Factor Trinomials using Trial and Error

What happens when the leading coefficient is not 1 and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method.

Let's factor the trinomial \(3 x^{2}+5 x+2\).

From our earlier work we expect this will factor into two binomials.

\( \begin{array}{c} 3 x^{2}+5 x+2 \\
(\quad)(\quad) \end{array} \)

We know the first terms of the binomial factors will multiply to give us \(3 x^{2}\). The only factors of \(3 x^{2}\) are \(1 x, 3 x\). We can place them in the binomials.

This figure has the polynomial 3 x^ 2 +5 x +2. Underneath there are two terms, 1 x, and 3 x. Below these are the two factors

Check. Does \(1 x \cdot 3 x=3 x^{2}\)?

We know the last terms of the binomials will multiply to 2. Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of 2 are 1 and 2. But we now have two cases to consider as it will make a difference if we write 1, 2, or 2, 1.

This figure demonstrates the possible factors of the polynomial 3x^2 +5x +2. The polynomial is written twice. Underneath both

Which factors are correct? To decide that, we multiply the inner and outer terms.

This figure demonstrates the possible factors of the polynomial 3 x^ 2 + 5 x +2. The polynomial is written twice. Underneath

Since the middle term of the trinomial is 5x, the factors in the first case will work. Let's FOIL to check.

\( \begin{array}{l} (x+1)(3 x+2)\\ 3 x^{2}+2 x+3 x+2\\ 3 x^{2}+5 x+2 \text{✓} \end{array} \)

Our result of the factoring is:

\( \begin{array}{l} 3 x^{2}+5 x+2 \\ (x+1)(3 x+2) \end{array} \)

Example 7.33

How to Factor Trinomials of the Form \(a x^{2}+b x+c\) Using Trial and Error

Factor completely: \(3 y^{2}+22 y+7\).

Solution
Step 1. Write the trinomial in descending order. The trinomial is already in descending order. \(3 y^{2}+22 y+7\) 
Step 2. Find all the factor pairs of the first term.

The only factors of \(3 y^{2}\) are \(1 y, 3 y\)

Since there is only one pair, we can put them in the parentheses.

\(3 y^{2}+22 y+7\\ 1y, 3y\)

\( 3 y^{2}+22 y+7 \\ 1y, 3y \\ (y \qquad )(3y \qquad) \)
Step 3. Find all the factor pairs of the third term. The only factors of 7 are \(1,7\).  

\(3 y^{2}+22 y+7 \\ 1y, 3y \qquad 1 ,7\\ (y \qquad )(3y \qquad) \)
Step 4. Test all the possible combinations of the factors until the correct product is found.

 

 

 

\(3 y^{2}+22 y+7\)
Possible factors
 Product
\((y+1)(3y+7)\) \(3 y^{2}+10y+7\)
\((y+7)(3y+1)\) \(3 y^{2}+22 y+7\)

\((y+7)(3 y+1)\)
Step 5. Check by multiplying.   \( \begin{array}{l} (y+7)(3 y+1) \\ 3 y^{2}+22 y+7 \text{✓} \end{array}\)
Try It 7.65

Factor completely: \(2 a^{2}+5 a+3\).

Try It 7.66

Factor completely: \(4 b^{2}+5 b+1\).

HOW TO

Factor trinomials of the form \(a x^{2}+b x+c\) using trial and error.

Step 1. Write the trinomial in descending order of degrees.

Step 2. Find all the factor pairs of the first term.

Step 3. Find all the factor pairs of the third term.

Step 4. Test all the possible combinations of the factors until the correct product is found.

Step 5. Check by multiplying.

When the middle term is negative and the last term is positive, the signs in the binomials must both be negative.

Example 7.34

Factor completely: \(6 b^{2}-13 b+5\).

Solution
The trinomial is already in descending order. \(6 b^{2}-13 b+5\)
Find the factors of the first term.
\(6b^2 -13b + 5\\
1b \cdot 6b \\
2b \cdot 3b\)
Find the factors of the last term. Consider the signs. Since the last term, 5 is positive its factors must both be positive or both be negative. The coefficient of the middle term is negative, so we use the negative factors. \( 6b^2 -13b + 5\\1b \cdot 6b \qquad -1, -5\\2b \cdot 3b \)

Consider all the combinations of factors.

\(6 b^{2}-13 b+5\)
Possible factors Product
\((b-1)(6 b-5)\) \(6 b^{2}-11 b+5\)
\((b-5)(6 b-1)\) \(6 b^{2}-31 b+5\)
\((2 b-1)(3 b-5)\) \(6 b^{2}-13 b+5^{*}\)
\((2 b-5)(3 b-1)\) \(6 b^{2}-17 b+5\)

 

The correct factors are those whose product s the original trinomial \((2 b-1)(3 b-5)\)

Check by multiplying

\((2 b-1)(3 b-5)\)
\(6 b^{2}-10 b-3 b+5\)
\(6 b^{2}-13 b+5 \text{✓}\)

  
Try It 7.67

Factor completely: \(8 x^{2}-13 x+3\).

Try It 7.68

Factor completely: \(10 y^{2}-37 y+7\).

When we factor an expression, we always look for a greatest common factor first. If the expression does not have a greatest common factor, there cannot be one in its factors either. This may help us eliminate some of the possible factor combinations.

Example 7.35

Factor completely: \(14 x^{2}-47 x-7\).

Solution
The trinomial is already in descending order. \(14 x^{2}-47 x-7\)
Find the factors of the first term. \(14x^2 -47x -7\\
1x \cdot 14x\\
2x \cdot 7x\)
Find the factors of the last term. Consider the signs. Since it is negative, one factor must be positive and one negative. \( 14x^2 -47x -7\\1x \cdot 14x \qquad 1, -7\\2x \cdot 7x \qquad -1, 7 \)


Consider all the combinations of factors. We use each pair of the factors of \(14 x^{2}\) with each pair of factors of \(−7\).

Factors of \(14 x^{2}\) Pair with Factors of \(-7\)
\(x, 14 x\)   \(1,-7\)
\(-7,1\)
(reverse order)
\(x, 14 x\)   \(-1,7\)
\(7,-1\)
(reverse order)
\(2 x, 7 x\)   \(1,-7\)
\(-7,1\)
(reverse order)
\(2 x, 7 x\)   \(-1,7\)
\(7,-1\)
(reverse order)

These pairings lead to the following eight combinations.

This table has the heading 14 x ^ 2 – 47 x minus 7. This table has two columns. The first column is labeled “possible factors

 

The correct factors are those whose product is the original trinomial. \((2 x-7)(7 x+1)\)
Check by multiplying.  
Try It 7.69

Factor completely: \(8 a^{2}-3 a-5\).

Try It 7.70

Factor completely: \(6 b^{2}-b-15\).

Example 7.36

Factor completely: \(18 n^{2}-37 n+15\).

Solution
The trinomial is already in descending order. \(18 n^{2}-37 n+15\)
Find the factors of the first term. \(18 n^{2}-37 n+15\\ 1n \cdot 18n\\2n \cdot 9n\\3n \cdot 6n\)
Find the factors of the last term. Consider the signs. Since 15 is positive and the coefficient of the middle term is negative, we use the negative factors. \( 18 n^{2}-37 n+15\\ 1n \cdot 18n \qquad -1(-15)\\2n \cdot 9n \qquad -3(-5)\\3n \cdot 6n \)

 

Consider all the combinations of factors.

This table has the heading 18 n ^ 2 – 37n + 15. This table has two columns. The first column is labeled possible factors and

The correct factors are those whose product is the original trinomial. \((2 n-3)(9 n-5)\)

Check by multiplying.

\((2 n-3)(9 n-5)\)

\(18 n^{2}-10 n-27 n+15\)

\(18 n^{2}-37 n+15 \text{✓}\)

  
Try It 7.71

Factor completely: \(18 x^{2}-3 x-10\).

Try It 7.72

Factor completely: \(30 y^{2}-53 y-21\).

Don't forget to look for a GCF first.

Example 7.37

Factor completely: \(10 y^{4}+55 y^{3}+60 y^{2}\).

Solution
  \(10 y^{4}+55 y^{3}+60 y^{2}\)
Notice the greatest common factor, and factor it first. \(15 y^{2}\left(2 y^{2}+11 y+12\right)\)
Factor the trinomial.

\( 5y^2(2y^2+11y+12)\\ \qquad y \cdot 2y \qquad 1 \cdot 12 \\ \qquad \qquad \qquad 2 \cdot 6 \\ \qquad \qquad \qquad 3 \cdot 4 \)

 

Consider all the combinations.

This table has the heading 2 y squared + 11 y + 12 This table has two columns. The first column is labeled “possible factors”

The correct factors are those whose product
is the original trinomial. Remember to include
the factor \(5 y^{2}\).		 \(5 y^{2}(y+4)(2 y+3)\)

Check by multiplying.

\(5 y^{2}(y+4)(2 y+3)\)

\(5 y^{2}\left(2 y^{2}+8 y+3 y+12\right)\)

\(10 y^{4}+55 y^{3}+60 y^{2} \text{✓}\)

  
Try It 7.73

Factor completely: \(15 n^{3}-85 n^{2}+100 n\).

Try It 7.74

Factor completely: \(56 q^{3}+320 q^{2}-96 q\).

Back to '9.9: Factoring General Polynomials'