6 Ways to Factor Second Degree Polynomials (Quadratic Equations)

Table of contents:

6 Ways to Factor Second Degree Polynomials (Quadratic Equations)
6 Ways to Factor Second Degree Polynomials (Quadratic Equations)
Anonim

A polynomial contains a variable (x) raised to a power, known as a degree, and several terms and/or constants. Factoring a polynomial means dividing the expression into smaller expressions that multiply. This knowledge is studied from Algebra I onwards, and can be difficult to understand if you don't have a foundation.

Steps

Starting

Factor Second Degree Polynomials (Quadratic Equations) Step 1

Step 1. Assemble the expression

The standard format for the quadratic equation is:

ax2 + bx + c = 0

Start by ordering the terms of the equation from greatest to least power, just as in the form above. For example, take;

6 + 6x2 + 13x = 0

The expression will be reordered so that it can be worked more easily by changing the location of the terms:

6x2 + 13x + 6 = 0

Factor Second Degree Polynomials (Quadratic Equations) Step 2

Step 2. Find the factored shape using one of the methods below

Factoring a polynomial results in two smaller expressions that can be multiplied to produce the original polynomial:

6x2 + 13x + 6 = (2x + 3)(3x + 2)

In this example, (2x +3) and (3x + 2) are factors of the original expression, 6x2 + 13x + 6.

Factor Second Degree Polynomials (Quadratic Equations) Step 3

Step 3. Check the result

Multiply the identified factors. Then just combine similar terms. Start with:

(2x + 3)(3x + 2)

Let's test it using the FOIL method (English for First Outside, Inside, Last – outside first, then inside), also called the distributive property of multiplication, getting:

6x2 + 4x + 9x + 6

It is now possible to add 4x and 9x as they are similar terms. You know the factors are correct because the original equation was obtained:

6x2 + 13x + 6

Method 1 of 6: Trial and Error

If you have a very simple polynomial, you may be able to figure out the factors yourself by looking at it. For example, after practice, many mathematicians are able to identify that the expression 4x2 + 4x + 1 has the factors (2x + 1) and (2x + 1) after having worked a lot with this expression previously. But of course it won't be so easy with the more complicated polynomials. In this example, we'll use a less common expression:

3x2 + 2x - 8

Factor Second Degree Polynomials (Quadratic Equations) Step 4

Step 1. List the factors for terms a and c

Using the standard ax format2 + bx + c = 0, identify the terms of a and c and list their factors. For 3x2 + 2x - 8, this means:

a = 3 and has a set of factors: 1 * 3

c = -8 and has four sets of factors: -2 * 4, -4 * 2, -8 * 1 and -1 * 8.

Factor Second Degree Polynomials (Quadratic Equations) Step 5

Step 2. Assemble two sets of empty parentheses

You will fill them with the constants of each expression:

(x)(x)

Factor Second Degree Polynomials (Quadratic Equations) Step 6

Step 3. Fill in the spaces in front of the x's with a couple of possible factors for the a value

For term a in the example used, 3x2, there is only one possibility:

(3x)(1x)

Factor Second Degree Polynomials (Quadratic Equations) Step 7

Step 4. Fill in the two spaces after the x's with a pair of factors for the constants

Suppose you choose the numbers 8 and 1. Write them down:

(3x

Step 8.)(

Step 1

Factor Second Degree Polynomials (Quadratic Equations) Step 8

Step 5. Decide which signs (addition or subtraction) should go between the variables of x and the numbers

Depending on the signs in the original expression, it is possible to figure out what the signs of the constants should be. Let's call the two constants for the two factors h and k:

if x2 + bx + c, then (x + h)(x + k)

if ax2 - bx - c or ax2 + bx - c, then (x - h)(x + k)

if x2 - bx + c, then (x - h)(x - k)

For example, 3x2 + 2x - 8, the signs must be: (x - h)(x + k), resulting in the two factors:

(3x + 8) and (x - 1)

Factor Second Degree Polynomials (Quadratic Equations) Step 9

Step 6. Test the choices using the distributive property

A quick first test to run is to see if the middle terms match the correct values. If not, you may have chosen the wrong factors for c. Let's test the answer:

(3x + 8)(x - 1)

When performing the multiplication, you will get:

3x2 - 3x + 8x - 8

By simplifying this expression by the sum of similar terms (-3x) and (8x), you get:

3x2 - 3x + 8x - 8 = 3x2 + 5x - 8

Now we know that we need to identify the wrong factors:

3x2 + 5x - 8 ≠ 3x2 + 2x - 8

Factor Second Degree Polynomials (Quadratic Equations) Step 10

Step 7. Change the factors if necessary

In the example used, let's try using 2 and 4 instead of 1 and 8:

(3x + 2)(x - 4)

Now the c term equals -8, but the outer/inner product (3x * -4) and (2 * x) equals -12x and 2x, which are not going to be combined to create the correct b term of +2x.

-12x + 2x = 10x

10x ≠ 2x

Factor Second Degree Polynomials (Quadratic Equations) Step 11

Step 8. Reverse the order if necessary

Let's try to move the 2 and the 4:

(3x + 4)(x - 2)

Now the c term (4 * 2 = 8) is still correct, but the outer/inner products are -6x and 4x. By combining them:

-6x + 4x = 2x

2x ≠ -2x We are close to 2x, but the signal is wrong.

Factor Second Degree Polynomials (Quadratic Equations) Step 12

Step 9. Check the signs if necessary

Keep the same order, but change the one with the minus sign:

(3x - 4)(x + 2)

Now the c term is still correct, but the outer/inner products are (6x) and (-4x). Like:

6x - 4x = 2x

2x = 2x It is now possible to recognize the positive term 2x from the original problem. These must be the right factors.

Method 2 of 6: Decomposition

This method identifies all the possible factors for the terms a and c and uses them to figure out what the factors should be. If the numbers are too big or the other methods seem more complicated, use this method. Let's use the example:

6x2 + 13x + 6

Factor Second Degree Polynomials (Quadratic Equations) Step 13

Step 1. Multiply the terms a and c

In this example, both equal 6.

6 * 6 = 36

Factor Second Degree Polynomials (Quadratic Equations) Step 14

Step 2. Find the value of term b by factoring and testing

You need to find two numbers that are factors of the product of a * c and are also equivalent to the term b (13) when added together.

4 * 9 = 36

4 + 9 = 13

Factor Second Degree Polynomials (Quadratic Equations) Step 15

Step 3. Substitute the two numbers obtained in the equation as the sum of the term b

Let's use k and h to represent the two numbers we get, 4 and 9:

ax2 + kx + hx + c

6x2 + 4x + 9x + 6

Factor Second Degree Polynomials (Quadratic Equations) Step 16

Step 4. Factor the polynomial through grouping

Arrange the equation so that you can factor out the greatest common factor of the first two and last two terms. Both factored groups must be the same. Add up the greatest common factors and place them in parentheses next to the factored group; the result will be the two factors:

6x2 + 4x + 9x + 6

2x(3x + 2) + 3(3x + 2)

(2x + 3)(3x + 2)

Method 3 of 6: Triple Match

Similar to decomposition, the "triple-start" method examines the possible factors of the products of the terms a and c, then uses them to find the value of b. As an example, consider the following equation:

8x2 + 10x + 2

Factor Second Degree Polynomials (Quadratic Equations) Step 17

Step 1. Multiply the terms a and c

This will help you identify the possibilities of the b term as well as the decomposition method. In this example, a equals 8 and c equals 2.

8 * 2 = 16

Factor Second Degree Polynomials (Quadratic Equations) Step 18

Step 2. Find two numbers with those numbers whose product and sum are equivalent to the term b

This step is identical to the decomposition method - you need to test and reject candidates for constants. The product of the terms a and c is 16, and the term c equals 10:

2 * 8 = 16

8 + 2 = 10

Factor Second Degree Polynomials (Quadratic Equations) Step 19

Step 3. Take these two numbers and test their substitution in the "triple match" formula

Take the two numbers from the previous step - let's call them h and k - and put them in this expression:

((ax + h)(ax + k)) / a

In this case, we will get:

((8x + 8)(8x + 2)) / 8

Factor Second Degree Polynomials (Quadratic Equations) Step 20

Step 4. See which of the two terms in the numerator is equally divisible by a

In this example, we're testing whether (8x + 8) or (8x + 2) can be divided by 8. (8x + 8) is divisible by 8, so let's divide this term by a and leave the others as they are.

(8x + 8) = 8(x + 1)

The term we are saving in this case is the remainder of the division by the term a: (x + 1)

Factor Second Degree Polynomials (Quadratic Equations) Step 21

Step 5. Take the greatest common factor of one or both terms, if any

In this example, the second term has the number 2 as its greatest common factor, since 8x + 2 = 2(4x + 1). Match this answer with the term identified in the previous step. These are the factors in the equation.

2(x + 1)(4x + 1)

Method 4 of 6: Difference of two roots

Some coefficients in polynomials can be identified as "roots", or the product of two numbers. Identifying these roots allows you to factor polynomials much more quickly. Consider the equation:

27x2 - 12 = 0

Factor Second Degree Polynomials (Quadratic Equations) Step 22

Step 1. Factor in the largest common factor if possible

In this case, we can see that 27 and 12 are both divisible by 3, so let's separate them:

27x2 - 12 = 3(9x2 - 4)

Factor Second Degree Polynomials (Quadratic Equations) Step 23

Step 2. Identify whether the equation's coefficients are square numbers

To use this method, you must be able to get the exact square root of terms. Note that the minus signs are left out, as these numbers are squares that can be products of two positive or negative numbers.

9x2 = 3x * 3x and 4 = 2 * 2

Factor Second Degree Polynomials (Quadratic Equations) Step 24

Step 3. Using the identified square roots, write down the factors

Take the values ​​of a and c from the step above (a = 9 and c = 4) and calculate their square roots – √ a = 3 and √ c = 2. They will be the factor coefficients of the expressions:

27x2 - 12 = 3(9x2 - 4) = 3(3x + 2)(3x - 2)

Method 5 of 6: Quadratic Formula

If the other methods fail and the equation is not evenly factored, use the quadratic formula. Consider the following example:

x2 + 4x + 1 = 0

Factor Second Degree Polynomials (Quadratic Equations) Step 25

Step 1. Substitute the corresponding values ​​into the quadratic formula:

x = -b ± √(b2 - 4c)

---------------------

2nd

We get the expression:

x = -4 ± √(42 - 4•1•1) / 2

Factor Second Degree Polynomials (Quadratic Equations) Step 26

Step 2. Calculate the value of x

You should get two values ​​for x. As shown above, we get two answers:

x = -2 + √(3) or x = -2 - √(3)

Factor Second Degree Polynomials (Quadratic Equations) Step 27

Step 3. Use the x values ​​to calculate the factors

Substitute the x values. They will be the factors. If we identify the two answers as h and k, we need to write the factors as follows:

(x - h) (x - k)

In this case, the final answer is:

(x - (-2 + √(3))(x - (-2 - √(3)) = (x + 2 - √(3))(x + 2 + √(3))

Method 6 of 6: Using a Calculator

If it is possible to use it, a graphing calculator makes the factoring process much easier, especially in tests. The following instructions are for a graphing calculator. Consider the following example:

y = x2 − x − 2

Factor Second Degree Polynomials (Quadratic Equations) Step 28

Step 1. Enter the equation into the calculator

You will use an equation solver, also known as a [Y =] screen.

Factor Second Degree Polynomials (Quadratic Equations) Step 29

Step 2. Graph the equation on the calculator

After typing in the equation, press the [GRAPH] key - you should see an arc representing the equation (and it will be an arc since we're dealing with polynomials).

Factor Second Degree Polynomials (Quadratic Equations) Step 30

Step 3. See where the arc intersects the x axis

Since polynomial equations are usually written as ax2 + bx + c = 0, these are the two values ​​of x that make the expression equal to zero:

(-1, 0), (2, 0)

x = -1, x = 2

If you cannot identify where the graph crosses the x-axis, press [2nd] and then [TRACE]. Press [2] or select "zero". Slide the cursor to the left of the intersection and press [ENTER]. Slide the cursor to the right of the intersection and press [ENTER]. Slide the cursor as close to the intersection and press [ENTER]. The calculator will find the value of x. Do the same for the other intersection

Factor Second Degree Polynomials (Quadratic Equations) Step 31

Step 4. Substitute the x values ​​obtained in the previous step into two factor expressions

When using the two values ​​of x (h and k), the expression used will be:

(x - h)(x - k) = 0

Therefore, the two factors must be:

(x - (-1))(x - 2) = (x + 1)(x - 2)

Tips

  • If you have a TI-84 (graphics) calculator, there is a program called "SOLVER" that solves a quadratic equation. It also solves polynomials of other degrees.
  • If a term does not exist, the coefficient is 0. It may be useful to rewrite the equation if it does, for example: x2 + 6 = x2 + 0x + 6.
  • If you factored a polynomial using the quadratic formula and got answers with radicals, convert the x values ​​to fractions to check them out.
  • If the term does not have a written coefficient, it will be 1, that is, x2 = 1x2.
  • After a lot of practice, you'll eventually be able to factor out polynomials in your head. Until then, write them down on paper.

Notices

Popular by topic