Main content

### Course: Algebra 1 > Unit 13

Lesson 5: Factoring quadratics intro- Factoring quadratics as (x+a)(x+b)
- Factoring quadratics: leading coefficient = 1
- Factoring quadratics as (x+a)(x+b) (example 2)
- More examples of factoring quadratics as (x+a)(x+b)
- Factoring quadratics intro
- Factoring quadratics with a common factor
- Factoring completely with a common factor
- Factoring quadratics with a common factor
- Factoring simple quadratics review

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 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

## Want to join the conversation?

- For question 5, why can't (x-10)(x+3) work?(27 votes)
- Here's the question: Factor x^2+7x-30x

The way Khan wants you to solve is this to find two numbers that add up to*positive*7 and multiply to negative 30. If you add those two numbers together, they add up to*negative*7. You were close, you just have to check that the numbers you get actually add and multiply correctly.

So the answer is (x+10)(x-3)

10+(-3) equals 7 and 10*(-3) equals 30.(78 votes)

- I can't be the only one who finds this the most confusing thing I've ever learned(26 votes)
- I can tell it needs some practice, don't worry(6 votes)

- On the "reflection question" there are two things wrong. The 2x^2 messes things up and also there are no two numbers that give you a sum of 3 and a product of 1.(10 votes)
- Good observation... You learn how to factor these types of quadratics in the next section: Factoring Quadratics 2.(15 votes)

- Reflection Q6: 2x^2+3x+1 = (2x+1)(x+1) and is = 2x(x+1) + 1(x+1) .(10 votes)
- It is a clarification, so it could be considered a question to clarify one's learning.(3 votes)

- The challenge problems are. Challenging.(17 votes)
- I mean, yeah, that's what they're there for, but if you don't challenge yourself, you'll never learn! 😊(1 vote)

- help me please(2 votes)
- I had a lot of trouble with this too, so don't worry. I'm going to use an expression as an example. Say we have the expression x^2+5x+6. We need to figure out what two integers will add up to 5, and when multiplied, will make 6. To figure this out, you have to guess and check. The best way is to find out what numbers make 6 when they are multiplied. 3 and 2 work here. When 2 and 3 are added together, they make five. Therefore, to correctly factor x^2+5x+6, you would say (x+3)(x+2), or you could do (x+3)(x+2). There are also times when the middle term or the third number is negative. For example, x^2+x-6. The first step would be to find what two numbers make 6 when they are multiplied. 2 and 3 do. And to make positive one with these two numbers, 2 has to be negative, so you would factor x^2+x-6 as (x-2)(x+3). Sometimes the middle term will be negative. Let's take another example. x^2-8x+16. First, what we would figure out is what to numbers make 8. Now, since the two numbers are both negative, because the middle term is negative and the constant is positive, we can disregard the negative in -8. Two numbers that would work are 4 and 4. Now we have to
**un**disregard the -8. So now, the numbers would be -4 and -4. -4*-4 is 16, so the factored form of x^2-8x+16 would be (x-4)(x-4). If both the middle term and the constant are negative, that means the greater number that makes the product is negative, and the smaller number is positive. For example, x^2-5x-14. The factored from would be (x-7)(x+2). I hope this helped, or at least made things a little bit more clear!(25 votes)

- For question 8 there can be another answer= (x^2-1)(-x^2-6)(3 votes)
- Sorry to tell you... your factors will not work.

Let's check them by multiplying:

(x^2-1)(-x^2-6) = -x^4 -6x^2 +x^2 + 6 = -x^4 -5x^2 + 6

Notice: You have the wrong sign on the x^4 term. So, your factors will not work.

Tip: If the leading term (the x^4) is positive, don't make one of them negative. A negative time a positive = a negative, not a positive.

Hope this helps.(21 votes)

- In practice question #3 would not (x-1)(x+9) work the same way as (x+1)(x-9)?(4 votes)
- No... your 2 examples are not the same. You can see this if you multiply the binomials.

(x-1)(x+9) = x^2 + 9x - x - 9 = x^2 + 8x - 9

(x+1)(x-9) = x^2 - 9x + x - 9 = x^2 - 8x - 9

Notice: the placement of the signs in your factors does matter. Practice question #3 wanted factors of x^2 - 8x -9, so the correct factors are (x+1)(x-9) because these would recreate the original polynomial.(10 votes)

- can there be more than one variable(4 votes)
- Yes! There can be more than one variable. We see this in challenge problem #7. When there are two variables, I prefer to solve by grouping as it makes more sense (at least to me). You can check out articles and videos under "Factoring Quadratics by Grouping". Assuming you took a look at those videos -

I solved number 7 by first setting up equations for which two numbers the middle term (5xy) will be split into: (f and g represent my two numbers - I randomly chose these variables)

f*g = 1*6 = 6

(the 1 comes from x^2 and the 6 comes from 6y^2 because our two numbers need to multiply to the coefficients of the two outer terms.)

f + g = 5 (our two numbers need to equal the coefficient of the middle term).

So, what two numbers multiply to 6 and add to 5? That's 3 and 2.

Now, we need to split up the 5xy term into two terms with coefficients 3 and 2. (3xy) and (2xy) will replace 5xy in the polynomial we are factoring. So now we have-

x^2 +**2xy**+**3xy**+ 6y^2

The order of these terms (2xy) and (3xy) does not matter in this particular problem, but it does in some (watch the videos on Factoring by Grouping for more details).

Now, we will separate the polynomial as shown and think of each part separately.**x^2 + 2xy**+*3xy + 6y^2*

Now we'll factor each separate part.

x(x+2y) +*3y(x+2y)*

We can then factor out the (x+2y) so that we are left with our finished product:

(x+2y) (3y+x)

Hope this helps! Don't forget to check out those videos if you are confused on any part of this explanation.(9 votes)

- By definition, what's a quadratic?(6 votes)
- quadratic polynomial is the one, who has unknown variable, x, raised to power of 2 and there is no higher term than x squared.(1 vote)