2.03 Polynomial division and remainder and factor theorems

Long division with polynomials

We have already discussed how to divide polynomials using factoring and laws of exponents. The video below explains how to divide polynomials using long division by linking the process to elementary long division. 

In algebra, we can divide one polynomial by another polynomial of the same or lower degree, using a process similar to long division. You can do it without a calculator because it breaks a complex-looking problem up into smaller ones.

Worked examples

Question 1

Simplify: $\frac{3x^3+2x^2-6}{x+2}$3x3+2x2−6x+2​ using long division

The dividend: $3x^3+2x^2-6$3x3+2x2−6

The divisor: $x+2$x+2

Think: It's helpful to write the dividend like this: $3x^3+2x^2+0x-6$3x3+2x2+0x−6.

Do: Then we can use long division to solve the problem.

1. Divide the first term of the dividend by the greatest term of the divisor (meaning the one with the greatest power of $x$x, which in this case is $x$x). Place the result above the bar ($\frac{3x^3}{x}=3x^2$3x3x​=3x2). 

2. Multiply the divisor by the number you just wrote above the top line. Write the result under the first two terms of the dividend. ($3x^2\times\left(x+2\right)=3x^3+6x^2$3x2×(x+2)=3x3+6x2). 

3. Subtract these terms from the from those in the original dividend, making sure you pay attention to the positive and negative signs ($3x^3+2x^2-\left(3x^3+6x^2\right)=-4x^2+0x$3x3+2x2−(3x3+6x2)=−4x2+0x). 

4. Repeat the previous three steps, except this time use the two terms that have just been written as the dividend (ie. divide $\frac{-4x^2+0x}{x+2}$−4x2+0xx+2​) 

. 

5. Repeat step 4. Keep going until there is nothing to "pull down".

The answer can be written: $3x^2-4x+8$3x2−4x+8 with remainder $-22$−22 or $3x^2-4x+8-\frac{22}{x+2}$3x2−4x+8−22x+2​

Reflect: This means that $3x^3+2x^2-6=\left(x+2\right)\left(3x^2-4x+8\right)-22$3x3+2x2−6=(x+2)(3x2−4x+8)−22

 

Practice questions

Question 2

Question 3

 

Synthetic division

To save time and space, and minimize error, an efficient method of polynomial division emerges, known as synthetic division. Synthetic division can only be used when the divisor is a linear expression.

Worked example

question 4

To see how it works we divide $3x^3+2x^2-6$3x3+2x2−6 by the linear divisor $x+2$x+2 .

Setting up the synthetic table: Top row contains the coefficients of the numerator, include any terms with coefficient of $0$0: $3x^3+2x^2+0x-6$3x3+2x2+0x−6

	3	2	0	-6
add
quotient

 Next, place the solution of the linear divisor equal to zero in the table: $x+2=0$x+2=0 ; $x=-2$x=−2

	$3$3	$2$2	$0$0	$-6$−6
$-2$−2
add
quotient

Take the first coefficient and carry it down, unchanged: 

	$3$3	$2$2	$0$0	$-6$−6
$-2$−2
add
quotient	$3$3

Multiply the carry down value by the result of the linear divisor: $-2\times3$−2×3

	$3$3	$2$2	$0$0	$-6$−6
$-2$−2
add		$-2\times3=-6$−2×3=−6
quotient	$3$3

Add this value to the coefficient in that column and repeat the process:

	$3$3	$2$2	$0$0	$-6$−6
$-2$−2
add		$-2\times3=-6$−2×3=−6	$-2\times\left(-4\right)=8$−2×(−4)=8	$-2\times8=-16$−2×8=−16
quotient	$3$3	$2+\left(-6\right)=-4$2+(−6)=−4	$8$8	$-22$−22

The last row are the coefficients of the answer in degree order (starting with one degree less than the dividend) with the last value being the remainder. 

The answer can be written: $3x^2-4x+8-\frac{22}{x+2}$3x2−4x+8−22x+2​or $3x^2-4x+8$3x2−4x+8 with remainder -22

QUESTION 3

Using synthetic division to find the quotient and remainder when $x^3-4x^2-5$x3−4x2−5 is divided by $x-3$x−3

Think: To set-up the synthetic table, find the coefficients of the dividend and the result of the divisor set equal to zero. 

Do: Create the synthetic table 

	$1$1	$-4$−4	$0$0	$-5$−5
$3$3
add		$3$3	$-3$−3	$-9$−9
quotient	$1$1	$-1$−1	$-3$−3	$-14$−14

Answer: $x^2-x-3-\frac{14}{x-3}$x2−x−3−14x−3​ or $x^2-x-3$x2−x−3with remainder $-14$−14

 

Practice question

Question 5

Question 6

 

The remainder theorem

To divide two polynomials, say $P\left(x\right)$P(x) divided by $D\left(x\right)$D(x), we need the degree of the divisor polynomial $D\left(x\right)$D(x) to be less than or equal to the degree of the dividend polynomial $P\left(x\right)$P(x). 

As an example, suppose we divide $P\left(x\right)=x^2-5x+6$P(x)=x2−5x+6 by the polynomial $D\left(x\right)=x-1$D(x)=x−1. We proceed in a manner similar to long division given as follows:

Thus we state:

$\frac{P\left(x\right)}{D\left(x\right)}=\frac{x^2-5x+6}{x-1}=\left(x-4\right)+\frac{2}{x-1}$P(x)D(x)​=x2−5x+6x−1​=(x−4)+2x−1​

We can express the result slightly differently by multiplying both sides by the divisor so that:

$P\left(x\right)=x^2-5x+6=\left(x-1\right)\left(x-4\right)+2$P(x)=x2−5x+6=(x−1)(x−4)+2

Stating the result like this is known as the division transformation.

In general, dividing $P\left(x\right)$P(x) by $\left(x-a\right)$(x−a) will always produce a result that looks like:

$P\left(x\right)=\left(x-a\right)Q\left(x\right)+R$P(x)=(x−a)Q(x)+R

Here, $Q\left(x\right)$Q(x) is the quotient polynomial of one less degree than $P\left(x\right)$P(x) and $R$R is the remainder.

This last equation holds the key to the remainder theorem. Because the polynomial holds true for all values of $x$x, by putting $x=a$x=a into this general result we see that:

$P\left(a\right)=\left(a-a\right)Q\left(a\right)+R=R$P(a)=(a−a)Q(a)+R=R

This means that substituting $x=a$x=a into $P\left(x\right)$P(x) before dividing will reveal the remainder. It's a little mathematical magic! We can actually know the remainder even before the division by $\left(x-a\right)$(x−a) is done. This nice result is known as the remainder theorem.

Let's try it with our example. With $P\left(x\right)=x^2-5x+6$P(x)=x2−5x+6 and $D\left(x\right)=x-1$D(x)=x−1, before dividing note that $P\left(1\right)=\left(1\right)^2-5\left(1\right)+6=2$P(1)=(1)2−5(1)+6=2 and this is indeed the remainder!

 

The factor theorem

The factor theorem is an extension of the remainder theorem. 

If a polynomial equation $P(x)=0$P(x)=0 has a root $x=a$x=a, meaning $P(a)=0$P(a)=0, then $x-a$x−a must be a factor of $P(x)$P(x). We could write $P(x)=(x-a)Q(x)$P(x)=(x−a)Q(x) where $Q$Q is a polynomial of degree one less than the degree of $P$P.

This means that if we can find by any means a number $a$a such that $P(a)=0$P(a)=0, then we know immediately that $x-a$x−a is a factor of $P$P.

 

Finding linear factors

We are often asked to find a linear factor of $p\left(x\right)$p(x). Using trial and error, we will substitute in values until we find a factor, that is a value such that $p\left(a\right)=0$p(a)=0.

For $p\left(x\right)=ax^n+...+c$p(x)=axn+...+c, we should start our trial and error with factors of $c$c and then move on to $\frac{\text{factors of }c}{\text{factors of }a}$factors of cfactors of a​.

For example, for the polynomial $p\left(x\right)=x^3-x^2-x-2$p(x)=x3−x2−x−2 we would want to try $x=1$x=1, $x=-1$x=−1, $x=2$x=2 and$x=-2$x=−2.

 

Worked example

Question 7

Without doing long division, what is the remainder when $p\left(x\right)=2x^3-4x^2+3x-1$p(x)=2x3−4x2+3x−1 is divided by $2x-1$2x−1? Is $\left(2x-1\right)$(2x−1) a factor of $p(x)$p(x)?

Think: The remainder theorem was stated with linear factors of the form $x-a$x−a and we substituted in $x=a$x=a, but the linear factor $\left(2x-1\right)$(2x−1) is of the form $bx-a$bx−a. We will substitute in $x=\frac{a}{b}$x=ab​, more specifically $x=\frac{1}{2}$x=12​ (the solution to the linear factor set to $0$0).

Do: 

$p\left(x\right)$p(x)	$=$=	$2x^3-4x^2+3x-1$2x3−4x2+3x−1
$p\left(\frac{1}{2}\right)$p(12​)	$=$=	$2\times\left(\frac{1}{2}\right)^3-4\times\left(\frac{1}{2}\right)^2+3\times\frac{1}{2}-1$2×(12​)3−4×(12​)2+3×12​−1
	$=$=	$2\times\frac{1}{8}-4\times\frac{1}{4}+3\times\frac{1}{2}-1$2×18​−4×14​+3×12​−1
	$=$=	$\frac{1}{4}-1+\frac{3}{2}-1$14​−1+32​−1
	$=$=	$\frac{1}{4}-\frac{4}{4}+\frac{6}{4}-\frac{4}{4}$14​−44​+64​−44​
	$=$=	$-\frac{1}{4}$−14​

The remainder is $-\frac{1}{4}$−14​, which is not $0$0, so $2x-1$2x−1 is not a factor of $p\left(x\right)$p(x).

 

Question 8

Factor $p(x)=x^3-x^2-x-2$p(x)=x3−x2−x−2.

Think: Using the factor theorem, we know we are looking for a value $a$a such that $p\left(a\right)=0$p(a)=0. By trial and error, we find that $p(2)=0$p(2)=0. Therefore, $p(x)=(x-2)q(x)$p(x)=(x−2)q(x).

Do: We can use the division algorithm to calculate $q(x)$q(x). That is, we divide $p(x)$p(x) by $x-2$x−2. In this way, we find that $q(x)=x^2+x+1$q(x)=x2+x+1 which is a prime polynomial. So, $p(x)=x^3-x^2-x-2=(x-2)(x^2+x+1)$p(x)=x3−x2−x−2=(x−2)(x2+x+1) is the fully factored form of $p\left(x\right)$p(x).

 

Practice questions

Question 9

Question 10

 

 

Applications of the remainder and factor theorems

Certain problems arise concerning polynomials where the remainder and factor theorem are required to ascertain unknowns - factors, roots or coefficients. We review three problems here:

 

Worked example

Question 11

Suppose $P\left(x\right)=5x^2-14x-3$P(x)=5x2−14x−3 and $Q\left(x\right)=2x^2-x+k$Q(x)=2x2−x+k (with $k$k unknown), both contain a common factor of the form $\left(x-a\right)$(x−a) where $a$a is an integer. Is it possible to determine $k$k?

Knowing $\left(x-a\right)$(x−a) is a factor of $P\left(x\right)$P(x), from the factor theorem we also know that $P\left(a\right)=0$P(a)=0.

Hence we can set $5a^2-14a-3=0$5a2−14a−3=0, and this factors to $\left(a-3\right)\left(5a+1\right)=0$(a−3)(5a+1)=0. 

So we now know that, because $a$a is an integer, the common factor must be $\left(x-3\right)$(x−3).

Therefore, from the second polynomial $Q\left(x\right)$Q(x), we apply the factor theorem again. Specifically, we know that $Q\left(3\right)=0$Q(3)=0. This means that $2\left(3\right)^2-\left(3\right)+k=0$2(3)2−(3)+k=0.

After simplifying we thus know that $15+k=0$15+k=0, and so $k=-15$k=−15. 

 

Question 12

When the polynomials $P\left(x\right)=x^4+5x^3-mx+n$P(x)=x4+5x3−mx+n and $Q\left(x\right)=mx^2+nx-1$Q(x)=mx2+nx−1 are both divided by $D\left(x\right)=x-1$D(x)=x−1, the remainders are $7$7 and $-6$−6 respectively. Can we find $m$m and $n$n?

Using the remainder theorem we can develop two equations from knowing that $P\left(1\right)=7$P(1)=7 and $Q\left(1\right)=-6$Q(1)=−6. Thus, after simplifying:

$m-n$m−n	$=$=	$-1$−1
$m+n$m+n	$=$=	$-5$−5

These simultaneous equations are easily solved. By addition, $2m=-6$2m=−6 and so $m=-3$m=−3. By subtracting, $2n=-4$2n=−4 and so $n=-2$n=−2. 

 

Question 13

Suppose we know that for $10x^3+23x^2+5x-2=0$10x3+23x2+5x−2=0, one of the roots is four times another root. If we know that one of the roots is an integer, can we solve the equation?

If we call one of the roots $x=a$x=a and the other root $x=4a$x=4a, then we know that $\left(x-a\right)$(x−a) and $\left(x-4a\right)$(x−4a) are factors of $P\left(x\right)=10x^3+23x^2+5x-2$P(x)=10x3+23x2+5x−2. 

Thus, by the factor theorem, we have:

                               $P\left(a\right)=10a^3+23a^2+5a-2=0$P(a)=10a3+23a2+5a−2=0

                              $P\left(4a\right)=640a^3+368a^2+20a-2=0$P(4a)=640a3+368a2+20a−2=0

If we multiply the first of these equations by $64$64 and then subtract the second equation from it, we determine the quadratic equation $1104a^2+300a-126=0$1104a2+300a−126=0.

This quadratic equation can be factored to $\left(2a+1\right)\left(92a-21\right)=0$(2a+1)(92a−21)=0 so that  $a=-\frac{1}{2}$a=−12​ and $a=\frac{1}{5}$a=15​ are two of the three real roots.

Note that for any cubic equation with real coefficients, knowing that two of the roots are real implies that the other root must also be real - this because if there is a consequence of the conjugate root theorem.  

Based on the information in the question, the integer root must be $x=-2$x=−2, because $4\times\left(-\frac{1}{2}\right)=-2$4×(−12​)=−2, whereas $4\times\left(\frac{1}{5}\right)=\frac{4}{5}$4×(15​)=45​. 

Hence we can write that $P\left(x\right)=\left(x+2\right)\left(2x+1\right)\left(5x-1\right)$P(x)=(x+2)(2x+1)(5x−1) with the roots of $P\left(x\right)=0$P(x)=0 given by $x=-2,-\frac{1}{2},\frac{1}{5}$x=−2,−12​,15​.

 

Practice questions

Question 14

Question 15

Question 16