Sum Product Factoring is a commonly used method of factoring on trinomials \( (ax^2 + bx + c)\) where the value of \(a\) is usually 1. In order to factor the trinomial properly, we need to find 2 integers whose Product is \(c\) and Sum is \(b\). These integers (better known as factors) will be referred to as \(r\) and \(s\). Once these values are found, create a factored expression in the form \((x+r)(x+s)\).
Factor \(x^2+4x+3\)
We identify that \(b\) = 4 and \(c\) = 3; as a result, we need to calculate 2 integers whose \(\text{sum} = 4\) and \(\text{product} = 3\).
We can create a table that outlines the possible combinations for what the sum and product can be based on the 2 integer values. In this case, we try to identify what the sum is based on the product of \(3\):
r value | 1 | -1 |
---|---|---|
s value | 3 | -3 |
Sum | 4 | -4 |
Product | 3 | 3 |
Based on these results, we can determine that the \(r\) and \(s\) values (or factors) are \(1\) and \(3\) respectively.
We can arrange the values in a factored expression, \((x+1)(x+3)\).
Enter in coefficients for the trinomial or click on the button to generate a new question. Try to factor the expression and then check your answer by clicking the "See Solution" button.
Factor \(2x^2+9x+10\)
We identify that \(a = 2\), \(b = 9\) and \(c = 10\); as a result, we need to calculate 2 integers whose \(\text{product} = 20\) and \(\text{sum} = 9\).
We can create a table that outlines the possible combinations for what the sum and product can be based on the 2 integer values. In this case, we try to identify what the sum is based on the product of \(20\):
r value | 1 | -1 | 2 | -2 | 4 | -4 |
---|---|---|---|---|---|---|
s value | 20 | -20 | 10 | -10 | 5 | -5 |
Sum | 21 | -21 | 12 | -12 | 9 | -9 |
Product | 20 | 20 | 20 | 20 | 20 | 20 |
Based on these results, we can determine that the \(r\) and \(s\) values (or factors) are \(4\) and \(5\) respectively.
We can arrange the values in a factored expression in the form \(ax^2+rx+sx+c\) which in this case will be \(2x^2+4x+5x+10\)
We can take out the GCF out of the 2 pairs of terms to get common expressions: \(2x(x+2)+5(x+2)\)
The GCF in the first expression was \(2x\), the GCF in the second expression was 5 and the common expression was \(x+2\)
Simplify by placing the GCF terms in its own expression: \((2x+5)(x+2)\)