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### Course: Algebra (all content) > Unit 10

Lesson 14: Factoring quadratics intro# Factoring quadratics: leading coefficient = 1

Learn how to factor quadratic expressions as the product of two linear binomials. For example, x²+5x+6=(x+2)(x+3).

#### What you need to know for this lesson

Factoring a polynomial involves writing it as a product of two or more polynomials. It reverses the process of polynomial multiplication. For more on this, check out our previous article on taking common factors.

#### What you will learn in this lesson

In this lesson, you will learn how to factor a polynomial of the form ${x}^{2}+bx+c$ as a product of two binomials.

## Review: Multiplying binomials

Let's consider the expression $(x+2)(x+4)$ .

We can find the product by applying the distributive property multiple times.

So we have that $(x+2)(x+4)={x}^{2}+6x+8$ .

From this, we see that $x+2$ and $x+4$ are factors of ${x}^{2}+6x+8$ , but how would we find these factors if we didn't start with them?

## Factoring trinomials

We can reverse the process of binomial multiplication shown above in order to factor a trinomial (which is a polynomial with $3$ terms).

In other words, if we start with the polynomial ${x}^{2}+6x+8$ , we can use factoring to write it as a product of two binomials, $(x+2)(x+4)$ .

Let's take a look at a few examples to see how this is done.

### Example 1: Factoring ${x}^{2}+5x+6$

To factor ${x}^{2}+{5}x+{6}$ , we first need to find two numbers that multiply to ${6}$ (the constant number) and add up to ${5}$ (the $x$ -coefficient).

These two numbers are ${2}$ and ${3}$ since ${2}\cdot {3}=6$ and ${2}+{3}=5$ .

We can then add each of these numbers to $x$ to form the two binomial factors: $(x+{2})$ and $(x+{3})$ .

In conclusion, we factored the trinomial as follows:

To check the factorization, we can multiply the two binomials:

The product of $x+2$ and $x+3$ is indeed ${x}^{2}+5x+6$ . Our factorization is correct!

### Check your understanding

Let's take a look at a few more examples and see what we can learn from them.

### Example 2: Factoring ${x}^{2}-5x+6$

To factor ${x}^{2}{-5}x+{6}$ , let's first find two numbers that multiply to ${6}$ and add up to ${-5}$ .

These two numbers are ${-2}$ and ${-3}$ since $({-2})\cdot ({-3})=6$ and $({-2})+({-3})=-5$ .

We can then add each of these numbers to $x$ to form the two binomial factors: $(x+({-2}))$ and $(x+({-3}))$ .

The factorization is given below:

**Factoring pattern:**Notice that the numbers needed to factor

In general, when factoring ${x}^{2}+bx+c$ , if $c$ is positive and $b$ is negative, then both factors will be negative!

### Example 3: Factoring ${x}^{2}-x-6$

We can write ${x}^{2}-x-6$ as ${x}^{2}-1x-6$ .

To factor ${x}^{2}{-1}x{-6}$ , let's first find two numbers that multiply to ${-6}$ and add up to ${-1}$ .

These two numbers are ${2}$ and ${-3}$ since $({2})\cdot ({-3})=-6$ and ${2}+({-3})=-1$ .

We can then add each of these numbers to $x$ to form the two binomial factors: $(x+{2})$ and $(x+({-3}))$ .

The factorization is given below:

**Factoring patterns:**Notice that to factor

In general, when factoring ${x}^{2}+bx+c$ , if $c$ is negative, then one factor will be positive and one factor will be negative.

## Summary

In general, to factor a trinomial of the form ${x}^{2}+{b}x+{c}$ , we need to find factors of ${c}$ that add up to ${b}$ .

Suppose these two numbers are $m$ and $n$ so that $c=mn$ and $b=m+n$ , then ${x}^{2}+bx+c=(x+m)(x+n)$ .

### Check your understanding

## Why does this work?

To understand why this factorization method works, let's return to the original example in which we factored ${x}^{2}+5x+6$ as $(x+2)(x+3)$ .

If we go back and multiply the two binomial factors, we can see the effect that the ${2}$ and the ${3}$ have on forming the product ${x}^{2}+5x+6$ .

We see that the coefficient of the $x$ -term is ${2}$ and ${3}$ , and the constant term is ${2}$ and ${3}$ .

*the sum*of*the product*of## The sum-product pattern

Let's repeat what we just did with $(x+{2})(x+{3})$ for $(x+{m})(x+{n})$ :

To summarize this process, we get the following equation:

This is called

**the sum-product pattern**.It shows why, once we express a trinomial ${x}^{2}+{b}x+{c}$ as ${x}^{2}+({m}+{n})x+{m}\cdot {n}$ (by finding two numbers ${m}$ and ${n}$ so ${b}={m}+{n}$ and ${c}={m}\cdot {n}$ ), we can factor that trinomial as $(x+{m})(x+{n})$ .

### Reflection question

## When can we use this method to factor?

In general, the sum-product method is only applicable when we can actually write a trinomial as $(x+m)(x+n)$ for some integers $m$ and $n$ .

This means that the leading term of the trinomial must be ${x}^{2}$ (and not, for instance, $2{x}^{2}$ ) in order to even consider this method. This is because the product of $(x+m)$ and $(x+n)$ will always be a polynomial with a leading term of ${x}^{2}$ .

However, not all trinomials with ${x}^{2}$ as a leading term can be factored. For example, ${x}^{2}+2x+2$ cannot be factored because there are no two integers whose sum is $2$ and whose product is $2$ .

In future lessons we will learn more ways of factoring more types of polynomials.

## Challenge problems

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