Dividing Trinomials by Binomials
Hello! Welcome to this video on dividing trinomials by binomials. When dividing trinomials by binomials, you often will need to use long division. This process may seem challenging at first, but it will get easier with practice.
Let’s jump right into an example.
\((x^{2}+4x-5)\div (x+1)\)
Remember, \(x^{2}+4x-5\) is a trinomial because it has 3 terms, and \(x+1\) is a binomial because it has 2 terms. Set up your long division by putting the trinomial under the “house,” and the binomial to the left.
\(\phantom{(0}\phantom{)}\phantom{)}\) \(x+1\) \(\phantom{(}x^{2}\) \(+\) \(\phantom{(}4x\phantom{)}-5\phantom{)}\)
Then, we want to figure out how many times the first term of our binomial goes into the first term of our trinomial. \(x\) goes into \(x^{2}\), \(x\) times, because \(x\cdot x=x^{2}\). So write an \(x\) right here, above our \(x\)-term in the trinomial.
\(\phantom{(0}x\phantom{)}\phantom{)}\) \(x+1\) \(\phantom{(}x^{2}\) \(+\) \(\phantom{(}4x\phantom{)}-5\phantom{)}\)
Then, multiply the value you just wrote down \((x)\) by the binomial \((x+1)\), and write this under the trinomial. Since \(x(x+1)=x^{2}+x\), we write this under the trinomial.
\(\phantom{(0}x\phantom{)}\phantom{)}\) \(x+1\) \(\phantom{(}x^{2}\) \(+\) \(\phantom{(}4x\phantom{)}-5\phantom{)}\) \(\phantom{x+}-\) \((x^{2}\) \(+\) \(\phantom{(0}x)\phantom{-}\phantom{)}\)
Then, subtract these two polynomials, and bring down the \(–5\).
\(\phantom{(0}x\phantom{)}\phantom{)}\) \(x+1\) \(\phantom{(}x^{2}\) \(+\) \(\phantom{(}4x\phantom{)}-5\phantom{)}\) \(\phantom{x+}-\) \((x^{2}\) \(+\) \(\phantom{(0}x)\phantom{-}\downarrow\phantom{)}\) \(\phantom{(}3x\phantom{)}-5\phantom{)}\)
From here, see how many times the first term of our divisor \((x+1)\) goes into the first term of the new binomial \((3x-5)\). So how many times does \(x\) fit into \(3x\)? \(x\) goes into \(3x\) three times, because \(x\cdot 3=3x\), so write this number after \(x\) in our answer. Since 5 is positive, write \(+3\).
Then, multiply the number you just wrote down \((3)\) by the divisor \((x+1)\), and write this binomial below the one we found a few steps earlier: \(3(x+1)=3x+3\).
\(\phantom{(0}x\phantom{)}+5\phantom{)}\) \(x+1\) \(\phantom{(}x^{2}\) \(+\) \(\phantom{(}4x\phantom{)}-5\phantom{)}\) \(\phantom{x+}-\) \((x^{2}\) \(+\) \(\phantom{(0}x)\phantom{-}\downarrow\phantom{)}\) \(\phantom{(}3x\phantom{)}-5\phantom{)}\)
And we’re going to subtract the two binomials, so \((3x-5)-(3x+3)=-8\).
\(\phantom{(0}x\phantom{)}+5\phantom{)}\) \(x+1\) \(\phantom{(}x^{2}\) \(+\) \(\phantom{(}4x\phantom{)}-5\phantom{)}\) \(\phantom{x+}-\) \((x^{2}\) \(+\) \(\phantom{(0}x)\phantom{-}\downarrow\phantom{)}\) \(\phantom{(}3x\phantom{)}-5\phantom{)}\) \(–\) \((3x\phantom{)}+3)\) \(\phantom{(0x)}-8\phantom{)}\)
Now we have a remainder here, so when you have a remainder in a division problem like this, you’re going to take the remainder and divide it by the divisor. And add it to the expression that we have up here \((x+3)\). So since we have –8, I’m going to put minus, 8 in our numerator, and then \(x+1\) in our denominator. If it was positive down here, we would put a plus sign, but since it’s negative, that’s why I put a minus sign right here. And then this is our final answer:
\(x+3-\frac{8}{x+1}\)
So when dividing you will sometimes have remainders like we did here, this is exactly how you would handle them and it’s okay that things don’t always divide out perfectly. So that’s what you do when it doesn’t.
Let’s try another one!
\(\frac{x^{2}+8x-9}{x+2}\)
This problem is written in fraction form. Remember, a fraction bar always represents division, so this is just asking us to divide \(x^{2}+8x-9\) by \(x+2\).
Let’s follow the same steps we did in the last problem. First, set up the long division problem.
\(\phantom{(x^{2}+8}x\phantom{)}\) \(x+2\) \(\phantom{(}x^{2}+8x\phantom{)}\) \(-\phantom{1}9\phantom{)}\)
Now, figure out how many times \(x\) goes into \(x^{2}\). Well, just like last time, \(x\cdot x=x^{2}\), so it goes into \(x^{2}\), \(x\) times.
Now, multiply \(x\) by \(x+2\) and write it below the trinomial. Since \(x(x+2)=x^{2}+2x\), we write this below the trinomial.
Then, subtract the two polynomials, and bring down the \(–9\).
\(\phantom{(x^{2}+8}x\phantom{)}\) \(x+2\) \(\phantom{(}x^{2}+8x\phantom{)}\) \(-\phantom{1}9\phantom{)}\) \(\phantom{x+}-\) \((x^{2}+2x)\) \(\phantom{-1}\downarrow \phantom{)}\) \(\phantom{(x^{2}+}6x\phantom{)}\) \(-\phantom{1}9\phantom{)}\)
Now, figure out how many times \(x\) goes into \(6x\). Since \(x\cdot 6=6x\), \(x\) goes into \(6x\) six times. Since 6 is positive, write \(+6\) after \(x\) in the solution.
\(\phantom{(x^{2}+8}x\phantom{)}\) \(+\phantom{1}6\phantom{)}\) \(x+2\) \(\phantom{(}x^{2}+8x\phantom{)}\) \(-\phantom{1}9\phantom{)}\) \(\phantom{x+}-\) \((x^{2}+2x)\) \(\phantom{-1}\downarrow \phantom{)}\) \(\phantom{(x^{2}+}6x\phantom{)}\) \(-\phantom{1}9\phantom{)}\) \(\phantom{(x^{2}}-(6x\phantom{)}\) \(+12)\) \(-21\phantom{)}\)
From here, multiply 6 by \(x+2\), and write this below the new binomial we just created: \(6(x+2)=6x+12\).
Subtract these two binomials.
So here we have a remainder again. And remember, when we have a remainder, we just take this number \((–21)\) and that will be the numerator of our fraction. So we’ll have minus 21 over, the denominator will be the divisor over here, so over \(x+2\). So our answer for this problem is:
\(\phantom{(x^{2}+8}x\phantom{)}\) \(+\phantom{1}6\phantom{)}\) \(-\frac{21}{x+2}\) \(x+2\) \(\phantom{(}x^{2}+8x\phantom{)}\) \(-\phantom{1}9\phantom{)}\) \(\phantom{x+}-\) \((x^{2}+2x)\) \(\phantom{-1}\downarrow \phantom{)}\) \(\phantom{(x^{2}+}6x\phantom{)}\) \(-\phantom{1}9\phantom{)}\) \(\phantom{(x^{2}}-(6x\phantom{)}\) \(+12)\) \(-21\phantom{)}\)
\(x+6-\frac{21}{x+2}\)
Before we go, I want to try one last example.
\((2x^{2}+4x-12)\div (x-6)\)
So again, we’re going to set up our long division problem, just like we have been.
\(\phantom{(2x^{2}+}2x\phantom{)}\) \(+16\phantom{)}\) \(+\frac{84}{x-6}\) \(x-6\) \(\phantom{(}2x^{2}+\phantom{1}4x\phantom{)}\) \(-12\phantom{)}\) \(\phantom{x-}-\) \((2x^{2}-12x)\) \(\phantom{-}\downarrow\phantom{)}\) \(\phantom{(2x^{2}+}16x\phantom{)}\) \(-12\phantom{)}\) \(\phantom{2x^{2}}-(16x\phantom{)}\) \(-96)\) \(\phantom{-}84\phantom{)}\)
Then, see how many times \(x\) goes into \(2x^{2}\). \(x\) goes into \(2x^{2} 2x\) times, because \(x\cdot 2x=2x^{2}\).
Now, multiply \(2x\) by \(x-6\).
From here, subtract the two polynomials, and bring down the \(–12\).
Now, figure out how many times \(x\) goes into \(16x\). It goes in 16 times because \(x\cdot 16=16x\), so we write plus 16 right here.
And then we multiply 16 by our expression \(x-6\). So that gives us \(16x-96\). And then from here, we’re just going to subtract our expressions.
And again, we’re going to do what we’ve been doing the last few times and add our remainder, so we’ll add 84x-6. So our answer for this question is:
\(2x+16+\frac{84}{x-6}\)
Now just as a disclaimer: for all of these examples we’ve had a remainder, but sometimes things will divide out nice and evenly and not have any remainder. So if you get a zero at the last step, you’re not going to have a remainder and you don’t have to worry about this fractional part. But if you do have a remainder, it’s nothing to be concerned about. It’s just a simple easy extra step that you now know how to do, and are hopefully comfortable with.
I hope that this video on dividing trinomials by binomials was helpful. Thanks for watching, and happy studying!
