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Updated on Oct 3, 2011

Life is good for only two things, discovering mathematics and teaching mathematics.

—Simeon Poisson (1781–1840) French Mathematician

In this lesson, you'll learn how to factor a trinomial into two binomials, and you'll learn how to factor the difference of perfect squares into two binomials.

Previously we learned how to factor a monomial out of a polynomial (Factoring Expressions Study Guide). We found the greatest common factor of the coefficients of each term, and we found the variables common to every term in the polynomial. Now, we will learn how to factor quadratic expressions into two binomials. Remember, a quadratic expression is an expression where the highest degree of a variable is 2.

We use FOIL to multiply two binomials. The result is often a trinomial. To break that trinomial back down into two binomials, we must do the reverse of FOIL. The expression x2 + 5x + 6 has three terms: an x2 term, an x term, and a constant. We start by setting up parentheses to show the two binomials that will be our factors:

(__ + __)(__ + __)

We must find the two terms that make up the first binomial and the two terms that make up the second binomial. Of the three terms in x2 + 5x + 6, it is the middle term, 5x, that is the trickiest, because it is likely the sum of two terms. Start by listing the factors of the first term and the last term.

Factors of x2: x
Factors of 6: 1, 2, 3, 6

The only way to form x2 is to multiply x by itself. Make the first term in each binomial x:

(x + __)(x + __)

The number 6 can be found by multiplying 1 and 6 or 2 and 3. When we list the factors of a number, we usually list only the positive factors, but when factoring a quadratic expression, it is just as likely that two negative numbers are the factors. The complete list of integer factors of 6 is –6, –3, –2, –1, 1, 2, 3, and 6.

We can use the middle term, 5x, to help us decide which factors to choose for 6. If the factors of 6 were negative, then when FOIL was used to make x2 + 5x + 6, the Outside and Inside products would be negative, and the middle term of x2 + 5x + 6 would be negative. Because 5x is positive, two positive numbers must have been multiplied. Let's try 1 and 6:

(x + 1)(x + 6)

Now that we've taken our first guess at the factors of x2 + 5x + 6, use FOIL to test if these factors are correct:

First: (x)(x) = x2
Outside: (x)(6) = 6x
Inside: (1)(x) = x
Last: (1)(6) = 6
x2 + 6x + x + 6 = x2 + 7x + 6, not x2 + 5x + 6

The middle term came out too large. Because the factors of 6 were both multiplied by x, which has a coefficient of 1, the sum of the factors of 6 are equal to the coefficient of the middle term of x2 + 5x + 6. We need two factors that multiply to 6 and add to 5. The factors 2 and 3 multiply to 6 and add to 5, so let's try those:

(x + 2)(x + 3)
Check with FOIL
First: (x)(x) = x2
Outside: (x)(3) = 3x
Inside: (2)(x) = 2x
Last: (2)(3) = 6
x2 + 3x + 2x + 6 = x2 + 5x + 6

#### Tip:

Try to begin by finding the first term of each binomial. Once you have limited the possibilities of those terms, you can begin to test different factors of the constant. Those factors will be the last term of each binomial. Keep trying different combinations of factors for the last terms until you find the pair that not only multiply to the constant of the polynomial, but also yield the correct middle term of the polynomial.

Once we found that 2 and 3 were the factors of 6 that we needed to use as the last terms for the binomials, it did not matter if we wrote (x + 2)(x + 3) or (x + 3)(x + 2). 2 could have been the last term of the first binomial or the second binomial. That's because the first term in each binomial was the same. If the first term in each binomial is different, we have to be more careful about where we place the last terms.

Factoring the polynomial 2x2 + x – 10 is a bit tougher, but we follow the same steps. List the factors of 2x2 and –10:

2x2: –2, –1, 1, 2, –2x, –x, x, 2x, 2x2
–10: –10, –5, –2, –1, 1, 2, 5, 10

In order for the first term of the polynomial 2x2 + x – 10 to be 2x2, the terms 2x and x must be multiplied. One of the terms cannot be x2, or else when that term is multiplied by another term that contains x, we would get an x3 term. There is no x3 term in 2x2 + x – 10. Make the first term of one binomial 2x and the first term of the other binomial x:

(2x + __)(x + __)

Now we need two constants that multiply to –10. The middle term of the polynomial is x, which means that the sum of the Outside and Inside products is 1x. The Outside product is made by multiplying 2x by one factor of –10. The Inside product is made by multiplying x by the other factor of –10. In other words, two times one factor of –10 plus one times the other factor of –10 add to 1. Let's try a few combinations:

(2x + 2)(x – 5)

These binomials multiply to 2x2 – 8x – 10. The middle term is too small. Try switching the signs:

(2x – 2)(x + 5)

These binomials multiply to 2x2 + 8x – 10. The middle term is too large. Try switching the constants:

(2x – 5)(x + 2)

These binomials multiply to 2x2x – 10. The middle term is too small, but we are getting close. Try changing the signs:

(2x + 5)(x – 2)

These binomials multiply to 2x2 + x – 10. We have found the right factors.

 Tip: It can take a little time to find the right combination of constants and signs when factoring a trinomial. Keep a list of the combinations you have tried, so that you do not try the same ones twice. Each time you try a wrong combination, try to learn from it. Ask yourself: How is the product of these factors different from the original polynomial? This can help you decide if you need to change constants or just change signs.

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