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
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.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
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