Adding and Subtracting Polynomials
What are Like-Terms?
For instance, the expressions \(5x^2\) and \(3.1x^2\) are like-terms because the only thing that is different about them is the coefficient out in front of the \(x^2\). Similarly, the expressions \(2(x^4-3x+2)\) and \(5(x^4-3x+2)\) are like-terms because, again, the only thing that is different about them is the coefficient out in front.
Adding and Subtracting Polynomials by Combining Like-Terms
Since we are not multiplying each of our quantities \((x^3-2x^2+x-5)\) and \( (2x^3+5x^2+3x+6)\) by anything, we can just drop the parentheses.
Protip: In general, any time you see subtraction in a polynomial expression turn all subtraction into “adding a negative.” For instance, the quantity \((x^3-2x^2+x-5)\) becomes \(x^3+(-2x^2)+x+(-5)\) (don’t need parentheses, but they might help).
Our like-terms are thus:
- \(x^3\) and \(2x^3\)
- \(-2x^2\) and \(5x^2\)
- \(x\) and \(3x\)
- \(-5\) and \(6\)
Rearrange the expression given in the problem so you have like-terms grouped together.
\((x^3-2x^2+x-5)+(2x^3+5x^2+3x+6)=x^3+2x^3+-2x^2+5x^2+x+3x+-5+6\)
Now add just the coefficients of the like-terms:
\( x^3+2x^3+-2x^2+5x^2+x+3x+-5+6=3x^3+3x^2+4x+1.\) Done.
Note that terms that look like they don’t have any coefficient technically have a coefficient of \(1\). Hence why \(x^3+2x^3=3x^3\); because \(x^3=1\cdot x^3\).
Since we are not multiplying each of our quantities \((x^3-2x^2+x-5)\) and \( (2x^3+5x^2+3x+6)\) by anything, we can just drop the parentheses. Just like in the last example, turn all subtraction into “adding a negative.”
$$\begin{align}(3x^4-2x^2+x)+(-5x^4+6x^3+2.5x^2-2x+7.22)&= 3x^4+-2x^2+x+-5x^4+6x^3+2.5x^2+-2x+7.22\end{align}$$
Now, group all like terms so they are beside one another. Note that some terms cannot be paired with another (such as \(6x^3\), and \(7.22\)); i.e. there are no other like-terms. In which case, just leave that term alone and do absolutely nothing to it. Combine the coefficients of all like-term pairs, as before.
$$\begin{align} 3x^4+-2x^2+x+-5x^4+6x^3+2.5x^2+-2x+7.22 &=3x^4+-5x^4+6x^3+-2x^2+2.5x^2+x+-2x+7.22\\&=-2x^4+6x^3+0.5x^2+-1x+7.22\end{align}$$
If you want, you can change all your “added negatives” back to subtraction, or you can leave things as you see them; up to you (or your teacher).
Recall that multiplication means “repeated addition,” indicating how many of something you have added together. For instance, the expression \(3x\) means you have \(3\) of those \(x\)’s (being added together). Similarly, an expression like \(5(x^2-4)\) means you have \(5\) of those \((x^2-4)\)’s.
So, when adding like-terms, we are simply counting the total number of a given term we have in an expression. For instance, if we were to add the expressions \(4x^2\) and \(7x^2\), the first says we have \(4\) of those \(x^2\)’s and the second says we have \(7\) of \(x^2\). So, adding asks “how many total \(x^2\)’s are there total?” In this case, we have \(11\) of those \(x^2\)’s, which is exactly what you’d get when adding the coefficients. Similar reasoning can be applied when you are subtracting like-terms.
Subtracting Polynomials
First, distribute the minus sign in front of \((2x^3+3x-5)\) to each term inside the parentheses, changing “\(+\)” to “\(-\)” and vice versa. Then drop the parentheses. This gives you:
$$\begin{align}(5x^3-3x^2+2x+1)-(2x^3+3x-5)&= 5x^3-3x^2+2x+1-2x^3-3x+5\end{align}$$
If you want to change subtraction to “adding a negative” and rearrange your terms, you can do that now. Then, combine like-terms.
$$\begin{align} 5x^3-3x^2+2x+1-2x^3-3x+5 &= 5x^3+-3x^2+2x+1+-2x^3+-3x+5\\&=5x^3+-2x^3+-3x^2+2x+-3x+5\\&=3x^3-3x^2-1x+5\end{align}$$
Note, in the last step, I changed all the negatives back to subtraction.
When asked to find the difference of polynomials, such as in \((2x^3-4x+5)-(5x^3-3x+2)\), we “distribute” the minus sign through the second quantity, changing the signs of each term in that quantity. This gives you
$$\begin{align} (2x^3-4x+5)-(5x^3-3x+2) &=(2x^3-4x+5)-2x^3+3x-2\end{align}$$
Think of \( (5x^3-3x+2) \) as “a box containing a \(5x^3\), a \(-3x\), and a \(2\).” So when you are subtracting this whole “box of stuff” away from the quantity \( (2x^3-4x+5) \), you are effectively subtracting each of the things in \( (5x^3-3x+2) \), i.e. you are subtracting a \(5x^3\), a \(-3x\), and a \(2\). That is precisely what the equation above expresses.
In summary, subtracting/taking away a “box full of stuff” is the same thing as subtracting/taking away “each thing in the box.”
Combining Multiples of Polynomials
When speaking of “multiples of polynomials,” we are talking about expressions like \(5(3x^2+2x+1)\), where we have five of the expression \((3x^2+2x+1)\).
First distribute the \(3\) to each term in the quantity \((x^3+2x^2+1)\) and distribute the \(4\) in front of \((2x^3-4x^2+5)\) to each term in that quantity, and multiply. This will give you
$$\begin{align} 3(x^3+2x^2+1)+4(2x^3-4x^2+5) &=(3x^3+3\cdot 2x^2+3\cdot 1)+(4\cdot 2x^3-4\cdot 4x^2+4\cdot 5)\\&=(3x^3+6x^2+3)+(8x^3-16x^2+20)\end{align}$$
Now all that’s left is to combine like-terms!
$$\begin{align}(3x^3+6x^2+3)+(8x^3-16x^2+20) &=11x^3-10x^2+23\end{align}$$
When asked to combine real-number-multiples of polynomials, like in the last example with \(3(x^3+2x^2+1)+4(2x^3-4x^2+5) \), we can think of the quantity \( (x^3+2x^2+1) \) as “a box containing \(x^3\), \(2x^2\), and a \(1\).” Similarly, we can think of \( (2x^3-4x^2+5) \) as “a box containing \(2x^3\), \(-4x^2\) and a \(5\).”
So, when we write things like \(3(x^3+2x^2+1)\) we are indicating that we have three “boxes” filled with \(x^3\), \(2x^2\), and \(1\). This means that we have three of each thing in the box (because we had three boxes containing each of \(x^3\), \(2x^2\), and \(1\)). Hence why we can distribute the \(3\) when we performed the addition. A similar reasoning can be applied when we distribute the \(4\) in front of \( (2x^3-4x^2+5) \); we had four “boxes” containing each of \(4x^2\) and a \(5\), so we therefore have four of each of those.
Solution: \(4x^2+6\)
Since you are simply adding two parentheses, you can combine like-terms directly. Add the coefficients on the \(x^2\) and the \(1\) and \(5\).
Solution: \(2x^2-8x+7\)
Since we are simply adding the two parentheses, the parentheses can be dropped and you can combine like-terms directly. There is nothing that can be combined with the \(2x^2\) so that gets left alone (since there aren’t any more \(x^2\)’s). The \(-5x\) and \(-3x\) can be combined to give you \(-8x\), and the numbers \(3\) and \(-4\) can be added.
Solution: \(5x^6-x^2+4\)
Since we are only adding the two sets of parentheses, we can add like-terms directly. The \(5x^6\) and \(-x^2\) can’t be combined with anything, so they stays the same. The first \(x\) and the second \(x\) cancel out (one \(x\) take away another gives you NO \(x\)’s). The numbers \(1\) and \(3\) can be added directly.
Solution: \(-x^2-4\)
Solution: \(x^3-1\)
Solution: \(2x^2+11x-3\)
Solution: \(-3x^3+8x^2-7\)
Solution: \(-21x^4+7x^3+5x^2-35x-12\)
Solution: \(-12x^7-x^3+8x^2+5\)