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Wize High School Grade 12 Pre-Calculus Textbook > Polynomial Functions

Solving General Polynomial Equations

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Applications of Polynomial FunctionsNext Section
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Solving General Polynomial Equations

A polynomial equation is an equation that can be expressed in the form:
axn+bxn−1+ ... +cx+d=0ax^n+bx^{n-1}+~...~+cx+d=0
where a,b,c,d∈Ra,b,c,d\in\mathbb{R} and n∈Wn\in\mathbb{W}.

A polynomial equation can be solved by factoring using the rational root, remainder & factor theorem.

Remainder Theorem

Let y=f(x) y=f(x)~ be a polynomial function of degree 'n'. If f(x) ÷ (x−a)f(x)~\div~(x-a), then the remainder, r, is:
f(a)=r\boxed{f(a)=r}

Factor Theorem

Let y=f(x) y=f(x)~ be a polynomial function of degree 'n'. If (x−a)(x-a)is a factor of y=f(x).y=f(x).Then:
f(a)=0\boxed{f(a)=0}


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Rational Root Theorem

If f(x)=axn+bxn−1+ ... +cx+df(x)=ax^n+bx^{n-1}+~...~+cx+d is divided by (x−a)(x-a), then the possible roots of f(x) f(x)~ are:
{Factors of dFactors of a}\Bigg\{\frac{\text{Factors of d}}{\text{Factors of a}} \Bigg\}

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Example

Solve f(x)=2x3−3x2−11x+6f(x)=2x^3-3x^2-11x+6

Step 1.
Use the Rational Root Theorem to determine all possible values of the roots.

Factors of 6: 1, 2, 3, 6
Factors of 2: 1, 2

All possible factors are:
±1,2,3,61,2=±12, 1, 32, 2, 3, 6\pm\frac{1, 2, 3, 6}{1, 2}=\pm\frac{1}{2},~1,~\frac{3}{2}, ~2, ~3,~6


Step 2.
Use the Remainder & Factor Theorem to test the possible factors for f(x).f(x).

x=12‾\underline{x=\frac{1}{2}}:
f(12)=2(12)3−3(12)2−11(12)+6=14−34−112+6=0\begin{array}{rcl} f\bigg(\frac{1}{2}\bigg)&=&2\bigg(\frac{1}{2}\bigg)^3-3\bigg(\frac{1}{2}\bigg)^2-11\bigg(\frac{1}{2}\bigg)+6\\\\&=&\frac{1}{4}-\frac{3}{4}-\frac{11}{2}+6\\\\ &=&0 \end{array}

Since x=12 x=\frac{1}{2}~ gives a remainder of 0, then (2x−1) (2x-1)~ is a factor of f(x)


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Step 3.
Use polynomial division (long or synthetic) to find the other factors of f(x).f(x).

(2x3−3x2−11x+6) ÷ (2x−1)(2x^3-3x^2-11x+6)~\div~(2x-1)
Let's use synthetic division:
2−3−11612↓↓1−1−6(2−2−12)÷20\begin{array}{r| c c c c} &2&-3&-11&6\\ \frac{1}{2}&\downarrow&&&\\ &\downarrow&1&-1&-6\\\hline &(2&-2&-12)\div2&0\\ \end{array}
The quotient becomes:
x2−x−6=(x−3)(x+2)x^2-x-6=(x-3)(x+2)

The division statement becomes:
(2x3−3x2−11x+6)=(2x−1)(x−3)(x+2)(2x^3-3x^2-11x+6)=(2x-1)(x-3)(x+2)
So,
2x3−3x2−11x+6=0and2x3−3x2−11x+6=(2x−1)(x−3)(x+2)∴ 0=(2x−1)(x−3)(x+2)x=−2, 12, 3\begin{array}{rclcc} 2x^3-3x^2-11x+6&=&0&\footnotesize\text{and}&2x^3-3x^2-11x+6=(2x-1)(x-3)(x+2)\\\\ \therefore~&&0=(2x-1)(x-3)(x+2)\\\\ &&x=-2,~\frac{1}{2},~3 \end{array}

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Example: Solving General Polynomial Equations


Solve 0=8x3+4x2−18x−90=8x^3+4x^2-18x-9

Step 1.
Use the Rational Root Theorem to determine all possible values of the roots.

Factors of 9: 1, 3, 9
Factors of 8: 1, 2, 4, 8

All possible factors are:
±1,3,91,2,4,8=±18,14,12,38,34,98,94,92,1,3,9\pm\frac{1, 3, 9}{1, 2, 4, 8}=\pm\frac{1}{8},\frac{1}{4},\frac{1}{2},\frac{3}{8},\frac{3}{4},\frac{9}{8},\frac{9}{4},\frac{9}{2},1, 3, 9


Step 2.
Use the Remainder & Factor Theorem to test the possible factors for f(x).f(x).

x=−12‾\underline{x=-\frac{1}{2}}:
f(−12)=8(−12)3+4(−12)2−18(−12)−9=8(−18)+4(14)−(−182)−9=−1+1+9−9=0\begin{array}{rcl} f\Big(-\frac{1}{2}\Big)&=&8\Big(-\frac{1}{2}\Big)^3+4\Big(-\frac{1}{2}\Big)^2-18\Big(-\frac{1}{2}\Big)-9\\\\ &=&8\Big(-\frac{1}{8}\Big)+4\Big(\frac{1}{4}\Big)-\Big(-\frac{18}{2}\Big)-9\\\\ &=&-1+1+9-9\\\\ &=&0 \end{array}

Since x=−12 x=-\frac{1}{2}~ gives a remainder of 0, then (2x+1) (2x+1)~ is a factor of f(x)


Step 3.
Use polynomial division (long or synthetic) to find the other factors of f(x).f(x).

(8x3+4x2−18x−9) ÷ (2x+1)(8x^3+4x^2-18x-9)~\div~(2x+1)
Let's use synthetic division:
84−18−9−12↓−409↓80−180\begin{array}{r| c c c c} &8&4&-18&-9\\ -\frac{1}{2}&\downarrow&-4&0&9\\ &\downarrow&&&\\\hline &8&0&-18&0\\ \end{array}
The quotient becomes:
(8x2−18)÷2=4x2−9=(2x−3)(2x+3)\begin{array}{rcl} (8x^2-18)\div2&=&4x^2-9\\\\ &=&(2x-3)(2x+3) \end{array}

The division statement becomes:
(8x3+4x2−18x−9)=(2x+1)(2x−3)(2x+3)(8x^3+4x^2-18x-9)=(2x+1)(2x-3)(2x+3)
So,
8x3+4x2−18x−9=0and8x3+4x2−18x−9=(2x+1)(2x−3)(2x+3)∴ 0=(2x+1)(2x−3)(2x+3)x=−12, ±32\begin{array}{rclcc} 8x^3+4x^2-18x-9&=&0&\footnotesize\text{and}&8x^3+4x^2-18x-9=(2x+1)(2x-3)(2x+3)\\\\ \therefore~&&0=(2x+1)(2x-3)(2x+3)\\\\ &&x=-\frac{1}{2},~\pm\frac{3}{2} \end{array}

Practice: Solving General Polynomial Equations

Solve: (x+3)(3x+2)(5x−1)(3x−1)=0(x+3)(3x+2)(5x-1)(3x-1)=0

Practice: Solving General Polynomial Equations

Solve: 6x4+x3−22x2−11x+6=06x^4+x^3-22x^2-11x+6=0.

(Select all that apply)

Practice: Solving General Polynomial Equations

Determine a function, f(x), f(x),~in factored form, that has the following properties:
  • xx-intercepts at −3, −1, 2, 4-3,\ -1,\ 2,\ 4
  • When f(x) ÷ (x+6)f(x)~\div~(x+6), the remainder is −96-96
Extra Practice
Solving General Polynomial Equations
Practice: Solving General Polynomial Equations
Determine a function, f(x), f(x),~in factored form, that has the following properties:
xx-intercepts at 12, 13, 5\dfrac{1}{2},~\dfrac{1}{3},~5
Solving General Polynomial Equations
Practice: Solving General Polynomial Equations
Determine a function, f(x), f(x),~in factored form, that has the following properties:
xx-intercepts at −2,3,4,7-2, 3, 4, 7
Solving General Polynomial Equations
Practice: Solving General Polynomial Equations
Solve: x4−5x2−6=0x^4-5x^2-6=0.
(Select all that apply)
Solving General Polynomial Equations
Practice: Solving General Polynomial Equations
Solve: 3x4+2x3−21x2+4x+6=03x^4+2x^3-21x^2+4x+6=0.
(Select all that apply)
Solving General Polynomial Equations
Practice: Solving General Polynomial Equations
Solve: (2x+5)(9x−1)(x−6)(x+8)=0(2x+5)(9x-1)(x-6)(x+8)=0
Solving General Polynomial Equations
Practice: Solving General Polynomial Equations
Solve: (x−2)(5x+4)(3x+7)=0(x-2)(5x+4)(3x+7)=0