8.1: Long division
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We now show how to divide two polynomials. The method is similar to the long division of natural numbers. Our first example shows the procedure in full detail.
Divide the following fractions via long division:
- 357111
- x3+5x2+4x+2x+3
Solution
- Recall the procedure for long division of natural numbers:
The steps above are performed as follows. First, we find the largest multiple of 11 less or equal to 35. The answer 3 is written as the first digit on the top line. Multiply 3 times 11, and subtract the answer 33 from the first two digits 35 of the dividend. The remaining digits 71 are copied below to give 271. Now we repeat the procedure, until we arrive at the remainder 7. In short, what we have shown is that:
3571=324⋅11+7 or alternatively, 357111=324+711
- We repeat the steps from part (a) as follows. First, write the dividend and divisor in the above format:
Next, consider the highest term x3 of the dividend and the highest term x of the divisor. Since x3x=x2, we start with the first term x2 of the quotient:
Step 1:
x2)x+3¯)x3+5x2+4x+2))
Multiply x2 by the divisor x+3 and write it below the dividend:
Step 2:
x2)x+3¯)x3+5x2+4x+2)x3+3x2)
Since we need to subtract x3+3x2, so we equivalently add its negative (don’t forget to distribute the negative):
Step 3:
x2)x+3¯)x3+5x2+4x+2)−(x3+3x2)_2x2)
Now, carry down the remaining terms of the dividend:
Step 4:
x2)x+3¯)x3+5x2+4x+2)−(x3+3x2)_2x2+4x+2)
Now, repeat steps 1-4 for the remaining polynomial 2x2+4x+2. The outcome after going through steps 1-4 is the following:
Since x can be divided into −2x, we can proceed with the above steps 1-4 one more time. The outcome is this:
Note that now x cannot be divided into 8 so we stop here. The final term 8 is called the remainder. The term x2+2x−2 is called the quotient. In analogy with our result in part (a), we can write our conclusion as:
x3+5x2+4x+2=(x2+2x−2)⋅(x+3)+8
Alternatively, we could also divide this by (x+3) and write it as:
x3+5x2+4x+2x+3=x2+2x−2+8x+3
Just as with a division operation involving numbers, when dividing f(x)g(x), f(x) is called the dividend and g(x) is called the divisor. As a result of dividing f(x) by g(x) via long division with quotient q(x) and remainder r(x), we can write
f(x)g(x)=q(x)+r(x)g(x)
If we multiply this equation by g(x), we obtain the following alternative version:
f(x)=q(x)⋅g(x)+r(x)
Divide the following fractions via long division.
- x2+4x+5x−4
- x4+3x3−5x+1x+1
- 4x3+2x2+6x+182x+3
- x3+x2+2x+1x2+3x+1
Solution
For part (a), we calculate:
Therefore, x2+4x+5=(x+8)⋅(x−4)+37.
Now, for part (b), there is no x2 term in the dividend. This can be resolved by adding +0x2 to the dividend:
Therefore, we showed:
x4+3x3−5x+1x+1=x3+2x2−2x−3+4x+1
For (c), calculate:
Since the remainder is zero, we succeeded in factoring 4x3+2x2+6x+18:
4x3+2x2+6x+18=(2x2−2x+6)⋅(2x+3)
d) The last example has a divisor that is a polynomial of degree 2. Therefore, the remainder is not a number, but a polynomial of degree 1.
Here, the remainder is r(x)=7x+3.
x3+x2+2x+1x2+3x+1=x−2+7x+3x2+3x+1
The divisor g(x) is a factor of f(x) exactly when the remainder r(x) is zero, that is:
f(x)=q(x)⋅g(x)⟺r(x)=0
For example, in the above Example 8.1.3, only part (c) results in a factorization of the dividend, since this is the only part with remainder zero.