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, start superscript, 2, end superscript, plus, b, x, plus, c as a product of two binomials.

Review: Multiplying binomials

Let's consider the expression left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 4, right parenthesis.
We can find the product by applying the distributive property multiple times.
Some people use the acronym start color blueD, F, end color blueD, start color greenD, O, end color greenD, start color goldD, I, end color goldD, start color purpleC, L, end color purpleC to help them multiply two binomials.
This is a way to help make sure that each term in the first binomial multiplies each term in the second binomial. Specifically, it reminds them to multiply the start color blueD, F, i, r, s, t, end color blueD terms in the binomial, then the start color greenD, O, u, t, e, r, end color greenD terms, next the start color goldD, I, n, n, e, r, end color goldD terms, and finally the start color purpleC, L, a, s, t, end color purpleC terms.
For example, using FOIL to multiply left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 4, right parenthesis would look like this:
(x+2)(x+4)=xx+x4+2x+24=x2+4x+2x+8=x2+6x+8\begin{aligned}(x+2)(x+4)&=\blueD{x}\cdot \blueD x+\greenD x\cdot \greenD4+\goldD 2\cdot \goldD x+\purpleC 2\cdot \purpleC 4\\ \\ &=x^2+4x+2x+8\\ \\ &=x^2+6x+8 \end{aligned}
While this is a correct method, FOIL sounds somewhat mysterious. It is easier to understand that each term in the first binomial multiplies each term in the second binomial when we apply the distributive property twice.
So we have that left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 4, right parenthesis, equals, x, start superscript, 2, end superscript, plus, 6, x, plus, 8.
From this, we see that x, plus, 2 and x, plus, 4 are factors of x, start superscript, 2, end superscript, plus, 6, x, plus, 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, start superscript, 2, end superscript, plus, 6, x, plus, 8, we can use factoring to write it as a product of two binomials, left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 4, right parenthesis.
Let's take a look at a few examples to see how this is done.

Example 1: Factoring x, start superscript, 2, end superscript, plus, 5, x, plus, 6

To factor x, start superscript, 2, end superscript, plus, start color goldD, 5, end color goldD, x, plus, start color purpleC, 6, end color purpleC, we first need to find two numbers that multiply to start color purpleC, 6, end color purpleC (the constant number) and add up to start color goldD, 5, end color goldD (the x-coefficient).
These two numbers are start color blueD, 2, end color blueD and start color greenD, 3, end color greenD since start color blueD, 2, end color blueD, dot, start color greenD, 3, end color greenD, equals, 6 and start color blueD, 2, end color blueD, plus, start color greenD, 3, end color greenD, equals, 5.
We can create a list of all numbers that multiply to 6. Then, within that list, we look for those that sum to 5.
Product: m, dot, n, equals, 6spaceSum: m, plus, n
1, dot, 6, equals, 61, plus, 6, equals, 7
start color blueD, 2, end color blueD, dot, start color greenD, 3, end color greenD, equals, 6start color blueD, 2, end color blueD, plus, start color greenD, 3, end color greenD, equals, 5
left parenthesis, minus, 1, right parenthesis, dot, left parenthesis, minus, 6, right parenthesis, equals, 6left parenthesis, minus, 1, right parenthesis, plus, left parenthesis, minus, 6, right parenthesis, equals, minus, 7
left parenthesis, minus, 2, right parenthesis, dot, left parenthesis, minus, 3, right parenthesis, equals, 6left parenthesis, minus, 2, right parenthesis, plus, left parenthesis, minus, 3, right parenthesis, equals, minus, 5
We can then add each of these numbers to x to form the two binomial factors: left parenthesis, x, plus, start color blueD, 2, end color blueD, right parenthesis and left parenthesis, x, plus, start color greenD, 3, end color greenD, right parenthesis.
In conclusion, we factored the trinomial as follows:
x, start superscript, 2, end superscript, plus, 5, x, plus, 6, equals, left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 3, right parenthesis
To check the factorization, we can multiply the two binomials:
(x+2)(x+3)=(x+2)(x)+(x+2)(3)=x2+2x+3x+6=x2+5x+6\begin{aligned}(x+2)(x+3)&=(x+2)(x)+(x+2)(3)\\ \\ &=x^2+2x+3x+6\\ \\ &=x^2+5x+6 \end{aligned}
The product of x, plus, 2 and x, plus, 3 is indeed x, start superscript, 2, end superscript, plus, 5, x, plus, 6. Our factorization is correct!

Check your understanding

1) Factor x, start superscript, 2, end superscript, plus, 7, x, plus, 10.
Choose 1 answer:
Choose 1 answer:

To factor x, start superscript, 2, end superscript, plus, start color goldD, 7, end color goldD, x, plus, start color purpleC, 10, end color purpleC, we need to find factors of start color purpleC, 10, end color purpleC that add up to start color goldD, 7, end color goldD.
The table below shows all possible factors of 10 and their sums.
Product: m, dot, n, equals, 10spaceSum: m, plus, n
1, dot, 10, equals, 101, plus, 10, equals, 11
start color blueD, 2, end color blueD, dot, start color greenD, 5, end color greenD, equals, 10start color blueD, 2, end color blueD, plus, start color greenD, 5, end color greenD, equals, 7
left parenthesis, minus, 1, right parenthesis, dot, left parenthesis, minus, 10, right parenthesis, equals, 10left parenthesis, minus, 1, right parenthesis, plus, left parenthesis, minus, 10, right parenthesis, equals, minus, 11
left parenthesis, minus, 2, right parenthesis, dot, left parenthesis, minus, 5, right parenthesis, equals, 10left parenthesis, minus, 2, right parenthesis, plus, left parenthesis, minus, 5, right parenthesis, equals, minus, 7
Only one pair of factors satisfies the conditions of the problem: start color blueD, 2, end color blueD and start color greenD, 5, end color greenD.
We can add each of these numbers to x to form the two binomial factors: left parenthesis, x, plus, start color blueD, 2, end color blueD, right parenthesis and left parenthesis, x, plus, start color greenD, 5, end color greenD, right parenthesis. The factorization of the polynomial is given below.
x, start superscript, 2, end superscript, plus, 7, x, plus, 10, equals, left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 5, right parenthesis
2) Factor x, start superscript, 2, end superscript, plus, 9, x, plus, 20.

To factor x, start superscript, 2, end superscript, plus, start color goldD, 9, end color goldD, x, plus, start color purpleC, 20, end color purpleC, we need to find factors of start color purpleC, 20, end color purpleC that add up to start color goldD, 9, end color goldD.
The table below shows all possible factors of 20 and their sums.
Product: m, dot, n, equals, 20spaceSum: m, plus, n
1, dot, 20, equals, 201, plus, 20, equals, 21
2, dot, 10, equals, 202, plus, 10, equals, 12
start color blueD, 4, end color blueD, dot, start color greenD, 5, end color greenD, equals, 20start color blueD, 4, end color blueD, plus, start color greenD, 5, end color greenD, equals, 9
left parenthesis, minus, 1, right parenthesis, dot, left parenthesis, minus, 20, right parenthesis, equals, 20left parenthesis, minus, 1, right parenthesis, plus, left parenthesis, minus, 20, right parenthesis, equals, minus, 21
left parenthesis, minus, 2, right parenthesis, dot, left parenthesis, minus, 10, right parenthesis, equals, 20left parenthesis, minus, 2, right parenthesis, plus, left parenthesis, minus, 10, right parenthesis, equals, minus, 12
left parenthesis, minus, 4, right parenthesis, dot, left parenthesis, minus, 5, right parenthesis, equals, 20left parenthesis, minus, 4, right parenthesis, plus, left parenthesis, minus, 5, right parenthesis, equals, minus, 9
Only one pair of factors satisfies the conditions of the problem: start color blueD, 4, end color blueD and start color greenD, 5, end color greenD.
We can add each of these numbers to x to form the two binomial factors: left parenthesis, x, plus, start color blueD, 4, end color blueD, right parenthesis and left parenthesis, x, plus, start color greenD, 5, end color greenD, right parenthesis. The factorization of the polynomial is given below.
x, start superscript, 2, end superscript, plus, 9, x, plus, 20, equals, left parenthesis, x, plus, 4, right parenthesis, left parenthesis, x, plus, 5, right parenthesis
Let's take a look at a few more examples and see what we can learn from them.

Example 2: Factoring x, start superscript, 2, end superscript, minus, 5, x, plus, 6

To factor x, start superscript, 2, end superscript, start color goldD, minus, 5, end color goldD, x, plus, start color purpleC, 6, end color purpleC, let's first find two numbers that multiply to start color purpleC, 6, end color purpleC and add up to start color goldD, minus, 5, end color goldD.
These two numbers are start color blueD, minus, 2, end color blueD and start color greenD, minus, 3, end color greenD since left parenthesis, start color blueD, minus, 2, end color blueD, right parenthesis, dot, left parenthesis, start color greenD, minus, 3, end color greenD, right parenthesis, equals, 6 and left parenthesis, start color blueD, minus, 2, end color blueD, right parenthesis, plus, left parenthesis, start color greenD, minus, 3, end color greenD, right parenthesis, equals, minus, 5.
We can create a list of all numbers that multiply to 6. Then, within that list, we look for those that sum to minus, 5.
Product: m, dot, n, equals, 6spaceSum: m, plus, n
1, dot, 6, equals, 61, plus, 6, equals, 7
2, dot, 3, equals, 62, plus, 3, equals, 5
left parenthesis, minus, 1, right parenthesis, dot, left parenthesis, minus, 6, right parenthesis, equals, 6left parenthesis, minus, 1, right parenthesis, plus, left parenthesis, minus, 6, right parenthesis, equals, minus, 7
left parenthesis, start color blueD, minus, 2, end color blueD, right parenthesis, dot, left parenthesis, start color greenD, minus, 3, end color greenD, right parenthesis, equals, 6left parenthesis, start color blueD, minus, 2, end color blueD, right parenthesis, plus, left parenthesis, start color greenD, minus, 3, end color greenD, right parenthesis, equals, minus, 5
We can then add each of these numbers to x to form the two binomial factors: left parenthesis, x, plus, left parenthesis, start color blueD, minus, 2, end color blueD, right parenthesis, right parenthesis and left parenthesis, x, plus, left parenthesis, start color greenD, minus, 3, end color greenD, right parenthesis, right parenthesis.
The factorization is given below:
x25x+6=(x+(2))(x+(3))=(x2)(x3)\begin{aligned}x^2-5x+6&=(x+(\blueD{-2}))(x+(\greenD{-3}))\\ \\ &=(x\blueD{-2})(x\greenD{-3}) \end{aligned}
We can check the factorization by multiplying the two binomials:
(x2)(x3)=x23x2x+6=x25x+6\begin{aligned}(x-2)(x-3)&=x^2-3x-2x+6\\ \\ &=x^2-5x+6 \end{aligned}
The product of x, minus, 2 and x, minus, 3 is indeed x, start superscript, 2, end superscript, minus, 5, x, plus, 6. Our factorization is correct!
Factoring pattern: Notice that the numbers needed to factor x, start superscript, 2, end superscript, minus, 5, x, plus, 6 are both negative left parenthesis, minus, 2 and minus, 3, right parenthesis. This is because their product needs to be positive left parenthesis, 6, right parenthesis and their sum negative left parenthesis, minus, 5, right parenthesis.
In general, when factoring x, start superscript, 2, end superscript, plus, b, x, plus, c, if c is positive and b is negative, then both factors will be negative!

Example 3: Factoring x, start superscript, 2, end superscript, minus, x, minus, 6

We can write x, start superscript, 2, end superscript, minus, x, minus, 6 as x, start superscript, 2, end superscript, minus, 1, x, minus, 6.
To factor x, start superscript, 2, end superscript, start color goldD, minus, 1, end color goldD, x, start color purpleC, minus, 6, end color purpleC, let's first find two numbers that multiply to start color purpleC, minus, 6, end color purpleC and add up to start color goldD, minus, 1, end color goldD.
These two numbers are start color blueD, 2, end color blueD and start color greenD, minus, 3, end color greenD since left parenthesis, start color blueD, 2, end color blueD, right parenthesis, dot, left parenthesis, start color greenD, minus, 3, end color greenD, right parenthesis, equals, minus, 6 and start color blueD, 2, end color blueD, plus, left parenthesis, start color greenD, minus, 3, end color greenD, right parenthesis, equals, minus, 1.
We can create a list of all numbers that multiply to minus, 6. Then, within that list, we look for those that sum to minus, 1.
Product: m, dot, n, equals, minus, 6spaceSum: m, plus, n
left parenthesis, minus, 1, right parenthesis, dot, 6, equals, minus, 6left parenthesis, minus, 1, right parenthesis, plus, 6, equals, 5
1, dot, left parenthesis, minus, 6, right parenthesis, equals, minus, 61, plus, left parenthesis, minus, 6, right parenthesis, equals, minus, 5
left parenthesis, minus, 2, right parenthesis, dot, 3, equals, minus, 6left parenthesis, minus, 2, right parenthesis, plus, 3, equals, 1
left parenthesis, start color blueD, 2, end color blueD, right parenthesis, dot, left parenthesis, start color greenD, minus, 3, end color greenD, right parenthesis, equals, minus, 6left parenthesis, start color blueD, 2, end color blueD, right parenthesis, plus, left parenthesis, start color greenD, minus, 3, end color greenD, right parenthesis, equals, minus, 1
We can then add each of these numbers to x to form the two binomial factors: left parenthesis, x, plus, start color blueD, 2, end color blueD, right parenthesis and left parenthesis, x, plus, left parenthesis, start color greenD, minus, 3, end color greenD, right parenthesis, right parenthesis.
The factorization is given below:
x2x6=(x+2)(x+(3))=(x+2)(x3)\begin{aligned}x^2-x-6&=(x+\blueD2)(x+(\greenD{-3}))\\ \\ &=(x+\blueD2)(x\greenD{-3}) \end{aligned}
We can check the factorization by multiplying the two binomials:
(x+2)(x3)=x23x+2x6=x2x6\begin{aligned}(x+2)(x-3)&=x^2-3x+2x-6\\ \\ &=x^2-x-6 \end{aligned}
The product of x, plus, 2 and x, minus, 3 is indeed x, start superscript, 2, end superscript, minus, x, minus, 6. Our factorization is correct!
Factoring patterns: Notice that to factor x, start superscript, 2, end superscript, minus, x, minus, 6, we need one positive number left parenthesis, 2, right parenthesis and one negative number left parenthesis, minus, 3, right parenthesis. This is because their product needs to be negative left parenthesis, minus, 6, right parenthesis.
In general, when factoring x, start superscript, 2, end superscript, plus, b, x, plus, 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, start superscript, 2, end superscript, plus, start color goldD, b, end color goldD, x, plus, start color purpleC, c, end color purpleC, we need to find factors of start color purpleC, c, end color purpleC that add up to start color goldD, b, end color goldD.
Suppose these two numbers are m and n so that c, equals, m, n and b, equals, m, plus, n, then x, start superscript, 2, end superscript, plus, b, x, plus, c, equals, left parenthesis, x, plus, m, right parenthesis, left parenthesis, x, plus, n, right parenthesis.

Check your understanding

3) Factor x, start superscript, 2, end superscript, minus, 8, x, minus, 9.

To factor x, start superscript, 2, end superscript, start color goldD, minus, 8, end color goldD, x, start color purpleC, minus, 9, end color purpleC, we need to find factors of start color purpleC, minus, 9, end color purpleC that add up to start color goldD, minus, 8, end color goldD.
Since the two numbers must multiply to minus, 9, one must be positive and one must be negative.
The factors of minus, 9 are listed in the table below.
Product: m, dot, n, equals, minus, 9spaceSum: m, plus, n
start color blueD, 1, end color blueD, dot, left parenthesis, start color greenD, minus, 9, end color greenD, right parenthesis, equals, minus, 9start color blueD, 1, end color blueD, plus, left parenthesis, start color greenD, minus, 9, end color greenD, right parenthesis, equals, minus, 8
left parenthesis, minus, 1, right parenthesis, dot, 9, equals, minus, 9minus, 1, plus, 9, equals, 8
left parenthesis, minus, 3, right parenthesis, dot, 3, equals, minus, 9minus, 3, plus, 3, equals, 0
Notice here that start color blueD, 1, end color blueD and start color greenD, minus, 9, end color greenD satisfy these requirements. We can add each of these numbers to x to form the two binomial factors: left parenthesis, x, plus, start color blueD, 1, end color blueD, right parenthesis and left parenthesis, x, plus, left parenthesis, start color greenD, minus, 9, end color greenD, right parenthesis, right parenthesis. The factorization of the polynomial is given below.
x28x9=(x+1)(x+(9))=(x+1)(x9)\begin{aligned}x^2-8x-9&=(x+\blueD1)(x+(\greenD{-9}))\\ \\ &=(x+1)(x-9) \end{aligned}
4) Factor x, start superscript, 2, end superscript, minus, 10, x, plus, 24.

To factor x, start superscript, 2, end superscript, start color goldD, minus, 10, end color goldD, x, plus, start color purpleC, 24, end color purpleC, we need to find factors of start color purpleC, 24, end color purpleC that add up to start color goldD, minus, 10, end color goldD.
Since the two numbers must multiply to 24 and add to minus, 10, both numbers must be negative.
The factors of 24 (where both are negative) are listed in the table below.
Product: m, dot, n, equals, 24spaceSum: m, plus, n
left parenthesis, minus, 1, right parenthesis, dot, left parenthesis, minus, 24, right parenthesis, equals, 24minus, 1, plus, left parenthesis, minus, 24, right parenthesis, equals, minus, 25
left parenthesis, minus, 2, right parenthesis, dot, left parenthesis, minus, 12, right parenthesis, equals, 24minus, 2, plus, left parenthesis, minus, 12, right parenthesis, equals, minus, 14
left parenthesis, minus, 3, right parenthesis, dot, left parenthesis, minus, 8, right parenthesis, equals, 24minus, 3, plus, left parenthesis, minus, 8, right parenthesis, equals, minus, 11
left parenthesis, start color blueD, minus, 4, end color blueD, right parenthesis, dot, left parenthesis, start color greenD, minus, 6, end color greenD, right parenthesis, equals, 24start color blueD, minus, 4, end color blueD, plus, left parenthesis, start color greenD, minus, 6, end color greenD, right parenthesis, equals, minus, 10
Notice here that start color blueD, minus, 4, end color blueD and start color greenD, minus, 6, end color greenD satisfy these requirements. We can add each of these numbers to x to form the two binomial factors: left parenthesis, x, plus, left parenthesis, start color blueD, minus, 4, end color blueD, right parenthesis, right parenthesis and left parenthesis, x, plus, left parenthesis, start color greenD, minus, 6, end color greenD, right parenthesis, right parenthesis. The factorization of the polynomial is given below.
x210x+24=(x+(4))(x+(6))=(x4)(x6)\begin{aligned}x^2-10x+24&=(x+(\blueD{-4}))(x+(\greenD{-6}))\\ \\ &=(x-4)(x-6) \end{aligned}
5) Factor x, start superscript, 2, end superscript, plus, 7, x, minus, 30.

To factor x, start superscript, 2, end superscript, plus, start color goldD, 7, end color goldD, x, start color purpleC, minus, 30, end color purpleC, we need to find factors of start color purpleC, minus, 30, end color purpleC that add up to start color goldD, 7, end color goldD.
Since the two numbers must multiply to minus, 30, one must be positive and one must be negative.
The factors of minus, 30 are listed in the table below.
Product: m, dot, n, equals, minus, 30spaceSum: m, plus, n
1, dot, left parenthesis, minus, 30, right parenthesis, equals, minus, 301, plus, left parenthesis, minus, 30, right parenthesis, equals, minus, 29
left parenthesis, minus, 1, right parenthesis, dot, 30, equals, minus, 30minus, 1, plus, 30, equals, 29
2, dot, left parenthesis, minus, 15, right parenthesis, equals, minus, 302, plus, left parenthesis, minus, 15, right parenthesis, equals, minus, 13
left parenthesis, minus, 2, right parenthesis, dot, 15, equals, minus, 30minus, 2, plus, 15, equals, 13
3, dot, left parenthesis, minus, 10, right parenthesis, equals, minus, 303, plus, left parenthesis, minus, 10, right parenthesis, equals, minus, 7
left parenthesis, start color blueD, minus, 3, end color blueD, right parenthesis, dot, start color greenD, 10, end color greenD, equals, minus, 30start color blueD, minus, 3, end color blueD, plus, start color greenD, 10, end color greenD, equals, 7
5, dot, left parenthesis, minus, 6, right parenthesis, equals, minus, 305, plus, left parenthesis, minus, 6, right parenthesis, equals, minus, 1
left parenthesis, minus, 5, right parenthesis, dot, 6, equals, minus, 30minus, 5, plus, 6, equals, 1
Notice here that start color blueD, minus, 3, end color blueD and start color greenD, 10, end color greenD satisfy these requirements. We can add each of these numbers to x to form the two binomial factors: left parenthesis, x, plus, start color blueD, minus, 3, end color blueD, right parenthesis and left parenthesis, x, plus, left parenthesis, start color greenD, 10, end color greenD, right parenthesis, right parenthesis. The factorization of the polynomial is given below.
x2+7x30=(x+3)(x+(10))=(x3)(x+10)\begin{aligned}x^2+7x-30&=(x+\blueD{-3})(x+(\greenD{10}))\\ \\ &=(x-3)(x+10) \end{aligned}

Why does this work?

To understand why this factorization method works, let's return to the original example in which we factored x, start superscript, 2, end superscript, plus, 5, x, plus, 6 as left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 3, right parenthesis.
If we go back and multiply the two binomial factors, we can see the effect that the start color blueD, 2, end color blueD and the start color greenD, 3, end color greenD have on forming the product x, start superscript, 2, end superscript, plus, 5, x, plus, 6.
(x+2)(x+3)=(x+2)(x)+(x+2)(3)=x2+2x+3x+23=x2+(2+3)x+23\begin{aligned}(x+\blueD 2)(x+\greenD3)&={(x+\blueD2)}(x)+(x+\blueD 2)(\greenD{3})\\ \\ &=x^2+\blueD2x+\greenD3x+\blueD2\cdot \greenD3\\ \\ &=x^2+(\blueD 2+\greenD 3)x+\blueD2\cdot \greenD3 \end{aligned}
We see that the coefficient of the x-term is the sum of start color blueD, 2, end color blueD and start color greenD, 3, end color greenD, and the constant term is the product of start color blueD, 2, end color blueD and start color greenD, 3, end color greenD.

The sum-product pattern

Let's repeat what we just did with left parenthesis, x, plus, start color blueD, 2, end color blueD, right parenthesis, left parenthesis, x, plus, start color greenD, 3, end color greenD, right parenthesis for left parenthesis, x, plus, start color blueD, m, end color blueD, right parenthesis, left parenthesis, x, plus, start color greenD, n, end color greenD, right parenthesis:
(x+m)(x+n)=(x+m)(x)+(x+m)(n)=x2+mx+nx+mn=x2+(m+n)x+mn\begin{aligned}(x+\blueD m)(x+\greenD n)&={(x+\blueD m)}(x)+(x+\blueD m)(\greenD{n})\\ \\ &=x^2+\blueD mx+\greenD nx+\blueD m\cdot \greenD n\\ \\ &=x^2+(\blueD m+\greenD n)x+\blueD m\cdot \greenD n \end{aligned}
To summarize this process, we get the following equation:
left parenthesis, x, plus, start color blueD, m, end color blueD, right parenthesis, left parenthesis, x, plus, start color greenD, n, end color greenD, right parenthesis, equals, x, start superscript, 2, end superscript, plus, left parenthesis, start color blueD, m, end color blueD, plus, start color greenD, n, end color greenD, right parenthesis, x, plus, start color blueD, m, end color blueD, dot, start color greenD, n, end color greenD
This is called the sum-product pattern.
It shows why, once we express a trinomial x, start superscript, 2, end superscript, plus, start color goldD, b, end color goldD, x, plus, start color purpleC, c, end color purpleC as x, start superscript, 2, end superscript, plus, left parenthesis, start color blueD, m, end color blueD, plus, start color greenD, n, end color greenD, right parenthesis, x, plus, start color blueD, m, end color blueD, dot, start color greenD, n, end color greenD (by finding two numbers start color blueD, m, end color blueD and start color greenD, n, end color greenD so start color goldD, b, end color goldD, equals, start color blueD, m, end color blueD, plus, start color greenD, n, end color greenD and start color purpleC, c, end color purpleC, equals, start color blueD, m, end color blueD, dot, start color greenD, n, end color greenD), we can factor that trinomial as left parenthesis, x, plus, start color blueD, m, end color blueD, right parenthesis, left parenthesis, x, plus, start color greenD, n, end color greenD, right parenthesis.

Reflection question

6) Can this factorization method be used to factor 2, x, start superscript, 2, end superscript, plus, 3, x, plus, 1?
Choose 1 answer:
Choose 1 answer:

In this method, we factor the trinomial by equating it to left parenthesis, x, plus, m, right parenthesis, left parenthesis, x, plus, n, right parenthesis for some integers m and n. This means that the leading term of this polynomial will always be x, start superscript, 2, end superscript. Since the leading term of this polynomial is 2, x, start superscript, 2, end superscript, it cannot be factored using this method.

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 left parenthesis, x, plus, m, right parenthesis, left parenthesis, x, plus, n, right parenthesis for some integers m and n.
This means that the leading term of the trinomial must be x, start superscript, 2, end superscript (and not, for instance, 2, x, start superscript, 2, end superscript) in order to even consider this method. This is because the product of left parenthesis, x, plus, m, right parenthesis and left parenthesis, x, plus, n, right parenthesis will always be a polynomial with a leading term of x, start superscript, 2, end superscript.
However, not all trinomials with x, start superscript, 2, end superscript as a leading term can be factored. For example, x, start superscript, 2, end superscript, plus, 2, x, plus, 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

7*) Factor x, start superscript, 2, end superscript, plus, 5, x, y, plus, 6, y, start superscript, 2, end superscript.

Recall the following factorization:
x, start superscript, 2, end superscript, plus, 5, x, plus, 6, equals, left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 3, right parenthesis
The polynomial x, start superscript, 2, end superscript, plus, 5, x, y, plus, 6, y, start superscript, 2, end superscript is very similar to x, start superscript, 2, end superscript, plus, 5, x, plus, 6. The only difference is that the middle term is multiplied by y and the last term is multiplied by y, start superscript, 2, end superscript.
We can obtain this by adding a factor of start color tealD, y, end color tealD to each binomial:
(x+2y)(x+3y)=x2+3xy+2xy+6y2=x2+5xy+6y2\begin{aligned}(x+2\tealD y)(x+3\tealD y) &=x^2+3x\tealD y +2x\tealD y+6\tealD{y^2}\\ \\ &=x^2+5x\tealD y+6\tealD{y^2} \end{aligned}
The factorization is left parenthesis, x, plus, 2, y, right parenthesis, left parenthesis, x, plus, 3, y, right parenthesis.
8*) Factor x, start superscript, 4, end superscript, minus, 5, x, start superscript, 2, end superscript, plus, 6.

This polynomial is of the form x, start superscript, 2, end superscript, plus, b, x, plus, c, although a substitution makes this easier to see.
Let start color tealD, Y, end color tealD, equals, start color tealD, x, start superscript, 2, end superscript, end color tealD. We can now write the polynomial as follows:
x45x2+6=(x2)25x2+6=Y25Y+6\begin{aligned}x^4-5x^2+6&=(\tealD{x^2})^2-5\tealD{x^2}+6\\ \\ &=\tealD{Y}^2-5\tealD{Y}+6\\ \end{aligned}
Since left parenthesis, minus, 2, right parenthesis, dot, left parenthesis, minus, 3, right parenthesis, equals, 6 and left parenthesis, minus, 2, right parenthesis, plus, left parenthesis, minus, 3, right parenthesis, equals, minus, 5, the polynomial factors as follows:
empty space, equals, left parenthesis, start color tealD, Y, end color tealD, minus, 2, right parenthesis, left parenthesis, start color tealD, Y, end color tealD, minus, 3, right parenthesis
Now, since start color tealD, Y, end color tealD, equals, start color tealD, x, start superscript, 2, end superscript, end color tealD, we can back substitute to find a factorization of the original polynomial.
x45x2+6=(x22)(x23)\begin{aligned}\phantom{x^4-5x^2+6}&= (\tealD{x^2}-2)(\tealD{x^2}-3) \end{aligned}
The factorization is left parenthesis, x, start superscript, 2, end superscript, minus, 2, right parenthesis, left parenthesis, x, start superscript, 2, end superscript, minus, 3, right parenthesis.