15.2 Synthetic Division
Steps to take when using synthetic division
- 1.
- Set the denominator equal to zero to find the number to put in the division box. Next, make sure the numerator is written in descending order and if any terms are missing you must use a zero to fill in the missing term, finally list only the coefficient in the division problem.
- 2.
- Once the problem is set up correctly, bring the leading coefficient (first number) straight down.
- 3.
- Multiply the number in the division box by the number just brought down, and put the result in the next column.
- 4.
- Add the two numbers in that column and write the result at the bottom.
- 5.
- Repeat the previous two steps until every column has been used.
- 6.
- Write the final answer. The final answer is made up of the numbers in the bottom row with
the last number being the remainder and the remainder must be written as a fraction.
The variables or \(x'\)s start off one power less than the original denominator and go down one
with each term.
Find the quotient and the remainder when the polynomial \(P(x)=x^3-7x^2+6x-2\) is divided by \(x-2\)
Solution.
Using synthetic division
\[\begin {array}{r|rrrr} 2 & 1 & -7 & 6 & -2\\ & & 2 & -10 & -8\\ \hline & 1 & -5 & -4 & -10 \end {array}\]
The top row holds the coefficients of \(x^{3}-7x^{2}+6x-2\). Each entry of the middle row is the previous bottom
entry multiplied by \(2\), and each bottom entry is the sum of the column above it. The last bottom
entry, \(-10\), is the remainder, and the others are the coefficients of the quotient \(x^{2}-5x-4\).
Using step 9, we obtain \(Q(x)=x^2-5x-4\) and \(R(x)=-10\).
Solution. The divisor is \(2x-1\), not \(x-a\), so a small adjustment is needed. It vanishes at \(x=\frac {1}{2}\), and the remainder theorem still applies at that value: \[R=f\left (\tfrac {1}{2}\right )=4\left (\tfrac {1}{8}\right )-6\left (\tfrac {1}{2}\right )+5 =\tfrac {1}{2}-3+5=\tfrac {5}{2} .\] Then \((2x-1)Q(x)=4x^{3}-6x+5-\frac {5}{2}=4x^{3}-6x+\frac {5}{2}\), and dividing, \[Q(x)=2x^{2}+x-\tfrac {5}{2} .\] Hence \[4x^{3}-6x+5=(2x-1)\left (2x^{2}+x-\tfrac {5}{2}\right )+\tfrac {5}{2} .\] Check by expanding: \((2x-1)\left (2x^{2}+x-\frac {5}{2}\right ) =4x^{3}+2x^{2}-5x-2x^{2}-x+\frac {5}{2}=4x^{3}-6x+\frac {5}{2}\), and adding \(\frac {5}{2}\) returns \(f(x)\).
Note 15.12. For a divisor \(bx-c\) the remainder is \(P\left (\frac {c}{b}\right )\), since that is where the divisor vanishes. The remainder is unchanged, but the quotient is divided by \(b\) compared with dividing by \(x-\frac {c}{b}\), because \(bx-c=b\left (x-\frac {c}{b}\right )\) and that factor of \(b\) has to come from somewhere.
Proof. By the remainder theorem the remainder on dividing by \(x-a\) is \(P(a)\). If \(P(a)=0\) the remainder vanishes and \(P(x)=(x-a)Q(x)\), so \(x-a\) is a factor. Conversely if \(x-a\) is a factor then \(P(x)=(x-a)Q(x)\) and substituting \(x=a\) gives \(P(a)=0\). □
A value \(a\) with \(P(a)=0\) is called a root or zero of \(P\). The factor theorem is therefore the bridge between the two
ways of looking at a polynomial: its factors and its roots are the same information written
differently.
\begin {align*} P(a) & =0.\\\\ P(x) & = a_nx^n+a_{n-1}x^{n-1}+\cdots +ax+a_0\\ \end {align*}
To factorise this \(P(x)\) we must find the roots of \(P(x)\).
Theorem 15.14 (Rational root theorem). If the polynomial \(a_nx^{n}+\cdots +a_1x+a_0\) with integer coefficients has a rational root \(\frac {p}{q}\) in lowest terms, then \(p\) divides \(a_0\) and \(q\) divides \(a_n\).
Note 15.15. In particular, if \(a_n=1\) — the usual case in this course — then \(q\) must divide \(1\), so any rational root is a whole number dividing the constant term. That is what makes the search below finite: instead of testing every number, only the divisors of \(a_0\) need trying, and there are few of them.
The condition on \(a_n\) cannot be dropped. \(2x^{2}-3x+1\) has the root \(\frac {1}{2}\), which is not an integer and does not divide the constant term \(1\); it is admissible because its denominator \(2\) divides the leading coefficient.
Note also what the theorem does not say. It lists the only possible rational roots; it does not promise any exist. \(x^{2}-2\) has no rational root at all, as chapter 2 proved.
Solution.
Here \(a_{0}=-15\) and \(a_{n}=1\), so any rational root is a whole number dividing \(15\). We list those and test them.
\(\pm 1, \pm 3, \pm 5, \pm 15\) are factors of 15
\(P(1)=1^2+2(1)-15\neq 0\)
\(P(-1)=(-1)^2+2(-1)-15\neq 0\)
\(P(3)=(3)^2+2(3)-15 = 0\)
\(P(-3) =(-3)^2+2(-3)-15\neq 0\)
\(P(5)=5^2+2(5)-15\neq 0\)
\(P(-5)=(-5)^2+2(-5)-15=0\)
\(P(15)=15^2+2(15)-15\neq 0\)
\(P(-15)=(-15)^2+2(-15)-15\neq 0\)
The two values giving zero are \(x=3\) and \(x=-5\), so these are the roots. (Note \(P(5)=20\neq 0\); it is \(-5\), not \(5\), that works.)
By the factor theorem both \(x-3\) and \(x+5\) are factors, and since the polynomial has degree \(2\) there are
no others:
\[x^{2}+2x-15=(x-3)(x+5).\]
Dividing by \(x-3\) confirms it.
\[\begin {array}{r@{\;}l} & x+5\\ x-3\,\big ) & \overline {x^{2}+2x-15}\\ & \underline {x^{2}-3x}\phantom {{}-15}\\ & \phantom {x^{2}}5x-15\\ & \phantom {x^{2}}\underline {5x-15}\\ & \phantom {x^{2}}0 \end {array}\] The remainder is zero, so \(x-3\) is a factor and the other factor is the quotient \(x+5\).
If \(P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +ax+a_0\) is a polynomial, any of rational roots of \(P(x)\) has factor \(x=\frac {b}{c}\) where \(c\neq 0\).
Solution.
\(f(x)=2x^3+3x^2-32x+15\), we first examine the factors of 15. Note and see if we can get a root.
\(\pm 1, \pm 3, \pm 5, \pm 15\)
\(f(-1)=-2+3+32+15\neq 0\)
\(f(1) =2+3-32+15\neq 0\)
\(f(3) = 2(3)^3+3(3)^2-32(3)+15=0\)
So \(x=3\) is a root.
Now, using synthetic division \[\begin {array}{r|rrrr} 3 & 2 & 3 & -32 & 15\\ & & 6 & 27 & -15\\ \hline & 2 & 9 & -5 & 0 \end {array}\] The remainder is \(0\), so \(x-3\) is a factor and the quotient is \(2x^{2}+9x-5\).
\[\therefore \quad f(x)=(x-3)(2x^2+9x-5)\]
We factorise \(2x^2+9x-5\) \begin {align*} & = 2x^2+9x-5\\ & = 2x^2+10x-x-5\\ & = 2x(x+5)-1(x+5)\\ & = (2x-1)(x+5) \end {align*}
\[\therefore \quad 2x^3+3x^2-32x+15=(x-3)(2x-1)(x+5)\]
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