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A1.7 · Factor simple trinomials and common-factor quadratics

Learn to factor simple trinomials and common-factor quadratics through clear examples and targeted practice.

Ontario Grade 11 Mathematics

Mathematical Models

MBF3C study topic A1.7

Factoring rewrites an expression as a product. It is the reverse of expanding brackets. For example, expanding 3(x+2)3(x+2) gives 3x+63x+6; factoring 3x+63x+6 returns 3(x+2)3(x+2). In this lesson, you will practise two kinds of factoring: taking out a common factor and factoring simple trinomials. These skills help you rewrite quadratic expressions in a useful form. The focus is on expressions, not on solving quadratic equations.

What you will learn

1. Review: factors, products, and quadratics

A factor is a number or expression that is multiplied by another number or expression. In 4⋅5=204\cdot 5=20, the numbers 44 and 55 are factors of the product 2020. A bracket can also be a factor: 2(x+3)2(x+3) means 22 multiplied by the entire expression in the bracket.
A quadratic expression has a variable term with an exponent of 22, along with any other terms. For example, x2+5x+6x^2+5x+6 is quadratic. In this lesson, the variable is usually xx. The exponent tells how many times xx is multiplied by itself, so x2=x⋅xx^2=x\cdot x.
To expand a product of two brackets, multiply each term in one bracket by each term in the other. For instance, (x+2)(x+3)(x+2)(x+3) expands to x2+5x+6x^2+5x+6. Factoring reverses this process: it looks for brackets whose product gives the original expression.
Before you factor, arrange the terms in descending powers of the variable. For a quadratic, this usually means writing the squared term first, then the term with xx, then the constant term. This makes it easier to see the structure.

2. Factor out a common factor

A common factor is a factor shared by every term in an expression. The greatest common factor, or GCF, is the largest factor shared by all the terms. To factor by the GCF, identify what divides every term, then write that factor outside a bracket.
For example, both terms in 6x2+9x6x^2+9x are divisible by 3x3x. Dividing each term by 3x3x leaves 2x2x and 33. Therefore, the expression becomes 3x(2x+3)3x(2x+3). The bracket contains what remains from each original term.
This method also works when the expression is quadratic. A common-factor quadratic often has two terms, such as 8x2−12x8x^2-12x, or three terms, such as 4x2+12x+84x^2+12x+8. Take out the GCF first whenever every term shares one. The expression left inside the bracket may itself be a trinomial that can be factored further.
The sign matters. If all terms are negative, factoring out a negative common factor can make the bracket easier to read. For example, −5x2−10x=−5x(x+2)-5x^2-10x=-5x(x+2). Expanding confirms that both terms have the original negative signs.

3. Factor simple trinomials

A simple trinomial in this lesson has the form x2+bx+cx^2+bx+c. A trinomial is an expression with three terms. The coefficient of x2x^2 is 11, so the leading term is just x2x^2. The numbers bb and cc can be positive or negative.
To factor x2+bx+cx^2+bx+c, look for two integers whose product is cc and whose sum is bb. The product gives the constant term, while the sum gives the coefficient of xx. If the two numbers are mm and nn, then the factors are (x+m)(x+n)(x+m)(x+n).
Why does the sum matter? Expanding (x+m)(x+n)(x+m)(x+n) gives x2+(m+n)x+mnx^2+(m+n)x+mn. The middle coefficient is the sum of the two numbers, and the constant is their product. So checking both conditions makes sure the brackets expand to the original trinomial.
Signs help narrow the search. If the constant is positive, the two numbers have the same sign. Their sum tells you whether both are positive or both are negative. If the constant is negative, the numbers have opposite signs. The larger absolute value determines the sign of their sum.
If no integer pair has both the required product and sum, do not force a factorization using an incorrect pair. For this method, the simple trinomial factors using integer brackets only when a suitable pair exists. Always expand your proposed factors to verify them.
(x+m)(x+n)=x2+(m+n)x+mn(x+m)(x+n)=x^2+(m+n)x+mn

4. Choose a method and check your result

When you see a quadratic expression, first check whether every term has a common factor. If it does, take out the GCF. Then look at the expression inside the bracket to see whether another factor can be taken out or whether the remaining expression is a simple trinomial.
If there is no common factor and the expression is a simple trinomial with leading coefficient 11, use the sum-and-product search. List factor pairs of the constant and test their sums. This is more reliable than guessing based only on the middle term.
After factoring, expand the result. Multiply the first terms, the outer terms, the inner terms, and the last terms. Combine like terms and compare with the starting expression. Like terms have the same variable part, such as 2x2x and 5x5x; they can be combined by adding or subtracting their coefficients.
Factoring changes the form of an expression but does not change its value. The factored and expanded forms represent the same expression. This is why expansion is a useful check, and why a missing sign or factor will show up when you compare the result.

Worked example

Example 1: Common-factor quadratic

Factor 12x2−18x12x^2-18x fully.
  1. Find the shared factor
    The coefficients 1212 and 1818 share a greatest numerical factor of 66. Each term also contains at least one xx, so the GCF is 6x6x.
    6x6x
  2. Divide each term
    Divide each original term by 6x6x. The first term leaves 2x2x, and the second leaves −3-3. Put these results inside the bracket.
    12x2−18x=6x(2x−3)12x^2-18x=6x(2x-3)
  3. Check by expanding
    Multiply 6x6x by both terms in the bracket. The result matches both terms of the original expression, including the negative sign.
    6x(2x−3)=12x2−18x6x(2x-3)=12x^2-18x
Answer: 6x(2x−3)6x(2x-3)
Check: Expanding gives 12x2−18x12x^2-18x, so the factorization is correct.

Worked example

Example 2: Simple trinomial

Factor x2−x−20x^2-x-20.
  1. Identify the target sum and product
    The coefficient of xx is −1-1, and the constant is −20-20. We need two integers that multiply to −20-20 and add to −1-1.
    mn=−20,m+n=−1mn=-20,\qquad m+n=-1
  2. Choose a pair
    The pair −5-5 and 44 has product −20-20 and sum −1-1. Since the product is negative, the pair has opposite signs, which fits the negative middle coefficient.
    (−5)(4)=−20,−5+4=−1(-5)(4)=-20,\qquad -5+4=-1
  3. Write the factors
    Place each selected integer with xx in a bracket. The brackets multiply to the original trinomial because their cross-products combine to give the coefficient −1-1.
    (x−5)(x+4)(x-5)(x+4)
  4. Check by expanding
    The cross-products are 4x4x and −5x-5x, which combine to −x-x. The constant is −20-20, so all three terms match.
    (x−5)(x+4)=x2−x−20(x-5)(x+4)=x^2-x-20
Answer: (x−5)(x+4)(x-5)(x+4)
Check: Expanding gives x2−x−20x^2-x-20, matching the original expression.

Common mistakes and how to avoid them

Taking out a factor that is not common to every term.
Correction: Check each term separately. The factor must divide all terms exactly.
Choosing numbers whose product is correct but whose sum is wrong.
Correction: For a simple trinomial, verify both the product and the sum before writing brackets.
Losing a negative sign when dividing a term or writing a bracket.
Correction: Keep the signs attached to their terms, then expand the proposed factors to check.
Stopping after taking out a common factor when the bracket can be factored further.
Correction: Look at the expression inside the bracket and check whether it has another common factor or is a factorable simple trinomial.

Lesson summary

Check your understanding

Question 1

Factor 10x2+15x10x^2+15x fully.
  1. 5x(2x+3)5x(2x+3)
  2. 5(2x2+3x)5(2x^2+3x)
  3. 10x(x+15)10x(x+15)
  4. 5x(2x−3)5x(2x-3)
Show answer and explanation
5x(2x+3)5x(2x+3)
The GCF is 5x5x. Dividing each term by 5x5x leaves 2x2x and 33, so 5x(2x+3)5x(2x+3) expands to the original expression.

Question 2

Which is the factorization of x2+2x−15x^2+2x-15?
  1. (x+5)(x−3)(x+5)(x-3)
  2. (x−5)(x+3)(x-5)(x+3)
  3. (x+15)(x−1)(x+15)(x-1)
  4. (x+3)(x−5)(x+3)(x-5)
Show answer and explanation
(x+5)(x−3)(x+5)(x-3)
The numbers 55 and −3-3 multiply to −15-15 and add to 22. Therefore, (x+5)(x−3)(x+5)(x-3) is correct.

Question 3

Which pair can be used to factor x2−7x+12x^2-7x+12?
  1. −3-3 and −4-4
  2. 33 and 44
  3. −2-2 and −6-6
  4. −1-1 and −12-12
Show answer and explanation
−3-3 and −4-4
The pair −3-3 and −4-4 has product 1212 and sum −7-7, giving (x−3)(x−4)(x-3)(x-4).

Key terms

Factor
A number or expression multiplied by another to make a product.
Quadratic expression
An expression that includes a variable term with exponent 22.
Trinomial
An algebraic expression with three terms.
Greatest common factor
The largest factor shared by every term in an expression.
Expand
Multiply factors and brackets to rewrite a product as a sum or difference of terms.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MBF3C), expectation A1.7. It is a study resource, not an official curriculum publication.

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