Explainer: Factoring Nonmonic Quadratics

In this explainer, we will learn how to factor quadratic expressions in the form π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨ where the coefficient of the leading term is greater than 1.

A quadratic is said to be monic if the coefficient of the leading term (the π‘₯ term) is equal to one. Therefore, any quadratic where the coefficient of the leading term is not one is called a nonmonic quadratic. An example of a nonmonic quadratic would be 6π‘₯βˆ’5π‘₯βˆ’4.

Now, we are going to look at how we factor a specific group of nonmonic quadratics. For example, the quadratic 5π‘₯+15π‘₯ is nonmonic, but this factors into the form π‘Žπ‘₯(𝑏π‘₯+𝑐) or, more explicitly, 5π‘₯(π‘₯+3). Here, we are going to look exclusively at quadratics that factor into the form (π‘Žπ‘₯+𝑏)(𝑐π‘₯+𝑑), where π‘Ž and 𝑐 are whole numbers greater than or less than one and 𝑏 and 𝑑 are nonzero whole numbers.

Before we look at factoring, let us recap expanding two binomials. This is the reverse process of factoring and can be used to better understand the process. We will demonstrate this in our first example.

Example 1: Expanding Two Binomials

Expand (3π‘₯+2)(4π‘₯βˆ’1).

Answer

We can use the distributive property to rewrite the question as follows: 3π‘₯(4π‘₯βˆ’1)+2(4π‘₯βˆ’1).

If we then expand the two expressions separately, we get 12π‘₯βˆ’3π‘₯+8π‘₯βˆ’2, and by simplifying we find that our solution is 12π‘₯+5π‘₯βˆ’2.

In Example 1, in expanding the binomials, it is worth noting that we have found the quadratic for which the binomials are factors. In fact, factoring a quadratic is the exact reverse of this process. Let us look at how to factor a nonmonic quadratic.

For a quadratic in the form π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, given that it factors into the form (𝛼π‘₯+𝛽)(𝛾π‘₯+𝛿), to find the factored form, we first need to find the product of π‘Ž and 𝑐. We then need to identify two factors of π‘Žπ‘ whose sum is equal to 𝑏. We then rewrite the 𝑏π‘₯ term as a sum of these factors and factor the expression to reveal a common binomial term which can subsequently be factored to reveal our solution. It is worth noting at this point that not all quadratic expressions are factorizable and the method will only work for expressions that are.

We will demonstrate the method for factoring nonmonic quadratics in Example 2.

Example 2: Factoring a Nonmonic Quadratic

Factor 6π‘₯βˆ’5π‘₯βˆ’4.

Answer

The quadratic is presented in the standard form, π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, so our first step is to multiply π‘Ž and 𝑐, list the factors, and identify the factor pair whose sum is 𝑏.

We then need to rewrite our expression using these factors: 6π‘₯βˆ’8π‘₯+3π‘₯βˆ’4.

It is worth noting here that you can also write this as 6π‘₯+3π‘₯βˆ’8π‘₯βˆ’4 which will ultimately yield the same result.

We then need to factor our expression to reveal a common binomial term: 2π‘₯(3π‘₯βˆ’4)+1(3π‘₯βˆ’4).

Finally, we factor out the common binomial: (3π‘₯βˆ’4)(2π‘₯+1).

Had we rewritten the quadratic as 6π‘₯+3π‘₯βˆ’8π‘₯βˆ’4 upon factoring, we would have got 3π‘₯(2π‘₯+1)βˆ’4(2π‘₯+1), which factors to (2π‘₯+1)(3π‘₯βˆ’4), which is equivalent.

We will look at another example to demonstrate the method further.

Example 3: Factoring a Nonmonic Quadratic

Factor 6π‘₯βˆ’19π‘₯+10.

Answer

We start by listing out the factors of π‘Žπ‘ and identify the pair that has a sum equal to 𝑏. Notice in this example that the product of π‘Žπ‘ is positive but 𝑏 is negative; thus, the factors must both be negative.

We then rewrite the expression: 6π‘₯βˆ’4π‘₯βˆ’15π‘₯+10.

Next, we factor to reveal a common binomial: 2π‘₯(3π‘₯βˆ’2)βˆ’5(3π‘₯βˆ’2).

Finally, we factor out the common binomial term: (3π‘₯βˆ’2)(2π‘₯βˆ’5).

For the sake of interest, it is worth noting that this method is identical to that of monic quadratics, the difference being that π‘Žπ‘ for a monic quadratic is the same as 𝑐 which simplifies the method. This is demonstrated in Example 4.

Example 4: Factoring a Monic Quadratic

Factor π‘₯βˆ’8π‘₯+15.

Answer

First list out the factors of π‘Žπ‘, which in this case is 15, and find the factor pair which has a sum equal to 𝑏, in this case, βˆ’8.

As before, we rewrite our expression: π‘₯βˆ’3π‘₯βˆ’5π‘₯+15.

Then, we factor to find the common binomial: π‘₯(π‘₯βˆ’3)βˆ’5(π‘₯βˆ’3).

Finally, we factor out the common binomial term: (π‘₯βˆ’3)(π‘₯βˆ’5).

Note that you can jump to the final step once you have identified the factors of 𝑐 for monic quadratics, if you are confident with the factored form of a quadratic.

Example 5: Factoring Nonmonic Quadratics

Completely factor 4π‘₯βˆ’32π‘₯+28.

Answer

We could apply the method for factoring nonmonic quadratics. However, it is usually worth looking, first, for any common factors. In this quadratic, we actually can factor out the 4 as follows: 4π‘₯βˆ’32π‘₯+28=4ο€Ήπ‘₯βˆ’8π‘₯+7.

The advantage of looking for common factors is that it often simplifies the process of factoring the resulting quadratic. In this case, we are left with a monic quadratic to factor, where the final term is a prime number so it only has two possible factors, in particular, 7 and 1. Since the coefficient of π‘₯ is negative and 𝑐 is positive, both factors of 7 must be negative. Hence, the factors are βˆ’7 and βˆ’1 which sum to βˆ’8 as required. Therefore, we can rewrite the expression as 4(π‘₯βˆ’7)(π‘₯βˆ’1).

We often use factoring as a method to solve quadratic equations of the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0.

Since we are able to rewrite this equation in the form (𝛼π‘₯βˆ’π›½)(𝛾π‘₯βˆ’π›Ώ)=0, we can conclude that the expression (𝛼π‘₯βˆ’π›½)(𝛾π‘₯βˆ’π›Ώ) is equal to zero if either of the factors are equal to zero. Hence, either 𝛼π‘₯βˆ’π›½=0 or 𝛾π‘₯βˆ’π›Ώ=0.

The solutions to these two equations are π‘₯=𝛽𝛼 and π‘₯=𝛿𝛾. In this way, we can use factoring to solve quadratic equations. When solving equations this way, we need to be careful about negative signs. It is a common mistake for someone to get an expression in the form (𝛼π‘₯+𝛽)(𝛾π‘₯+𝛿)=0 and mistakenly conclude that the solutions are π‘₯=𝛽𝛼 and π‘₯=𝛿𝛾, whereas the actual solutions are π‘₯=βˆ’π›½π›Ό and π‘₯=βˆ’π›Ώπ›Ύ.

In the last example, we will demonstrate how we can use the technique of factoring nonmonic quadratics to solve quadratic equations.

Example 6: Solving Nonmonic Quadratic Equations by Factoring

Find the solution set of 5π‘₯+12π‘₯=βˆ’7 in ℝ.

Answer

We begin by adding 7 to both sides of the equation to get it in the standard form for a quadratic equation where we have a quadratic expression equal to zero: 5π‘₯+12π‘₯+7=0.

We can now consider the factors of π‘Žπ‘=35.

Since 5+7=12, we can rewrite the left-hand side using these factors as follows: 5π‘₯+5π‘₯+7π‘₯+7=0.

We can now factor the first two and last two terms to reveal a common factor: 5π‘₯(π‘₯+1)+7(π‘₯+1)=0.

We can now factor out π‘₯+1 to get (π‘₯+1)(5π‘₯+1)=0.

Since we have a product of two expressions equal to zero, we can conclude that this happens if either of the expressions is equal to zero. Therefore, either π‘₯+1=0 or 5π‘₯+1=0.

Solving each equation, we find that the two solutions of the quadratic equation are π‘₯=βˆ’1 and π‘₯=βˆ’15. Since the question asked for the solution set of the equation, we should express our answer as ο¬βˆ’1,βˆ’15.

Key Points

  1. A quadratic is an expression in the form π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, where π‘Žβ‰ 0, 𝑏, and 𝑐 are constants. We call a quadratic monic if π‘Ž=1; otherwise, we call it nonmonic.
  2. It is not always possible to factor quadratics with simple factors of the form 𝛼π‘₯+𝛽, where 𝛼 and 𝛽 are integer constants.
  3. If a nonmonic quadratic in the form π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨ can be factored, we can find the factors using the following technique:
    1. List the factors of π‘Žπ‘. If π‘Žπ‘ is positive, both factors will be positive or both factors will be negative. If π‘Žπ‘ is negative, one factor will be positive and the other will be negative.
    2. Find a pair of factors 𝑛 and π‘š whose sum (taking into account the necessary signs as per the first step) equals 𝑏.
    3. Rewrite the expression in the form π‘Žπ‘₯+𝑛π‘₯+π‘šπ‘₯+𝑐. We can equally write this as π‘Žπ‘₯+π‘šπ‘₯+𝑛π‘₯+π‘οŠ¨; it will result in the same answer.
    4. Factor the first two terms and the last two terms to find a common factor.
    5. Factor out the common factor to get a fully factored form of the expression.
  4. Using factoring, we can solve quadratic equations of the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0 by writing them in factored form and then setting each factor equal to zero to solve for the solutions.

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