term is possitive, we can disregard this possibility).

\[(x+1)(3x+2)\text{ or }(x+2)(3x+1)\nonumber\]

Which factors are correct? To decide that, we multiply the inner and outer terms.

\[3x^2+5x+2\text{or }3x^2+7x+2.\nonumber\]

Since the middle term of the trinomial is \(5x\), the factors in the first case will work. Let’s use FOIL to check.

\[(x+1)(3x+2)\nonumber\]\[3x^2+2x+3x+2\nonumber\]\[3x^2+5x+2\checkmark\nonumber\]

Our result of the factoring is:

\[3x^2+5x+2\nonumber\]\[(x+1)(3x+2)\nonumber\]

Remember, when the middle term is negative and the last term is positive, the signs in the binomials must both be negative.

Example \(\PageIndex{22}\)

Factor \(6b^2−13b+5\)using trial and error.

Solution

The trinomial is already in descending order.	\(6b^2−13b+5\)
Find the factors of the first term.
Find the factors of the last term. Consider the signs. Since the last term, \(5\), is positive its factors must both be positive or both be negative. The coefficient of the middle term is negative, so we use the negative factors.

Consider all the combinations of factors.

\(6b^2−13b+5\)
Possible factors	Product
\((b−1)(6b−5)\)	\(6b^2−11b+5\)
\((b−5)(6b−1)\)	\(6b^2−31b+5\)
\((2b−1)(3b−5)\)	\(6b^2−13b+5^∗\)
\((2b−5)(3b−1)\)	\(6b^2−17b+5\)

\(\begin{array} {ll} \text{The correct factors are those whose product} & \\ \text{is the original trinomial.} &(2b−1)(3b−5) \\ \text{Check by multiplying:} & \\ \hspace{50mm} (2b−1)(3b−5) & \\ \hspace{47mm} 6b^2−10b−3b+5 & \\ \hspace{50mm} 6b^2−13b+5\checkmark & \end{array} \)

When we factor an expression, we always look for a greatest common factor first. If the expression does not have a greatest common factor, there cannot be one in its factors either. This may help us eliminate some of the possible factor combinations.

Example \(\PageIndex{25}\)

Factor \(18x^2−37xy+15y^2\)using trial and error.

Solution

The trinomial is already in descending order.	\(18x^2−37xy+15y^2\)
Find the factors of the first term.
Find the factors of the last term. Consider the signs. Since 15 is positive and the coefficient of the middle term is negative, we use the negative factors.

Consider all the combinations of factors.

\(\begin{array} {ll} \text{The correct factors are those whose product is} & \\ \text{the original trinomial.} &(2x−3y)(9x−5y) \\ \text{Check by multiplying:} & \\ & \\ & \\ & \\ \hspace{50mm} (2x−3y)(9x−5y) & \\ \hspace{45mm}18x^2−10xy−27xy+15y^2 & \\ \hspace{47mm}18x^2−37xy+15y^2\checkmark & \end{array} \)

Don’t forget to look for a GCF first and remember if the leading coefficient is negative, so is the GCF.

Example \(\PageIndex{28}\)

Factor \(−10y^4−55y^3−60y^2\)using trial and error.

Solution

	\(\quad −10y^4−55y^3−60y^2\)
Notice the greatest common factor, so factor it first.	\(-5y^2(2y^2+11y+12)\)
Factor the trinomial.

Consider all the combinations.

\(\begin{array} {ll} \text{The correct factors are those whose product} & \\ \text{is the original trinomial. Remember to include} & \\ \text{the factor }−5^y2. &−5y^2(y+4)(2y+3) \\ \text{Check by multiplying:} & \\ \hspace{50mm} −5y^2(y+4)(2y+3) & \\ \hspace{45mm} −5y^2(2y^2+8y+3y+12) & \\ \hspace{47mm}−10y^4−55y^3−60y^2\checkmark & \end{array} \)

Factoring Trinomials of the Form \(ax^2+bx+c\) Using the “\(ac\)” Method

Another way to factor trinomials of the form \(ax^2+bx+c\) is the “\(ac\)” method. (The “\(ac\)” method is sometimes called the grouping method.) The “\(ac\)” method is actually an extension of the methods you used in the last section to factor trinomials with leading coefficient one. This method is very structured (that is step-by-step), and it always works!

The “\(ac\)” method is summarized here.

Don’t forget to look for a common factor!

Factoring Using Substitution (optional)

Sometimes a trinomial does not appear to be in the \(ax^2+bx+c\) form. However, we can often make a thoughtful substitution that will allow us to make it fit the \(ax^2+bx+c\) form. This is called factoring by substitution. It is standard to use \(u\) for the substitution.

In the \(ax^2+bx+c\), the middle term has a variable, \(x\), and its square, \(x^2\), is the variable part of the first term. Look for this relationship as you try to find a substitution.

Sometimes the expression to be substituted is not a monomial.

Key Concepts

• How to factor trinomials of the form \(x^2+bx+c\).
  1. Write the factors as two binomials with first terms x. \(\quad \begin{array}{l}x^2+bx+c \\ (x\quad)(x\quad)\end{array}\)
  2. Find two numbers \(m\) and \(n\) that
     \(\begin{array} {ll} \text{multiply to} &c,\space m·n=c \\ \text{add to} &b,\space m+n=b\end{array}\)
  3. Use \(m\) and \(n\) as the last terms of the factors. \(\qquad (x+m)(x+n)\)
  4. Check by multiplying the factors.
• Strategy for Factoring Trinomials of the Form \(x^2+bx+c\): When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors.

  For trinomials of the form: \(x^2+bx+c = (x+m)(x+n)\)

  When \(c\) is positive, \(m\) and \(n\) must have the same sign (and this will be the sign of \(b\)).

  Examples: \(x^2+5x+6=(x+2)(x+3)\), \(x^2−6x+8 = (x−4)(x−2)\)

  When \(c\) is negative, \(m\) and \(n\) have opposite signs. The larger of \(m\) and \(n\) will have the sign of \(b\).

  Examples: \(x^2+x−12=(x+4)(x−3)\), \(x^2−2x−15=(x−5)(x+3)\)

  Notice that, in the case when \(m\) and \(n\) have opposite signs, the sign of the one with the larger absolute value matches the sign of \(b\).

• How to factor trinomials of the form \(ax^2+bx+c\) using trial and error.
  1. Write the trinomial in descending order of degrees as needed.
  2. Factor any GCF.
  3. Find all the factor pairs of the first term.
  4. Find all the factor pairs of the third term.
  5. Test all the possible combinations of the factors until the correct product is found.
  6. Check by multiplying.
• How to factor trinomials of the form \(ax^2+bx+c\) using the “\(ac\)” method.
  1. Factor any GCF.
  2. Find the product \(ac\).
  3. Find two numbers \(m\) and \(n\) that:
     \(\begin{array} {ll} \text{Multiply to }ac. &m·n=a·c \\ \text{Add to }b. &m+n=b \\ &ax^2+bx+c\end{array}\)
  4. Split the middle term using \(m\) and \(n\). \(\quad ax^2+mx+nx+c\)
  5. Factor by grouping.
  6. Check by multiplying the factors.