Learn on PengienVision, Algebra 1Chapter 7: Polynomials and Factoring

Lesson 5: Factoring x² + bx + c

In this Grade 11 enVision Algebra 1 lesson, students learn to factor quadratic trinomials of the form x² + bx + c by identifying factor pairs of c whose sum equals b. The lesson covers all sign cases — when b and c are positive, when b is negative and c is positive, and when c is negative — as well as factoring two-variable trinomials of the form x² + bxy + cy². Students connect the factoring process to multiplying binomials, reinforcing the relationship between these inverse operations.

Section 1

Factoring quadratic trinomials

Property

To factor $x^2 + bx + c$ , we look for two numbers $p$ and $q$ so that

$pq = c$ and $p + q = b$

When we expand the factored form $(x + p)(x + q)$ , we get $x^2 + (p + q)x + pq$ .

Section 2

Sign patterns for factoring

Property

Assume that $b$ , $c$ , $p$ , and $q$ are positive integers.

$x^2 + bx + c = (x + p)(x + q)$ . If all the coefficients of the trinomial are positive, then both $p$ and $q$ are positive.
$x^2 - bx + c = (x - p)(x - q)$ . If the linear term of the trinomial is negative and the other two terms positive, then $p$ and $q$ are both negative.
$x^2 \pm bx - c = (x + p)(x - q)$ . If the constant term of the trinomial is negative, then $p$ and $q$ have opposite signs.

Examples

To factor $x^2 - 7x + 10$ , the constant term is positive and the middle term is negative. So, we need two negative numbers that multiply to 10 and add to -7. The factors are $(x - 2)(x - 5)$ .
In $y^2 + 4y - 12$ , the constant term is negative, so the factors have opposite signs. We need numbers that multiply to -12 and add to 4. The factorization is $(y + 6)(y - 2)$ .
For $z^2 - z - 30$ , the negative constant term means opposite signs. We need numbers that multiply to -30 and add to -1. The correct pair is -6 and 5, so we get $(z - 6)(z + 5)$ .

Explanation

The signs in the trinomial give you huge clues! A positive last term means the signs in your factors are the same. A negative last term means the signs are different. Use this to factor faster!

Section 3

Factor Trinomials of the Form x²+bx+c

Property

To factor a trinomial of the form $x^2 + bx + c$ means to start with the product and end with the factors, $(x+m)(x+n)$ . To get the correct factors, we find two numbers $m$ and $n$ whose product is $c$ and sum is $b$ .

How to factor trinomials of the form $x^2 + bx + c$ :

Write the factors as two binomials with first terms $x$ : $(x \quad)(x \quad)$ .
Find two numbers $m$ and $n$ that multiply to $c$ ( $m \cdot n = c$ ) and add to $b$ ( $m+n = b$ ).
Use $m$ and $n$ as the last terms of the factors: $(x+m)(x+n)$ .
Check by multiplying the factors.

Examples

To factor $x^2 + 9x + 20$ , we need two numbers that multiply to 20 and add to 9. The numbers are 4 and 5. So, the factors are $(x+4)(x+5)$ .
To factor $a^2 - 8a + 15$ , we need two numbers that multiply to 15 and add to -8. The numbers are -3 and -5. Thus, the factors are $(a-3)(a-5)$ .
To factor $p^2 + 3p - 10$ , we need two numbers that multiply to -10 and add to 3. The numbers are 5 and -2. Therefore, the factors are $(p+5)(p-2)$ .

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Factoring quadratic trinomials

Property

To factor $x^2 + bx + c$ , we look for two numbers $p$ and $q$ so that

$pq = c$ and $p + q = b$

When we expand the factored form $(x + p)(x + q)$ , we get $x^2 + (p + q)x + pq$ .

Section 2

Sign patterns for factoring

Property

Assume that $b$ , $c$ , $p$ , and $q$ are positive integers.

$x^2 + bx + c = (x + p)(x + q)$ . If all the coefficients of the trinomial are positive, then both $p$ and $q$ are positive.
$x^2 - bx + c = (x - p)(x - q)$ . If the linear term of the trinomial is negative and the other two terms positive, then $p$ and $q$ are both negative.
$x^2 \pm bx - c = (x + p)(x - q)$ . If the constant term of the trinomial is negative, then $p$ and $q$ have opposite signs.

Examples

To factor $x^2 - 7x + 10$ , the constant term is positive and the middle term is negative. So, we need two negative numbers that multiply to 10 and add to -7. The factors are $(x - 2)(x - 5)$ .
In $y^2 + 4y - 12$ , the constant term is negative, so the factors have opposite signs. We need numbers that multiply to -12 and add to 4. The factorization is $(y + 6)(y - 2)$ .
For $z^2 - z - 30$ , the negative constant term means opposite signs. We need numbers that multiply to -30 and add to -1. The correct pair is -6 and 5, so we get $(z - 6)(z + 5)$ .

Explanation

The signs in the trinomial give you huge clues! A positive last term means the signs in your factors are the same. A negative last term means the signs are different. Use this to factor faster!

Section 3

Factor Trinomials of the Form x²+bx+c

Property

How to factor trinomials of the form $x^2 + bx + c$ :

Write the factors as two binomials with first terms $x$ : $(x \quad)(x \quad)$ .
Find two numbers $m$ and $n$ that multiply to $c$ ( $m \cdot n = c$ ) and add to $b$ ( $m+n = b$ ).
Use $m$ and $n$ as the last terms of the factors: $(x+m)(x+n)$ .
Check by multiplying the factors.

Examples

To factor $x^2 + 9x + 20$ , we need two numbers that multiply to 20 and add to 9. The numbers are 4 and 5. So, the factors are $(x+4)(x+5)$ .
To factor $a^2 - 8a + 15$ , we need two numbers that multiply to 15 and add to -8. The numbers are -3 and -5. Thus, the factors are $(a-3)(a-5)$ .
To factor $p^2 + 3p - 10$ , we need two numbers that multiply to -10 and add to 3. The numbers are 5 and -2. Therefore, the factors are $(p+5)(p-2)$ .

Overview

Lesson overview

Review the lesson summary and core properties without leaving the current page.

Overview

Section 1

Factoring quadratic trinomials

Property

To factor $x^2 + bx + c$ , we look for two numbers $p$ and $q$ so that

$pq = c$ and $p + q = b$

When we expand the factored form $(x + p)(x + q)$ , we get $x^2 + (p + q)x + pq$ .

Section 2

Sign patterns for factoring

Property

Assume that $b$ , $c$ , $p$ , and $q$ are positive integers.

$x^2 + bx + c = (x + p)(x + q)$ . If all the coefficients of the trinomial are positive, then both $p$ and $q$ are positive.
$x^2 - bx + c = (x - p)(x - q)$ . If the linear term of the trinomial is negative and the other two terms positive, then $p$ and $q$ are both negative.
$x^2 \pm bx - c = (x + p)(x - q)$ . If the constant term of the trinomial is negative, then $p$ and $q$ have opposite signs.

Examples

To factor $x^2 - 7x + 10$ , the constant term is positive and the middle term is negative. So, we need two negative numbers that multiply to 10 and add to -7. The factors are $(x - 2)(x - 5)$ .
In $y^2 + 4y - 12$ , the constant term is negative, so the factors have opposite signs. We need numbers that multiply to -12 and add to 4. The factorization is $(y + 6)(y - 2)$ .
For $z^2 - z - 30$ , the negative constant term means opposite signs. We need numbers that multiply to -30 and add to -1. The correct pair is -6 and 5, so we get $(z - 6)(z + 5)$ .

Explanation

The signs in the trinomial give you huge clues! A positive last term means the signs in your factors are the same. A negative last term means the signs are different. Use this to factor faster!

Section 3

Factor Trinomials of the Form x²+bx+c

Property

How to factor trinomials of the form $x^2 + bx + c$ :

Write the factors as two binomials with first terms $x$ : $(x \quad)(x \quad)$ .
Find two numbers $m$ and $n$ that multiply to $c$ ( $m \cdot n = c$ ) and add to $b$ ( $m+n = b$ ).
Use $m$ and $n$ as the last terms of the factors: $(x+m)(x+n)$ .
Check by multiplying the factors.

Examples

To factor $x^2 + 9x + 20$ , we need two numbers that multiply to 20 and add to 9. The numbers are 4 and 5. So, the factors are $(x+4)(x+5)$ .
To factor $a^2 - 8a + 15$ , we need two numbers that multiply to 15 and add to -8. The numbers are -3 and -5. Thus, the factors are $(a-3)(a-5)$ .
To factor $p^2 + 3p - 10$ , we need two numbers that multiply to -10 and add to 3. The numbers are 5 and -2. Therefore, the factors are $(p+5)(p-2)$ .