3. Polynomials

Lesson

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`)=`x`2−5`x`+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`)=`x`2−5`x`+6`x`−1=(`x`−4)+2`x`−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`)=`x`2−5`x`+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`)=`x`2−5`x`+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 remainder theorem

If a polynomial $P\left(x\right)$`P`(`x`) is divided by $x-a$`x`−`a`, the remainder is a constant $R$`R`, and

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

where $Q\left(x\right)$`Q`(`x`) is a polynomial with degree one less than $P\left(x\right)$`P`(`x`).

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`.

The factor theorem

The binomial $x-a$`x`−`a` is a factor of the polynomial $P(x)$`P`(`x`) if and only if $P(a)=0$`P`(`a`)=0.

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`)=`a``x``n`+...+`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 `c`factors of `a`.

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

Without doing long division, what is the remainder when $p\left(x\right)=2x^3-4x^2+3x-1$`p`(`x`)=2`x`3−4`x`2+3`x`−1 is divided by $2x-1$2`x`−1? Is $\left(2x-1\right)$(2`x`−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 $\left(2x-1\right)$(2`x`−1) is of the form $bx-a$`b``x`−`a`. We will substitute in $x=\frac{a}{b}$`x`=`a``b`, 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$2`x`−1 is **not **a factor of $p\left(x\right)$`p`(`x`).

Factor $p(x)=x^3-x^2-x-2$`p`(`x`)=`x`3−`x`2−`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`)=`x`2+`x`+1 which is a prime polynomial. So, $p(x)=x^3-x^2-x-2=(x-2)(x^2+x+1)$`p`(`x`)=`x`3−`x`2−`x`−2=(`x`−2)(`x`2+`x`+1) is the fully factored form of $p\left(x\right)$`p`(`x`).

Christa wants to test whether various linear expressions divide exactly into $P\left(x\right)$`P`(`x`), or whether they leave a remainder. For each linear expression below, state the value of $x$`x` that needs to be substituted into $P\left(x\right)$`P`(`x`) to find the remainder.

$x+3$

`x`+3$8-x$8−

`x`$5+4x$5+4

`x`$6-x$6−

`x`

Using the remainder theorem, find the remainder when $P\left(x\right)=-4x^4+6x^3+4x^2-7x+7$`P`(`x`)=−4`x`4+6`x`3+4`x`2−7`x`+7 is divided by $A\left(x\right)=3x-1$`A`(`x`)=3`x`−1.

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:

Suppose $P\left(x\right)=5x^2-14x-3$`P`(`x`)=5`x`2−14`x`−3 and $Q\left(x\right)=2x^2-x+k$`Q`(`x`)=2`x`2−`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$5`a`2−14`a`−3=0, and this factors to $\left(a-3\right)\left(5a+1\right)=0$(`a`−3)(5`a`+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.

When the polynomials $P\left(x\right)=x^4+5x^3-mx+n$`P`(`x`)=`x`4+5`x`3−`m``x`+`n` and $Q\left(x\right)=mx^2+nx-1$`Q`(`x`)=`m``x`2+`n``x`−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$2`m`=−6 and so $m=-3$`m`=−3. By subtracting, $2n=-4$2`n`=−4 and so $n=-2$`n`=−2.

Suppose we know that for $10x^3+23x^2+5x-2=0$10`x`3+23`x`2+5`x`−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`=4`a`, then we know that $\left(x-a\right)$(`x`−`a`) and $\left(x-4a\right)$(`x`−4`a`) are factors of $P\left(x\right)=10x^3+23x^2+5x-2$`P`(`x`)=10`x`3+23`x`2+5`x`−2.

Thus, by the factor theorem, we have:

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

$P\left(4a\right)=640a^3+368a^2+20a-2=0$`P`(4`a`)=640`a`3+368`a`2+20`a`−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$1104`a`2+300`a`−126=0.

This quadratic equation can be factored to $\left(2a+1\right)\left(92a-21\right)=0$(2`a`+1)(92`a`−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)(2`x`+1)(5`x`−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.

When $3x^3-2x^2-4x+k$3`x`3−2`x`2−4`x`+`k` is divided by $x-3$`x`−3, the remainder is $47$47. Find the value of $k$`k`.

The polynomials $4x^2-7x-15$4`x`2−7`x`−15 and $5x^2+13x+k$5`x`2+13`x`+`k` have a common factor of $x+p$`x`+`p`, where $p$`p` is an integer.

Using the fact that $x+p$

`x`+`p`is a factor of $4x^2-7x-15$4`x`2−7`x`−15, solve for the value of $p$`p`.Using the fact that $x+p$

`x`+`p`is a factor of $5x^2+13x+k$5`x`2+13`x`+`k`, solve for $k$`k`.

The polynomials $P\left(x\right)=x^3+4x^2-5x+n$`P`(`x`)=`x`3+4`x`2−5`x`+`n` and $Q\left(x\right)=x^3+2x+17$`Q`(`x`)=`x`3+2`x`+17 leave the same remainder when divided by $x+1$`x`+1.

Solve for the value of $n$`n`.

Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x â€“ a is p(a), so p= 0 if and only if (x â€“ is a factor of p(x).