# Applications of Synthetic Division of Polynomials

 

The synthetic division of polynomials is applied to factor polynomials and simplify rational functions using examples and questions and their solutions.
An online synthetic division calculator may be used to check results of synthetic divisions done by hand.

## Factor Polynomials Using Synthetic Division

Example 1
a) Show that $x - 1$ and $x + 3$ are factors of the polynomial $P(x) = x^4+2x^3-7x^2-8x+12$ .
b) use synthetic division to factor $P(x)$ completeley.

Solution
a)
Check the factor $x - 1$ by setting $x - 1 = x - k$ which gives $k = 1$
$P(k) = P(1) = (1)^4+2(1)^3-7(1)^2-8(1)+12 = 1 + 2 - 7 - 8 + 12 = 0$
Hence according to the factor theorem $x - 1$ is a factor of $P(x)$.
Check the factor $x + 3$ by setting $x + 3 = x - k$ which gives $k = -3$
$P(k) = P(-3) = (-3)^4+2(-3)^3-7(-3)^2-8(-3)+12 = 81 - 54 - 63 + 24 + 12 = 0$
Hence according to the factor theorem $x + 3$ are factors of the polynomial $P(x)$

b)
We use the synthetic division to divide $P(x)$ by $x - 1$. Hence from the results of the synthetic division, we can write $\quad \dfrac{P(x)}{x - 1} = x^3 + 3x^2 -4 x - 12$
The above division may be written as a multiplication
$P(x) = (x - 1)( x^3 + 3x^2 -4 x - 12) \quad \quad (I)$
We now divide the quotient obtained above which is $\quad x^3 + 3x^2 -4 x - 12 \quad$ by $\quad x + 3 \quad$ using synthetic division The above division may be written as $\dfrac{x^3 + 3x^2 -4 x - 12}{x+3} = x^2 - 4$
and also as a multiplication
$x^3 + 3x^2 -4 x - 12 = (x+3)(x^2 - 4)$
We now substitute $x^3 + 3x^2 -4 x - 12$ in $(I)$ by its factored form $(x+3)(x^2 - 4)$ obtained and write the given polynomial in factored form as follows
$P(x) = (x - 1)(x+3)(x^2 - 4) \quad \quad (II)$
We use difference of squares to factor $x^2 - 4$ as $(x-2)(x+2)$ and substitute it in $(II)$ above to completely factor the given polynomial.
$P(x) = (x - 1)(x+3)(x - 2)(x + 2)$

## Simplify Rational Functions Using Synthetic Division

A rational function of the form $\dfrac{P(x)}{D(x)}$ may be simplified if $P(x)$ and $D(x)$ have common factors.
Example 6
a) Show that $x - 2$ is a factor of the polynomials $P(x) = x^3 + 2x^2 - x - 14$ and $D(x) = x^2 - 4$.
b) Use synthetic division to factor $P(x)$ and $D(x)$ and then simplify the rational function $\dfrac{P(x)}{D(x)} = \dfrac{x^3 + 2x^2 - x - 14}{x^2 - 4}$.

Solution
a)
Check the factor $x - 2$ by setting $x - 2 = x - k$ which gives $k = 2$
$P(k) = P(2) = (2)^3 + 2(2)^2 - (2) - 14 = 8 + 8 - 2 - 14 = 0$
$D(k) = D(2) = (2)^2 - 4 = 4 - 4 = 0$
Hence $x - 2$ is a factor to both $P(x)$ and $D(x)$

b)
Use synthetic division to divide $P(x)$ by $x - 2$ The above division may be written as a multiplication. Hence $P(x)$ may be written as
$P(x) = (x - 2)(x^2 + 4x + 7)$
We now factor $x^2 - 4$ using the difference of squares
$x^2 - 4 = (x - 2)(x + 2)$
We now substitute $P(x)$ and $D(x)$ in the rational function by their factored form

$\dfrac{P(x)}{D(x)} = \dfrac{ (x - 2)(x^2 + 4x + 7)}{(x - 2)(x + 2)} \quad \quad (III)$

Divide both numerator and denominator by the common factor $x - 2$ and hence simplify

$\dfrac{P(x)}{D(x)} = \dfrac{ x^2 + 4x + 7}{x + 2}$

In order to further simplify, the numerator must have $x + 2$ as factor.
Use the factor theorem to find out if $x + 2$ is a factor of the numerator $N(x) = x^2 + 4x + 7$ in $(III)$ by evaluating $N(-2)$
$N(-2) = (-2)^2 + 4(-2) + 7 = 4 - 8 + 7 = 3$
$x + 2$ is not a factor of the numerator in (III) and we therefore cannot further simplify the rational function in (III).
Hence , $\dfrac{P(x)}{D(x)} = \dfrac{ x^2 + 4x + 7}{x + 2}$

.

## Questions

( with solutions )

Part A
Which of the following
a)   $x + 2$     b)   $x - 7$     c)   $x - 3$     d)   $x + 5$
is (are) factor of the polynomial $P(x) = x^4-4x^3-31x^2+70x$

Part B
Find
a) Find $r$ so that the polynomial $P_1(x) = x^3 + r x^2 - x - 4 \quad$ has the factor $x - 1$.
b) Find $r$ and $s$ so that the polynomial $P_2(x) = x^5- r x^4-3x^3+3x + s \quad$ has the factors $x + 1$ and $x - 2$.

Part C
Use the factor theorem and the synthetic division to completely factor each of the polynomials given some factors
a) $P_1(x) = x^3+3x^2-10x-24 \quad$ , Factor: $x + 4$
b) $P_2(x) = x^4-x^3-22x^2+16x+96 \quad$ , Factors: $x + 2$ and $x - 3$

Part D
Simplify the given rational function given a factor that is common to both numerator and denominator.

a) $F_1(x) = \dfrac{x^3+5x^2-7x-35}{x^2+4x-5} \quad$ , Common Factor: $x + 5$

b) $F_2(x) = \dfrac{x^4-3x^3-16x^2-6x-36}{x^2-3x-18} \quad$ , Common Factors: $x - 6$ and $x + 3$

## Solutions to the Above Questions

Part A
Given $P(x) = x^4-4x^3-31x^2+70x$,
For a linear term of the form $x - k$ to be a factor of $P(x)$, we need to have $P(k) = 0$
a)
check if $x + 2$ is a factor of $P(x)$.
$x + 2 = x - k$ gives $k = - 2$
$P(-2) = (-2)^4-4(-2)^3-31(-2)^2+70(-2) = -126$
Hence $x + 2$ IS NOT a factor of $P(x)$

b)
check if $x - 7$ is a factor of $P(x)$.
$x - 7 = x - k$ gives $k = 7$
$P(7) = (7)^4-4(7)^3-31(7)^2+70(7) = 0$
Hence $x - 7$ IS a factor of $P(x)$

c)
check if $x - 3$ is a factor of $P(x)$.
$x - 3 = x - k$ gives $k = 3$
$P(3) = (3)^4-4(3)^3-31(3)^2+70(3) = -96$
Hence $x - 3$ IS NOT a factor of $P(x)$

d)
check if $x + 5$ is a factor of $P(x)$.
$x + 5= x - k$ gives $k = - 5$
$P(-5) = (-5)^4-4(-5)^3-31(-5)^2+70(-5) = 0$
Hence $x + 5$ IS a factor of $P(x)$

Part B
a)
For the polynomial $P_1(x) = x^3 + r x^2 - x - 4$ to have $x - 1$ as a factor and according to the factor theorem we need to have $P_1(1) = 0$
Hence the equation
$(1)^3 + r (1)^2 - (1) - 4 = 0$
Simplify
$r - 4 = 0$
$r = 4$

b)
For the polynomial $P_2(x) = x^5- r x^4-3x^3+3x + s$ to have $(x + 1)$ and $x - 2$ as factors and according to the factor theorem we need to have $P_2(-1) = 0$ and $P_2(2) = 0$.
Hence the system of two equations
$(-1)^5- r (-1)^4-3(-1)^3+3(-1) + s = 0$ and $(2)^5- r (2)^4-3(2)^3+3(2) + s = 0$
Simplify
$- r + s - 1 = 0$ and $- 16 r + s + 14 = 0$
Solve the above system of linear equations to obtain: $r = 1$ and $s = 2$

Part C
a)
Use synthetic division to divide $P_1(x)$ by $(x + 4)$ The result of the division may be written as a multiplication, hence factoring $P_1(x)$ as follows
$P_1(x) = (x^2-x-6)(x+4)$
Factor the quadratic term $x^2-x-6$ and hence completely factor $P_1(x)$ as follows
$P_1(x) = (x^2-x-6)(x+4) = (x-3)(x+2)(x+4)$

b)
Use synthetic division to divide $P_2(x)$ by $(x + 2)$ The result of the division may be written as a multiplication, hence factoring $P_2(x)$ as follows
$P_2(x) = (x^3-3x^2-16x+48)(x+2) \quad \quad (I)$
Use synthetic division to divide the quotient in the above division $\quad x^3-3x^2-16x+48 \quad$ by $(x - 3)$ The result of the division may be written as a multiplication, hence factoring $x^3-3x^2-16x+48$ as follows
$x^3-3x^2-16x+48 = (x^2-16)(x-3)$
factor completely the above and write as
$x^3-3x^2-16x+48 = (x-4)(x+4)(x-3)$
Substitute the above in $(I)$ and factor $P_2(x)$ completely as follows
$P_2(x) = (x-4)(x+4)(x-3)(x+2)$

Part D
a)
Use synthetic division to divide the numerator of $F_1(x)$ by $(x + 5)$ Factor the numerator, using the result of the above synthetic division, and denominator, which is a quadratic expression, of $F_1(x)$.
$F_1(x) = \dfrac{(x^2-7)(x+5)}{(x -1)(x+5)}$
Divide numerator and the denominator by $(x + 5)$ and simplify
$F_1(x) = \dfrac{x^2-7}{x -1}$
It is easy to check that $x - 1$ is not a factor of $x^2-7$ and therefore we cannot further simplify $F_1(x)$.

b)
Use synthetic division to divide the numerator of $F_2(x)$ by $(x - 6)$ hence factor the numerator $F_2(x) = \dfrac{(x^3+3x^2+2x+6)(x-6)}{x^2-3x-18} \quad$ Use synthetic division to divide $x^3+3x^2+2x+6$ in the numerator above by $(x + 6)$ hence factor the numerator
$F_2(x) = \dfrac{(x^2+2)(x+3)(x-6)}{x^2-3x-18} \quad$
Factor the quadratic expression $x^2-3x-18$
$F_2(x) = \dfrac{(x^2+2)(x+3)(x-6)}{(x+3)(x-6)} \quad$
Divide numerator and the denominator by $(x+3)(x-6)$ and simplify
$F_2(x) = x^2+2 \quad$