Probably the most common thing you will be doing with polynomials is "combining like terms". This is the process of adding together whatever terms you can, but not overdoing it by trying to add together terms that can't actually be combined. So which terms can be combined, and why?
Terms can be combined only if they have the exact same variable portion. And, by "exact same", I mean "the same variable(s), raised to the same power(s)". Here is a rundown of what's what:
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These are not like terms...
4x and 3
...because the first term contains a variable and the second term does not.
These are not like terms...
4x and 3y
...because the two terms contain different variables.
These are not like terms...
4x and 3x^{2}
...because the variables are the same, but the powers on those variables are not.
These are like terms...
4x and 3x
...because the variables are the same, and so are their powers.
But these are not like terms...
4x and 3xy
...because the second term has an additional variable, so the variable portions don't match.
So, to decide if two terms are "like" terms that can be combined, we look at the variable portion of those terms. The numerical portion of two terms is what will get combined (as we'll see shortly), but it's the variable portion of those two terms that determines if the terms are combine-able. To be able to be combined, the terms' variable portions must contain the exact same variable(s) with the exact same power(s).
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Once you have determined that two terms are indeed "like" terms and can indeed therefore be combined, you can then deal with the terms in a manner similar to what you did in grammar school. When you were first learning to add, you would do "five apples and six apples is eleven apples". You have since learned that, as they say, "you can't add apples and oranges". That is, "five apples and six oranges" is just a big pile of fruit; it isn't something like "eleven applanges". Combining like terms works much the same way; we add the numerical parts, while carrying along the variable parts, almost like a unit, or like the "apples" we'd added.
Looking at these two terms, I see that each contains the variable x, and the variable has the same (understood) power of 1 in each term. So these are like terms, and I can combine them.
Back in grade-school arithmetic, "three apples plus four apples" got combined into "seven apples" by adding the three and the four to get seven, and bringing the "apples" along for the ride. In the same way, I will combine these two like terms by adding the numerical portion of each term, while carrying the x along for the ride:
3x + 4x
(3 + 4) x
(7) x
I showed every step above in order to highlight how the terms are being combined. I'm adding the 3 and the 4, and carrying the x along with the numerical result. My answer is:
7x
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It is often best to group like terms together first, and then simplify:
2x^{2} + 3x – 4 – x^{2} + x + 9
(2x^{2} – x^{2}) + (3x + x) + (–4 + 9)
(2 – 1) x^{2} + (3 + 1) x + (5)
(1) x^{2} + (4) x + 5
x^{2} + 4x + 5
In the second line above, many students find it helpful to write in the understood coefficient of 1 in front of any variable expressions having no written coefficient, as is shown in red below:
(2x^{2} – x^{2}) + (3x + x) + (–4 + 9)
(2x^{2} – 1x^{2}) + (3x + 1x) + (–4 + 9)
(2 – 1) x^{2} + (3 + 1) x + (5)
1x^{2} + 4x + 5
x^{2} + 4x + 5
It is not required that the understood 1 be written in when simplifying expressions like this, but many students find this technique to be very helpful, at least when they're starting out. Whatever method helps you consistently complete the simplification correctly is the method you should use.
I will start by grouping the terms according to their degree.
10x^{3} – 14x^{2} + 3x – 4x^{3} + 4x – 6
(10x^{3} – 4x^{3}) + (–14x^{2}) + (3x + 4x) – 6
6x^{3} – 14x^{2} + 7x – 6
Warning: When moving the terms around, remember that the terms' signs move with them. Don't mess yourself up by leaving orphaned "plus" and "minus" signs behind.
If it helps you keep things straight, rewrite the expression like this:
10x^{3} + (–14x^{2}) + 3x + (–4x^{3}) + 4x + (–6)
By turning the subtractions into additions of negatives, it is clear where the "minus" signs belong, and it's easier to move them correctly:
10x^{3} + (–4x^{3}) + (–14x^{2}) + 3x + 4x + (–6)
(10 – 4) x^{3} – 14x^{2} + (3 + 4) x – 6
...and so forth. Do what works for you.
The first thing I need to do is take the negative through the parentheses:
25 – (x + 3 – x^{2})
25 – x – 3 + x^{2}
Okay; these terms are not only not in descending order, they're almost completely backwards! I'll put them the right way 'round, and then simplify by combining the two constants, which are the only like terms:
x^{2} – x + 25 – 3
x^{2} – x + 22
Many students, especially when starting out, experience difficulties with the "minus" signs, including when taking them through a parenthetical like I just did above. If it helps you to keep track of what's going on, try putting the "understood" 1 in front of the parentheses (highlighted in red below), and then taking that through onto the terms within:
25 – (x + 3 – x^{2})
25 – 1(x + 3 – x^{2})
25 – 1(x) – 1(+3) – 1(–x^{2})
25 – 1x – 3 + 1x^{2}
1x^{2} – 1x + 25 – 3
1x^{2} – 1x + 22
x^{2} – x + 22
While the first format (without the 1's being written in) is the more "standard" format, either format is mathematically valid. You should use the format that works most successfully for you. Don't be shy about inserting the understood 1 when you're starting out; don't feel bound to continue using it once you get to feeling confident without it.
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This is just an order of operations problem with a variable in it. If I work carefully from the inside out, paying careful attention to my "minus" signs, then I should be fine:
x + 2(x – [3x – 8] + 3)
x + 2(x – 1[3x – 8] + 3)
x + 2(x – 1[3x] – 1[–8] + 3)
x + 2(x – 3x + 8 + 3)
x + 2(–2x + 11)
x + 2(–2x) + 2(+11)
x – 4x + 22
1x – 4x + 22
–3x + 22
Just so you know, this is the kind of problem that us math teachers love to put on tests (yes, some of us are kinda sick puppies), so you should expect to need to be able to deal with nested grouping symbols like this.
I'll work from the inside out:
[(6x – 8) – 2x] – [(12x – 7) – (4x – 5)]
[6x – 8 – 2x] – [12x – 7 – 1(4x) – 1(–5)]
[6x – 2x – 8] – [12x – 7 – 4x + 5]
[4x – 8] – [12x – 4x – 7 + 5]
4x – 8 – [8x – 2]
4x – 8 – 1[8x] – 1[–2]
4x – 8 – 8x + 2
4x – 8x – 8 + 2
–4x – 6
As always, I'll start at the innermost grouping, and simplify my way out to the answer.
–4y – [3x + (3y – 2x + {2y – 7} ) – 4x + 5]
–4y – [3x + (3y – 2x + 2y – 7) – 4x + 5]
–4y – [3x + (–2x + 3y + 2y – 7) – 4x + 5]
–4y – [3x + (–2x + 5y – 7) – 4x + 5]
–4y – [3x – 2x + 5y – 7 – 4x + 5]
–4y – [3x – 2x – 4x + 5y – 7 + 5]
–4y – [3x – 6x + 5y – 7 + 5]
–4y – [–3x + 5y – 2]
–4y – 1[–3x] – 1[+5y] – 1[–2]
–4y + 3x – 5y + 2
3x – 4y – 5y + 2
3x – 9y + 2
The simplification above involved two different variables. It is customary (though not technically required) to list the terms so that their variables are in alphabetical order. That's why I listed the 3x before the 9y. As always, the constant term goes at the very end.
(If you think you need more practice with this last type of problem (with all the brackets and the negatives and the parentheses, then you might want to review the "Simplifying with Parentheses" lesson.)
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When we're talking apples and oranges, it's usually pretty easy to see what should be combined with what else, and in what manner. But once we start dealing with variables, things can get confusing. Don't get careless and confuse multiplication and addition. This may sound like a silly thing to me to say, but it may be the most commonly-made student mistake when combining like terms (after messing up the order of operations).
This is multiplication...
(x)(x) = (x^{1})(x^{1})
= x^{1+1} = x^{2}
...so it is simplified using rules for combining exponents.
This is addition...
x + x = 1x + 1x
= (1 + 1) x = 2x
...so it is simplified using rules for combining coefficients.
Let me be very clear:
(x)(x) does not equal x + x
x^{2} does not equal 2x
So if you have something like x^{3} + x^{2}, do not try to say that this somehow equals something like x^{5} or 5x. The terms are unlike, and cannot be combined. If you have something like 2x + x, do not try to say that this somehow equals something like 2x^{2}. These terms are like terms, and are combined by adding their coefficients.
Take your time, and make sure you are keeping straight in your head how multiplication works, versus how addition works. In fact, "combining like terms" is a topic for which it would be difficult to do too much practice. Do as many practice problems as you can!
URL: https://www.purplemath.com/modules/polydefs2.htm
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