Unit 9: Factoring Polynomials

9a. Rewrite polynomial as a product of two or more polynomials

  • How can you rewrite a polynomial as a product of two polynomials?

Some polynomials have a common factor, which all the terms can be divided by evenly. This common factor can be factored out, creating the product of the common factor and what remains of the polynomial. This is commonly referred to as factoring out a common factor. 

For example, consider the polynomial \(6x^3+12x^2-3x\). First, we must inspect the three terms and find the greatest common factor (GCF) of the coefficients. In this case, the GCF is 3. Similarly, we look for the GCF of the variable, which is the lowest degree of any of the terms. In this case, it is \(x\). We will now factor out the \(3x\), by dividing each term and writing what is left inside the parentheses:

\(6x^3+12x^2-3x=3x(2x^2+4x-1)\)

This is now the factored expression. You can verify that the answer is correct by multiplying the factor outside to the polynomial inside and checking that you get what you started with. 

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9b. Identify polynomials that cannot be factored

  • How do you know if a polynomial can be factored?

Not all polynomials can be factored. The first type of factoring to check for is always a common factor. If there is no common factor, there may be no other way to factor the polynomial. For quadratic trinomials, you should attempt one of the methods for factoring into two binomials and exhaust all possibilities before deciding the polynomial doesn't factor. A polynomial that does not factor is called a prime polynomial.

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9c. Choose an appropriate factoring strategy for a given polynomial

  • What factoring strategies are appropriate for polynomials with 4 terms?
  • What factoring strategies are appropriate for polynomials with 3 terms?
  • What factoring strategies are appropriate for polynomials with 2 terms?

The strategy with which you attempt to factor a polynomial will depend on the number of terms in the polynomial. Regardless of the type of polynomial, you should always attempt to find and factor out a common factor before attempting any of these methods, as factoring out a common factor will always make the process easier. 

When a polynomial has 4 terms, the only method we know how to use is factor by grouping. Group the first two terms together, and the last two terms together, then factor out the common factor for each group. If the polynomial can be factored by grouping, the factors that remain in the parentheses after taking out the common factor will be identical. If they are not, either you didn't factor completely/correctly, or the polynomial can't be factored by grouping. 

For example, consider \(3x^3+x^2+6x+2\). First we group our terms, and factor out the common factor from each: \((3x^3+x^2)+(6x+2)=x^2(3x+1)+2(3x+1)\). Now, notice the binomial in each set of parentheses is the same. We then factor that common binomial out, leaving us with the factored polynomial: 

\((3x+1)(x^2+2) \)

When we have a quadratic trinomial, we can factor either by guessing and checking the factors in two binomials, or using the "AC" method. 

When factoring a binomial, the available methods depend on the degree. If the binomial is of the 1st degree, then only a common factor method can be used. If the binomial is of the 2nd degree (a quadratic), then check if it can be factored by difference of squares. If not, it cannot be factored. 

If the binomial is of the 3rd degree, check if it can be factored by the sum or difference of cubes. The sum of cubes factors as \((x^3+a^3)=(x+a)(x^2-ax+a^2)\) and the difference of cubes factors as \((x^3-a^3)=(x-a)(x^2+ax+a^2)\). For example, the binomial \(x^3+27\) can be factored as \(x^3+27=x^3+3^3=(x+3)(x^2-3x+6)\).

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9d. Solve quadratic equations by factoring

  • How do you factor quadratics with a leading coefficient of 1?
  • How do you factor quadratics with a leading coefficient that is not 1?

When factoring a quadratic expressions of the form \(ax^2+bx+c\), the approach you take may depend on the values for a, b, and c. First we will discuss factoring when \(a = 1\).

When factoring a quadratic with a leading coefficient of 1, you just need to focus on finding two numbers that add to give the value of b, and the same two numbers that multiply to give the value of c. For example, when factoring the quadratic expression \(x^2-2x-8\), you must find two numbers that add to make -2, and multiply to make -8. Those two numbers are 2 and -4. Therefore, \(x^2-2x-8\) can be factored as \((x-4)(x+2)\). You can always multiply your answer using the distributive property to check that it is correct. 

When factoring a quadratic in which a does not equal 1, we have to use a different approach. One approach is to use a guess and check method, where the leading coefficients in each binomial multiply to equal the leading coefficient in the quadratic, and the constants in each binomial multiply to make the constant in the quadratic. For example, when we factor \(2x^2-13x-45\), we can try different combinations of \(2x\) and \(x\) for our leading term in each binomial, along with pairs of factors of -45, such as 1 and 45, 3 and 15, and 5 and 9. By trial and error we can see that \((2x-3)(x+15)=2x^2+27x-45\), which means this factoring is incorrect. Eventually you will find the correct factoring, which is \((2x+5)(x-9)\).

The other method for factoring a quadratic in which a does not equal 1, is to use the "ac method". This method uses factor by grouping strategies to make factoring more standardized. For this approach, let's consider the same problem, \(2x^2-13x-45\). To begin, we will find ac, which is \((2)(-45)=-90\). Therefore we must now find two numbers that multiply to make -90, and also add to make b, which is -13. The two numbers that satisfy this condition are (-18) and (5). Now we will replace b with these two numbers, making our expression \(2x^2-18x+5x-45\). This can now be factored by grouping, giving us \(2x(x-9)+5(x-9)\). By factoring out the \((x-9)\), we get the final answer, \((x-9)(2x+5)\).

To review, see:

 

Unit 9 Vocabulary

This vocabulary list includes terms you will need to know to successfully complete the final exam.

  • "AC" method
  • common factor
  • difference of cubes
  • factoring out
  • greatest common factor (GCF)
  • grouping
  • guess and check method
  • prime polynomial
  • quadratic
  • quadratic trinomial
  • sum of cubes
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