How to Factor Quadratic Equations, Trinomials & Polynomials

Factoring is the process of writing a number or algebraic expression as a product of factors. For a monomial (one term) expression, it is relatively easy to find common factors. Take the expression 4ax + 4ay^3 – 4az^2, it is easy to spot 4a as a common factor. 4a(x + y^3 –z^2). For trinomials, (three term expressions), you may be familiar with the FOIL method. If you have (2x + 1)(3x + 4), multiply the First terms, the Outer terms, the Inner terms, and the Last terms. 6x^2 + 6x + 3x + 4 = 6x^2 + 9x + 4. Use the FOIL method in reverse to factor trinomials or polynomials (many term expressions).

Instructions

Factoring trinomials of the form x^2 + ax + b

1. Say you have a trinomial in the form x^2 + ax + b. For example, x^2 – 7x + 12.

2. Write x as the first term in each set of parentheses. (x __) (x __).

3. Look at the signs of the middle and last numbers, -7 means that at least one sign is negative, while +12 means both are negative. This is true since multiplying two negatives gives a positive answer.

4. Find factors of the last number (12). You have 1 and 12, 2 and 6, 3 and 4.

5. Look at the middle number (-7) and determine which factors add up. Here, -3 + -4 = -7. (x – 3)( x – 4).

6. Check your work by using FOIL. (x – 3)(x – 4) = x^2 – 4x – 3x + 12 = x^2 – 7x + 12.

Trinomials in the form ax^2 + bx + c

1. To factor trinomials in the form ax^2 + bx + c, there is an added step. Take 6x^2 + 11x + 3. To factor out this trinomial expression, first find factors of 6. Two of your choices (2x + 3)(3x + 1) or (6x + 3)(x + 1). In each case, the outer terms multiply to 3.

2. Since you want the middle term to add up to 11, (2x + 3)(3x + 1) is the correct pair of factors.

3. Check your work by using FOIL. (2x + 3)(3x + 1) = 6x^2 + 2x + 9x + 3 = 6x^2 + 11x + 3.

Solve Quadratic Equations by Finding the Values for x

1. Quadratic equations take the form ax^2 + bx + c = 0. To factor quadratic equations, find values for x which make the equation work.

2. Take an example problem: x^2 + 3x = 28. First rewrite it in the form ax^2 + bx + c. x^2 + 3x – 28 = 0.

3. Find factors for -28: 1 and 28, 2 and 14, 4 and 7. Since 7 – 4 is 3, 4 and 7 is the correct pair. (x – 4)(x + 7).

4. x – 4 = 0 and x + 7 = 0. So x = 4 and x = -7. This method only works if there is no coefficient in front of x^2 in the original equation.

Quadratic Equations with a coefficient in front of x^2

1. If the quadratic equation has a coefficient (number) in front of x^2, you would use a different method. Take an example: 2x^2 – 3x – 5 = 0. First move the constant (-5) to the right side: 2x^2 – 3x = 5. Then divide both sides by 2: x^2 – (3/2)x = 5/2.

2. Add to the left side a number that produces a complete square. Half of -3/2 is -3/4. Add (-3/4)^2 to both sides. X^2 – (3/2)x + (-3/4)^2 = 5/2 + 9/16. (x – ¾)^2 = 40/16 + 9/16). (x – 3/4_)^2 = 29/16.

3. Take the square root of both sides. X – ¾ = +/- 7/4. x = (3 +/- 7) / 4. x = (3 + 7) / 4 = 10/4 = 5/2. Or x = (3-7) / 4 = -4/4 = -1.

4. So x = 5/2 or -1.

Resources

• Purple Math: Solving Quadratic Equations
• The Math Page: Quadratic Equations