Solving Equations II

Free Math Help Online presents Solving Equations II.  My last post, Solving Equations, covered linear equations, and polynomials that factor to linear factors (if you need help understanding how to do the factoring, please look back at prior factoring posts).  Today I will cover Completing the Square, and the Quadratic Formula for solving quadratic equations.  Further posts will cover more complicated equations like rational equations, radical equations, exponential and logarithmic equations, and the cubic and quartic formulas.

1.     Completing the Square: This is a method involves rewriting a part of a quadratic equation as a perfect square and then taking the square root of both sides to find the solution.  To begin with, lets look at a few examples the fairly easy quadratic equations to solve.  The first one is x² = 121.  It may be tempting to just say that the answer is just 11, but this would be incomplete.  To start with I will use the factoring method from the Solving Equations post:
x² = 121
x² - 121 = 0
(x - 11)(x + 11) = 0
x = 11, or -11

Another way to write this is x = ±11, read “x equals plus or minus 11”.  Checking is very easy because 11² = 121 and (-11) ² = 121.  As you can see, 11 was only part of the solution.  This is because squaring has the nice property that both negative and positive values turn out positive when the value is squared.  Thus, whenever you take the square root of both sides of an equation you have to include both the positive and negative values so you must always introduce the ± in your answer or write two separate equations.  To solve this equation without factoring you would write it as follows:
x² = 121
√(x²) = ±√(121)
x = ±11

Now lets do another one:
(2x + 7)² = 25
2x + 7 = ±5
2x = -7 ± 5
x = (-7 ± 5)/2
x = (-7 + 5)/2 or x = (-7 - 5)/2
x = -1, -6

Check: x = -1
(2(-1) + 7)² = 25?
5² = 25? YES

Check: x = -6
(2(-6) + 7)² = 25?
(-5)² = 25? YES

So that is not so bad as long as the quadratic is written as a perfect square.  But now we need to figure out how to solve an equation like 4x² + 24 = -14x.  We could, of course, add 14x to both sides and factor, but this only works if the equation does factor.  Not all will.  This is why we now will learn the Completing the Square method.  This method makes use of the following formula:
(x + y)² = x² + 2xy + y²

You can verify this formula with the FOIL method, or there is a beautiful geometric explanation that involves a big square, a little square, and two rectangles.  I will not cover these here, but I will write a future post on multiplying polynomials that covers this in detail.  For now, I ask that you either verify it for yourself or take my word for it.

To do the Completing the Square method, first rearrange the terms so that the variable terms are all on one side and the constant term is on the other side.  Then, if the leading coefficient is not 1, divide by that number on both sides:
4x² + 24 = -28x
4x² + 28x = -24
x² + 7x = -6

To use the equations above, the coefficient of the middle term will be 2y.  Take half of this to find y, and then add y² to both sides of the equation.
y = 7/2.  Add (7/2)² to both sides of the equation.
x² + 7x = -6
x² + 7x + (7/2)² = -6 + (7/2)²
(x + 7/2)² = -24/4 + 49/4
(x + 7/2)² = 25/4
x + 7/2 = ±5/2
x = -7/2 ± 5/2
x = -7/2 + 5/2 or x = -7/2 - 5/2
x = -1, -6

Check: x = -1
4(-1)² + 24 = -28(-1)?
4 + 24 = 28? YES

Check: x = -6
4(-6)² + 24 = -28(-6)?
144 + 24 = 168? YES

Now that we have been through the process once, lets boil down the actual mechanics.
- Rearrange the terms if needed
- Divide by the leading coefficient
- Add (half the coefficient of x)² to both sides
- Write the variable side as a perfect square
- Take the square root of both sides
- Solve the linear equations

Lets try this on another:
9x² - 29x + 6 = 0
9x² - 29x = -6
x² - 29/9 x = -2/3
x² - 29/9 x + (-29/18)² = -2/3 + 841/324
(x - 29/18)² = -216/324 + 841/324
(x - 29/18)² = 625/324
x - 29/18 = ±25/18
x = 29/18 ± 25/18
x = 54/18 or x = 4/18
x = 3 or x = 2/9

Check: x = 3
9(3)² - 29(3) + 6 = 0?
81 - 87 + 6 = 0? YES

Check: x = 2/9
9(2/9)² - 29(2/9) + 6 = 0?
4/9 - 58/9 + 54/9 = 0? YES

Granted, this was tedious and could have been solved faster factoring by the AC method, but it is a good example to see that even when there are fractions involved the method still works.  The final two examples in this section will be quadratic equations that do not have rational solutions:
x² - 5x + 3 = 0
x² - 5x = -3
x² - 5x + (-5/2)² = -3 + 25/4
(x - 5/2)² = -12/4 + 25/4
(x - 5/2)² = 13/4
x - 5/2 = ± √(13/4)
x - 5/2 = ± √(13)/2
x = 5/2 ± √(13)/2 * My preferred solution expression
x = (5 ± √(13))/2 * Another popular solution expression

Check: 5/2 + √(13)/2
(5/2 + √(13)/2)² - 5(5/2 + √(13)/2) + 3 = 0?
25/4 + 5√(13)/2 + 13/4 - 25/2 - 5√(13)/2 + 3 = 0?
38/4 - 25/2 + 3 = 0?
19/9 - 25/2 + 3 = 0?
-6/2 + 3 = 0? YES

Check: 5/2 - √(13)/2
(5/2 - √(13)/2)² - 5(5/2 - √(13)/2) + 3 = 0?
25/4 - 5√(13)/2 + 13/4 - 25/2 + 5√(13)/2 + 3 = 0?
38/4 - 25/2 + 3 = 0?
19/9 - 25/2 + 3 = 0?
-6/2 + 3 = 0? YES

Final Completing the square example:
x² + 6x + 11 = 0
x² + 6x = -11
x² + 6x + (3)² = -11 + 9
(x + 3)² = -2
x + 3 = √(-2)

The square root of a negative number is not a real number.  It is a complex number, and I will deal with complex numbers in a future post.  For the purposes of this post, any time I get a square root of a negative I will stop and report the answer as “No real solutions”

2.     Quadratic Formula: Completing the square can be tedious and boring.  It would be nice if we did not have to wade through all the steps and could just jump to the answer.  This is exactly what the quadratic formula does.  To derive the quadratic formula, start with a generic quadratic equation and solve it by completing the square:
ax² + bx + c = 0
ax² + bx = -c
x² + b/a x = -c/a
x² + b/a x + (b/(2a))² = -c/a + b2/(4a²)
(x + b/(2a))² = b²/(4a²) - 4ac/(4a²)
(x + b/(2a))² = (b² - 4ac)/(4a²)
x + b/(2a) = ±√((b² - 4ac)/(4a²))
x + b/(2a) = ±√(b² - 4ac)/(2a)
x = -b/(2a) ± √(b² - 4ac)/(2a) * My preferred expression
x = (-b ± √(b² - 4ac))/(2a) * Another popular expression

The quadratic formula, the way I like to write it, is x = -b/(2a) ± √(b² - 4ac)/(2a).  I prefer to write it as two separate fractions because I have seen students make fewer simplification mistakes when using it this way, and because it will be more informative when you learn other topics such as the axis of symmetry of a parabola (another future post).  Most text books write it as a single fraction with -b ± √(b² - 4ac all in the numerator and 2a in the denominator.  Be aware that both are valid.

Now that we have the quadratic formula, what do we do with it? c For any quadratic equation, all you need to do to solve it is to first put it in standard form, ax² + bx + c = 0, identify the coefficients a, b, and c, and use these values in the quadratic formula.  As examples I will re-solve all the quadratic equations that I solved with completing the square (checking will not be doe since I already checked them all):
4x² + 24 = -28x
4x² + 28x + 24 = 0
a = 4, b = 28, c = 24
x = -28/(2*4) ± √(28² - 4*4*24)/(2*4)
x = -7/2 ± √(784 - 384)/8
x = -7/2 ± √(400)/8
x = -7/2 ± 20/8
x = -7/2 ± 5/2
x = -1, -6

9x² - 29x + 6 = 0
a = 9, b = -29, c = 6
x = 29/(2*9) ± √((-29)² - 4*9*6)/(2*9)
x = 29/18 ± √(841 - 216)/18
x = 29/18 ± √(625)/18
x = 29/18 ± 25/18
x = 3, 2/9

x² - 5x + 3 = 0
a = 1, b = -5, c = 3
x = 5/(2*1) ± √((-5)² - 4*1*3)/(2*1)
x = 5/2 ± √(25 - 12)/2
x = 5/2 ± √(13)/2

x² + 6x + 11 = 0
a = 1, b = 6, c = 11
x = -6/(2*1) ± √(6² - 4*1*11)/(2*1)
x = -3 ± √(36 - 44)/2
x = -3 ± √(-8)/2
No real solutions

The quadratic formula can even be used to solve very simple quadratic equations:
x² = 121
x² - 121 = 0
a = 1, b = 0, c = -121
x = 0/(2*1) ± √(0² - 4*1*(-121))/(2*1)
x = ±√(484)/2
x = ±22/2
x = ±11

Another application of the quadratic formula is to factor a quadratic expression for which it is difficult for you to find the factors or even one that does not have rational factors.  For example lets factor the expression 60x² - 17x - 21.  First I will set it equal to 0, factor out any common constant factor if any, and use the quadratic formula solve the equation.  Next I will write each answer separately.  Finally, I will rearrange the two answer equations to be equal to zero with no fractions.  This will give the factors.
60x² - 17x - 21 = 0
a = 60, b = -17, c = -21
x = 17/(2*60) ± √((-17)² - 4*60*(-21))/(2*60)
x = 17/120 ± √(5329)/120
x = 17/120 ± 73/120
x = 90/120, x = -56/120
x = 3/4, x = -7/15
4x = 3, 15x = -7
4x - 3 = 0, 15x + 7 = 0
60x² - 17x - 21 = (4x - 3)(15x + 7)

You can check this with the FOIL method.  One final example will show how to write a quadratic expression in factored form even if the expression has no rational factors:
x² + 2x - 4
x² + 2x - 4 = 0
a = 1, b = 2, c = -4
x = -2/(2*1) ± √(2² - 4*1*(-4))/(2*1)
x = -1 ± √(20)/2
x = -1 ± 2√(5)/2
x = -1 ± √(5)
x = -1 + √(5), x = -1 - √(5)
x + 1 - √(5) = 0, x + 1 + √(5) = 0
x² + 2x - 4 = (x + 1 - √(5))(x + 1 + √(5))

In conclusion, I’d like to mention that although we only used the Completing the Square method as a stepping stone for deriving the Quadratic Formula, there are many future uses for completing the square such as finding important points for circles, parabolas, ellipses, and hyperbolas.

In my next post I will discuss solving various other kinds of equations, many of which will give rise to quadratic equations in the solutions.  Some of these methods will also give rise to extraneous solutions, which means a solution that was introduced during the solving of the problem that is not really a solution of the original equation.

Solving Equations

Free Math Help Online presents Solving Equations.  My first three posts were all about factoring: Factoring Polynomials, Factoring Polynomials II, and Factoring Polynomials III.  Now I will start to cover Solving Equations.  Today I will cover linear equations, and polynomials that factor to linear factors (if you need help understanding how to do the factoring, please look back at prior posts).  The next post, Solving Equations II, will cover completing the square and the quadratic formula for solving quadratic equations.  Further posts will cover more complicated equations like rational equations, radical equations, exponential and logarithmic equations, and the cubic and quartic formulas.

1.     Linear Equations: A linear equation is an equation that only has a variable with no powers and constants.  Linear equations are the easiest polynomial equations to solve because they only require addition, subtraction, multiplication and division.  Paying attention to the mechanics can be valuable to understanding what to do with other equations that are less familiar.  I consider equations to be balances where the weight on one side is equal to the weight in the other side.  One way to look at solving linear equations is to “undo” what ever has been done to the variable in reverse order of operations, but keep the equation balanced at all times.  This means that whatever is done to one side of the equation must also be done to the other side of the equation.  Consider the following linear equation 3x + 5 = 14.  On the left we have a variable that has been multiplied by three and then had 5 added to it.  To solve this, I will first “undo” the added 5 by subtracting 5 from both sides of the equation.  Then I will “undo” the multiplication by dividing by 3 on both sides of the equation:
3x + 5 = 14
3x + 5 - 5 = 14 - 5
3x = 9
3x/3 = 9/3
x = 3

When solving an equation you should always check your answer:
3*3 + 5 = 14? YES

Here is another slightly more complicated equation, 6(4 - 7x) - 3 = 14.  There is more than one approach that can be used here.  I could undo the subtracted 3, undo the multiplied 6, undo the added 4, and undo the multiplied -7.  Another solution method could be to distribute the 6 through the parentheses, combine like terms, and the solve in a method similar to the first.  I will do both here.  Solution 1:
6(4 - 7x) - 3 = 14
6(4 - 7x) - 3 + 3 = 14 + 3
6(4 - 7x) = 17
6(4 - 7x)/6 = 17/6
4 - 7x = 17/6
4 - 7x - 4 = 17/6 - 4
-7x = 17/6 - 24/6 = -7/6
-7x/(-7) = (-7/6)/(-7)
x = 1/6

Solution 2:
6(4 - 7x) - 3 = 14
24 - 42x - 3 = 14
21 - 42x = 14
21 - 42x - 21 = 14 - 21
-42x = -7
-42x/(-42) = -7/(-42)
x = 1/6

Check:
6(4 - 7*1/6) - 3 = 14?
6(4 - 7/6) - 3 = 14?
6(24/6 - 7/6) - 3 = 14?
6(17/6) - 3 = 14?
17 - 3 = 14? YES

Things can get a more complicated still when there are multiple expressions with both the variable and constants.  In these cases it is usually best to distribute numbers through all parentheses, combine all like terms, move terms around (always keeping the equation balanced) to get all terms with the variable on one side of the equation, and finally solve as before.  This is not really difficult, it just requires attention to detail.  The most important thing to do in looking at the solution below is to pay attention to haw the negative values are handled with the distributions because many students make mistakes in distributing negatives.  Lets do the following:
3(2x + 1) - 4(2 - 5x) = -(x + 1) + 7(x - 2) + 11(x - 4)
6x + 3 - 8 + 20x = -x - 1 + 7x - 14 + 11x - 44
26x - 5 = 17x - 59
26x - 5 - 17x + 5 = 17x - 59 - 17x + 5
9x = -54
x = -6

Check:
3(2(-6) + 1) - 4(2 - 5(-6)) = -(-6 + 1) + 7(-6 - 2) + 11(-6 - 4)?
3(-12 + 1) - 4(2 + 30) = -(-5) + 7(-8) + 11(-10)?
3(-11) - 4(32) = 5 - 56 - 110?
-33 - 128 = -51 - 110?
-161 = -161? YES

When the expressions contain fractions my approach is to multiply both sides of the equation (to keep it balanced) by the least common multiple of all the denominators involved.  This is my approach regardless of whether the denominators are constants or contain variables, but the example here has only constant denominators.  When there are variables in the denominators the equations are called rational equations and I am only dealing with linear equations here.  Rational equations will be examined in a later post.
3x/5 + 7x/10 - 2 = x/20
20*3x/5 + 20*7x/10 - 20*2 = 20*x/20
4*3x + 2*7x - 40 = x
12x + 14x - 40 = x
26x - 40 = x
26x - x = 40
25x = 40
x = 40/25 = 8/5

Check:
(3*8/5)/5 + (7*8/5)/10 - 2 = (8/5)/20?
24/25 + 56/50 - 2 = 8/100?
24/25 + 28/25 - 50/25 = 2/25?
52/25 - 50/25 = 2/25? YES

2.     Polynomials that Factor: The factoring method for solving polynomials always starts with setting a polynomial equal to zero, factoring, and using the Zero Product Rule to solve the equation.  The Zero Product Rule says that if there is a product of factors equal to zero than one of the factors must be zero.  This is because you can never multiply non-zero numbers and get a zero as the answer.  Here is a typical equation that can be solved by factoring, x² - 9x = 22.  First we will subtract the 22 from both sides (still keep the equation balanced) to get it equal to zero.  We will then factor, set both factors equal to zero, and find two solutions:
x² - 9x = 22
x² - 9x - 22 = 0
(x - 11)(x + 2) = 0
x - 11 = 0 or x + 2 = 0
x = 11 or x = -2

Check: x = 11
11² - 9*11 = 22?
121 - 99 = 22? YES

Check: x = -2
(-2)² - 9(-2) = 22?
4 + 18 = 22? YES

The degree of the polynomial is the maximum number of solutions there are to the equation.  This was a quadratic equation (degree 2) and there were 2 solutions.  Even if the polynomial factors completely into linear factors, there may be fewer solutions than the degree.  For example, the next example is a 4th degree polynomial that has only 2 solutions:
2x⁴ + 50x² = 20x³
2x⁴ - 20x³ + 50x² = 0
2x²(x² - 10x + 25) = 0
2x²(x - 5)² = 0
2x*x(x - 5)(x - 5) = 0
2x = 0, x = 0, x - 5 = 0, or x - 5 = 0
x = 0, 0, 5, or 5
x = 0, 5

Check: x = 0
2*0⁴ + 50*0² = 20*0³? YES

Check: x = 5
2*5⁴ + 50*5² = 20*5³?
2*625 + 50*25 = 20*125?
1250 + 1250 = 2500? YES

Even though there were four variable factors, there were only two unique equations when the factors were set equal to 0.  You probably noticed this when you saw that the factors were squared.  The power of the factor is referred to as the multiplicity of the solution.  This is an important property once we start trying to use the solutions to polynomial equations to understand what the graph of a polynomial looks like.  For this equation, the solutions are x = 0 with a multiplicity of 2 and x = 5 with a multiplicity of 2.

Now lets try another:
9x³ + 9x² - 4x - 4 = 0
9x²(x + 1) - 4(x + 1) = 0
(x + 1)(9x² - 4) = 0
(x + 1)(3x - 2)(3x + 2) = 0
x + 1 = 0, 3x - 2 = 0, 3x + 2 = 0
x = -1, 2/3, or -2/3, each with multiplicity of 1

Check: x = -1
9(-1)³ + 9(-1)² - 4(-1) - 4 = 0?
-9 + 9 + 4 - 4 = 0? YES

Check: x = 2/3
9(2/3)³ + 9(2/3)² - 4(2/3) - 4 = 0?
9*8/27 + 9*4/9 - 8/3 - 4 = 0?
8/3 + 4 - 8/3 - 4 = 0? YES

Check: x = -2/3
9(-2/3)³ + 9(-2/3)² - 4(-2/3) - 4 = 0?
-9*8/27 + 9*4/9 + 8/3 - 4 = 0?
-8/3 + 4 + 8/3 - 4 = 0? YES

Now, does it get boring doing all the checking? Yes.  Do you still have to do it?  That depends entirely on how correct you want your answers to be.  For example, as I typed this I accidentally missed the “-“ key when typing the answer above and wrote x = 1, 2/3, or -2/3.  I caught my mistake when I did the check because the check did not work.

Check: x = 1
9(1)³ + 9(1)² - 4(1) - 4 = 0?
9 + 9 - 4 - 4 = 0? NO, wrong answer

As you may have noticed, these linear equations that we solve at the end are very easy.  Most people do not even take the time to write out these individual equations and just go directly from the factored form to the final answer.
(x + 1)(3x - 2)(3x + 2) = 0
x = -1, 2/3, or -2/3

As a final example, lets solve a 10th degree polynomial equation.  Generally speaking, this would be very hard because 10th degree polynomials are very hard to factor if they do factor, but the example here will be given in factored form already:
5x(2x - 3)³(x - 2)²(7x + 6)(6x - 5)(x + 812)(x - 1) = 0
These solutions have multiplicity of 1: x = 0, -6/7, 5/6, -812, or 1
x = 2 has multiplicity of 2, and x = 3/2 has multiplicity of 3.

Checking here is really only verifying that one of the factors is 0.  The rest of the factors do not need to be calculated because if there is one 0 in the string of multiplications then the whole thing is 0.  I will show three of the checks and leave the rest to you.

Checks:
x = 0
5*0*(-3)³*(-2)²*6*(-5)*812*(-1) = 0? YES

x = 5/6
5*5/6(2*5/6 - 3)³(5/6 - 2)²(7*5/6 + 6)(6*5/6 - 5)(5/6 + 812)(5/6 - 1) = 0?
5*5/6(2*5/6 - 3)³(5/6 - 2)²(7*5/6 + 6)(0)(5/6 + 812)(5/6 - 1) = 0? YES

x = 2
5*2(2*2 - 3)³(2 - 2)²(7*2 + 6)(6*2 - 5)(2 + 812)(2 - 1) = 0?
5*2(2*2 - 3)³(0)²(7*2 + 6)(6*2 - 5)(2 + 812)(2 - 1) = 0? YES

Once you have digested this go on to the next post, Solving Equations II, in which I cover Completing the Square and the Quadratic Formula for solving quadratic equations.

Factoring Polynomials III

Free Math Help Online presents Factoring Polynomials Part III.  In my first post, Factoring Polynomials, I covered factoring out a common factor from all terms, factoring quadratics, and factoring using the basics of the special rules for perfect squares, difference of squares, and sum and difference of cubes.  In Factoring Polynomials II I covered factoring by grouping, and recognizing special forms.  This post will cover factoring negative and fractional exponents, and putting it all together.  My next post will cover Solving Equations.


If you get lost in any of this, please go back to Factoring Polynomials and Factoring Polynomials II first and make sure you understand these before moving on.

1. Negative and Fractional Exponents: Once we have either negative or fractional exponents, we are actually no longer dealing with polynomials, but these can still often be factored using the same methods as polynomials.  You may or may not see questions like these in a high school algebra class, but you will in a pre-calculus class because expressions like this arise commonly in calculus problems.  First, lets consider how to factor out a common factor:
   7x^-5 - 3x^-3 + 4x^-2
   
   The common thing to do here is to factor out the most negative exponent, in this case, x^-5.  To do this, use the rules of exponents to rewrite each term with a factor of x^-5
   7x^-5 - 3x^-3 + 4x^-2
   7x^-5 - 3x^-5x² + 4x^-5x³
   x^-5(7 - 3x² + 4x³)
   
   Factoring fractional common factors is done similarly by using the rules of exponents to rewrite each term with the fractional exponent factor:
   8x^13/3 + x^7/3 - 5x^4/3
   8x^4/3x³ + x^4/3x - 5x^4/3
   x^4/3(8x³ + x - 5)
   
   You can also factor out negative fractional exponents:
   x^1/2 + x^-1/2
   x^-1/2x + x^-1/2
   x^-1/2(x + 1)
   
   Expressions with negative and fractional exponents can also often be written in quadratic or special forms:
   x^-2/3 - 8x^-1/3 - 20
   (x^-1/3 - 10)(x^-1/3 +2)
   
   x^-2 - y^-4
   (x^-1 + y^-2)(x^-1 - y^-2)
   
2. Putting It All Together: Sometimes you must put two or more of the factoring techniques together to completely factor an expression:
   3x⁶ - 192y¹² (factor out 3)
   3(x⁶ - 64y¹²) (use difference of squares***)
   3(x³ + 4y⁶)(x³ - 4y⁶) (use sum and difference of cubes)
   3(x + 2y²)(x² - 2xy² + 4y⁴)(x - 2y²)(x² + 2xy² + 4y⁴)
   
   ***Note: It is possible to use either use difference of squares or use difference of cubes on x⁶ - 64y¹².  When this is the case, always use difference of squares.  This is because the difference of cubes formula will leave two factors that both do factor, but the second one is not factorable by any of the methods I have shown in any of these posts, it is much harder.  To see this, lets factor x⁶ - 64y¹² by difference of cubes first:
   (x² - 4y⁴)(x⁴ + 4x²y⁴ + 16y⁸) (use difference of squares)
   (x + 2y²)(x - 2y²)(x⁴ + 4x²y⁴ + 16y⁸)
   
   Now x⁴ + 4x²y⁴ + 16y⁸ does factor to (x² - 2xy² + 4y⁴)(x² + 2xy² + 4y⁴) but you cannot figure that out by any of the methods I have shown.  You can verify it is true by multiplying the factors.  For this reason, always use difference of squares rather than difference of cubes when there is a choice.
   
   Here is another expression to factor I recently saw:
   x² - 12x + 36 - 49y² (group the first three terms)
   (x² - 12x + 36) - 49y² (use perfect square)
   (x - 6)² - 49y² (use difference of squares)
   (x - 6 + y)(x - 6 - y)
   
   Finally, here is an expression I just made up:
   (x + 12)(x - 1)(x - 3)^-1 + (x - 3)
   (x + 12)(x - 1)(x - 3)^-1 + (x - 3)^-1(x - 3)²
   (x - 3)^-1((x + 12)(x - 1) + (x - 3)²)
   (x - 3)^-1(x² + 11x - 12 + x² - 6x + 9)
   (x - 3)^-1(2x² + 5x - 3)
   (x - 3)^-1(2x - 1)(x + 3)

Well, I hope you enjoyed reading this as much as I enjoyed writing it, or at least I hope it has helped you.  My next post covers Solving Equations, with more to come soon.

Factoring Polynomials II

Free Math Help Online presents Factoring Polynomials Part II.  In my first post, Factoring Polynomials, I covered factoring out a common factor from all terms, factoring quadratics, and factoring using the basics of the special rules for perfect squares, difference of squares, and sum and difference of cubes.  In this post I will cover factoring by grouping, and recognizing special forms.  In Factoring Polynomials III I will cover factoring negative and fractional exponents, and putting it all together.  My next post will cover Solving Polynomial Equations.


If you get lost in any of this, please go back to Factoring Polynomials first and make sure you understand the basics before moving on.

1. Factoring by Grouping: Consider the following expression:
   6x² - 15x + 14x - 35
   
   If asked to factor this, the most common approach would be to first add like terms and then factor the quadratic like we did in Factoring Polynomials.  Doing this we could find that
   6x² - 15x + 14x - 35 = (3x + 7)(2x - 5)
   
   Being very conscientious, we would then check our work by multiplying with the FOIL method
   (3x + 7)(2x - 5) = 3x∙2x + 3x∙(-5) + 7∙2x + 7∙(-5)
   (3x + 7)(2x - 5) = 6x² - 15x + 14x - 35
   
   Notice that before combining the Inside and Outside terms, this is exactly what we started with.  To factoring by grouping you first group the terms in a convenient way so that each group is factorable, then factor each group, and finally look to see if you have created a new whole expression that can be factored itself.  For this example:
   6x² - 15x + 14x - 35
   (6x² - 15x) + (14x - 35)
   (3x∙2x + 3x∙(-5)) + (7∙2x + 7∙(-5))
   3x(2x - 5) + 7(2x - 5)
   (2x - 5)(3x + 7)
   
   Notice in the last step that the binomial (2x - 5) is now a factor in each term of the whole expression so it can be factored out leaving the answer (2x - 5)(3x + 7).  This is the same answer I got before, just with the factors written in reverse order.  This is, of course, perfectly OK.  Also, given that the problem was written with the - 15x + 14x rather than - x, factoring by grouping is much faster than the trial and error approach used before.
   
   Now, how can we figure out how to rewrite a quadratic binomial into an expression that can be factored by grouping? The answer, called the AC method, is very much like factoring a quadratic trinomial with leading coefficient of 1. Generically speaking, quadratics can be written in the form ax² + bx + c. When a is 1 all we had to do was find factors of c that add to b. When a is not 1 all we have to do is find factors of ac that add to b. Lets look again at 6x² - 15x + 14x - 35 = 6x² - x - 35. What are the factors of 6(-35) = -210 that add to -1? The factors of 210 are
   1, 210
   2, 105
   3, 70
   5, 42
   6, 35
   7, 30
   10, 21
   14, 15
   
   Since we actually need factors of -210 then we must consider each pair above with one positive and one negative. It is now easy to see that the factors we want are 14 and -15. The only question that remains is the order in which to put them. Fortunately it does not matter, but I recommend putting a positive number last if possible to reduce the likelihood of making a mistake factoring out a negative. I will do it the opposite way here, but pay attention to how the negative is handled.
   6x² + 14x - 15x - 35
   6x² + 14x + (-15x - 35)
   2x(3x + 7) - 5(3x + 7)
   (3x + 7)(2x - 5)
   
   Unfortunately, most cubic expressions are not easy factor by grouping, but in a High School Algebra course you are very likely to see cubic expressions that can be factored by grouping such as the following:
   x³ + 2x² - 7x - 14
   
   Again, there is one part that is tricky to most students, so pay close attention to how negative values are handled.  Notice that -7x - 14 is NOT the same thing as -(7x - 14) because when you distribute the negative in the second expression you change the sign of both terms on the inside: -(7x - 14) = -7x + 14 ≠ -7x - 14.  When we group here we will group the entire (-7x - 14) and then factor out the -7.
   x³ + 2x² - 7x - 14
   (x³ + 2x²) + (-7x - 14)
   x²(x + 2) - 7(x + 2)
   (x + 2)(x² - 7)
   
   Technically, the (x² - 7) can be factored further, but not with rational coefficients.  I will discuss this more in Factoring Polynomials III
   
   
2. Recognizing Special Forms: Here we will be using the quadratic form and special forms introduced in Factoring Polynomials.  Consider this expression:
   5x⁴ + 34x² + 24
   
   This is not a quadratic expression, it is a 4^th degree polynomial.  With a convenient substitution, however, it can be written in quadratic form.  Let t = x².
   
   5x⁴ + 34x² + 24
   5(x²)² + 34(x²)+ 24
   5t² + 34t+ 24
   (5t + 4)(t + 6)
   (5x² + 4)(x² + 6)
   
   Once you see how it is done, you probably won’t want to actually write the three step in the middle.  Instead it will look more like:
   5x⁴ + 34x² + 24
   (5x² + 4)(x² + 6)
   
   There is still trial and error going on to find the factors of 24 that work with the factors of 5 to give the 34, I just am not showing that part here.  Now lets try some that are of the other special forms:
   16x¹⁰ - 121z⁴ (Difference of Squares form, let t = 4x⁵ and u = 11z²)
   (4x⁵)² - (11z²)²
   t² - u²
   (t + u)(t - u)
   (4x⁵ + 11z²)(4x⁵ - 11z²)
   
   a¹² + 6a⁶b² + 9b⁴ (Perfect Square form, let t = a⁶ and u = 3b²)
   (a⁶)² + 2(a⁶)(3b²) + (3b²)²
   t² + 2tu + u²
   (t + u)²
   (a⁶ + 3b²)²
   
   27w⁹ - 64y¹² (Difference of Cubes form, let t = 3w³ and u = 4y⁴)
   (3w³)³ - (4y⁴)³
   t³ - u³
   (t - u)(t² + tu + u²)
   (3w³ - 4y⁴)((3w³)² + (3w³)(4y⁴) + (4y⁴)²)
   (3w³ - 4y⁴)(9w⁶ + 12w³y⁴ + 16y⁸)
   
   The more you practice these kinds of problems, the fewer of the substitution steps you will need to show.

If you need more factoring help, go on to the next post, Factoring Polynomials III as soon as it is available.

Factoring Polynomials

Free Math Help Online presents Factoring Polynomials.  In this post I will covered factoring out a common factor from all terms, factoring quadratics, and factoring using the basics of the special rules for perfect squares, difference of squares, and sum and difference of cubes.  In my next post, Factoring Polynomials II, I will cover factoring by grouping and recognizing special forms.  Factoring Polynomials III will cover factoring negative and fractional exponents, and putting it all together.  The following post will cover Solving Polynomial Equations.

Factoring is a skill you will need starting in high school algebra and will continue using in just about every other math course you will ever take even past calculus.  The more complicated the math course becomes, the more complicated factoring will be required.

Factors are just terms that multiply together.  For example, 63 can be written as 3·21 or 7·9.  This means 3, 7, 9, and 21 are all factors of 63 (so are 1 and 63, but those are boring factors).

Factoring just means finding factors.  Factoring polynomials means finding polynomial factors.  For example:

x² + 5x + 4 = (x + 1)(x + 4)

This is because multiplying the factors on the right gives you the quadratic on the left.

Now, without further ado, here are some of the common things to know for factoring:

1. Common Factors: This is the easiest thing to do in factoring, but it is often overlooked after more complicated factoring methods are covered.  Let's start with this expression:
   63x - 42
   
   Notice that the coefficient of each term has a common factor of 21.  This can be written as
   63x - 42 = 21·3x - 21·2 = 21(3x - 2)
   
   We have now factored out the 21.  The factors of 63x - 42 are 21 and 3x - 2.
   
   Now consider this expression:
   21x⁵ + 3x⁴ - 6x²
   
   Again, the coefficients each have a common factor (3 this time), but now there is also a common factor of x²
   21x⁵ + 3x⁴ - 6x² = 3x²·7x³ + 3x²·x² - 3x²·2 = 3x²(7x³ + x² - 2)
   
   
2. Factoring Quadratics: The most common type of factoring question in a high school algebra class is to factor a trinomial (a polynomial with three terms).  Polynomials with highest power 2 are called quadratics.  Quadratics with all non-zero coefficients are tinomials.  A very simple quadratic is the following:
   x² + 5x + 4
   
   We already know how to factor this one because it is written in factored form above.  The question is, how do we figure it out when the answer is not already given.  To solve this problem you need to remember how to multiply binomials (polynomials with two terms), often called the FOIL method.  When multiplying two linear binomials (an x term and a constant term), the First terms always give the x² term in the trinomial, the Outside and Inside terms always add to give the x term of the trinomial, and the Last terms always give the constant term of the trinomial.  This tells us the possible terms to try when attempting to factor a trinomial.  For our example, we start with factoring the First terms:
   (x ___)(x ___)
   
   This part was really easy because there is only one way to factor x².  It gets harder when the coefficient of the x² term is not a 1.  Next, we consider the possible factors of 4 for the Last terms.  There are more choices than you might think at first.  Naturally, 4 = 2·2, so 2 and 2 are a possibility, but 4 = 1·4 also.  Not only that, but 4 = (-2)·(-2) and 4 = (-1)·(-4).  These four different factorizations of 4 result in four different trinomials when used to fill in the blanks above, but only one of then will result in the trinomial we want.  The key is to also consider that the Outside and Inside terms always add to give the x term of the trinomial.  Again, when the coefficient of the x² term is a 1 this is easy because all we need to do is add the two factors.  What factors of 4 add to 5? They are 1 and 4!  We now have the answer:
   x² + 5x + 4 = (x + 1)(x + 4)
   
   Here is another easy one, this time without so many words involved.  Factor:
   x² - 4x - 12
   
   What are the factors of -12 that add to -4? They are -6 and 2
   x² - 4x - 12 = (x - 6)(x + 2)
   
   This process takes more trial and error when the problems get harder.  Factor:
   6x² - 23x + 15
   
   First we need factors of 6x² and 15, then we need to find the combination where the Outside and Inside terms add to -23.  Possible factors of 6x² are 3x·2x and x·6x.  Possible factors of 15 are (-3)·(-5) and (-1)·(-15).  Notice that I ignored the positive factors of 15.  This is because the positive factor will never give me a -23.  Next we need to try all the combinations of our 6x² and 15 factors to find the right set:
   3·(-3) + 2·(-5) = -23? NO
   3·(-5) + 2·(-3) = -23? NO
   3·(-1) + 2·(-15) = -23? NO
   3·(-15) + 2·(-1) = -23? NO
   1·(-3) + 6·(-5) = -23? NO
   1·(-5) + 6·(-3) = -23? YES These make up our Outside and Inside terms
   
   So now we know the correct numbers, all that is left is to write them in the correct order.  Notice that the 1x and -5 are NOT together, nor are the 6x and -3, otherwise they would not get multiplied in FOIL:
    6x² - 23x + 15 = (x - 3)(6x - 5)
   
   
3. There are many special forms that will crop up in your homework, quizzes, and tests.  Some of these are:
   a² + 2ab + b² = (a + b)² Perfect Square (sum)
   a² - 2ab + b² = (a - b)² Perfect Square (difference)
   a² - b² = (a + b)(a - b) Difference of Squares
   a³ + b³ = (a + b)(a² - ab + b²) Sum of Cubes
   a³ - b³ = (a - b)(a² + ab + b²) Difference of Cubes
   
   Some examples of these are:
   25x² - 49 = (5x + 7)(5x - 7)
   27t³ + 8 = (3t + 2)(9t² - 6t + 4)
   9z² - 24x + 16 = (3z - 4)²
   
   These can also be more complicated, but I am just sticking to some simple ones today.

This is really just touching the surface of polynomial factoring.  Go to Factoring Polynomials II to see some of the more complicated scenarios commonly found in algebra courses.