State the possible rational roots for each function
\(\textbf{1)}\) \(f(x)=x^5+3x^3+2x^2+5x-12\) All possible rational roots are \(\displaystyle \pm1, \pm2, \pm3, \pm4, \pm6, \pm12\)
\(\textbf{2)}\) \(f(x)=4x^5+3x^3+2x^2+5x-12\) All possible rational roots are \(\displaystyle \pm1,\pm2,\pm3,\pm4,\pm6,\pm12,\pm\frac{1}{4},\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{3}{4},\pm\frac{1}{4}\)
\(\textbf{3)}\) \(f(x)=7x^8+9x^4+3x^3-2x+4\) All possible rational roots are \(\displaystyle \pm1,\pm2,\pm4,\pm\frac{1}{7},\pm\frac{2}{7},\pm\frac{4}{7}\)
\(\textbf{4)}\) \(f(x)=x^3+9x^2+2x+16\) All possible rational roots are \(\displaystyle \pm1,\pm2,\pm4,\pm8,\pm16\)
\(\textbf{5)}\) \(f(x)=-8x^4+-2x^2+x-1\) All possible rational roots are \(\displaystyle \pm1,\pm\frac{1}{2},\pm\frac{1}{4},\pm\frac{1}{8}\)
Find all zeroes of each function
\(\textbf{6)}\) \(f(x)=x^3+2x^2-11x-12\) The zeroes are \(x=-4,-1,3\)
\(\text{We use synthetic division with } x = -4 \text{ and the coefficients of } x^3 + 2x^2 – 11x – 12.\)
\(
\begin{array}{r|rrrr}
-4 & 1 & 2 & -11 & -12 \\
& & -4 & 8 & 12 \\
\hline
& 1 & -2 & -3 & 0 \\
\end{array}
\)
\(\text{This yields a quotient of } x^2 – 2x – 3.\)
\(\,\,\,\,\, x^2 – 2x – 3 = (x + 1)(x – 3)\)
\(\,\,\,\,\, x = -4, -1, 3\)
\(\text{We use synthetic division with } x = -4 \text{ and the coefficients of } x^3 + 2x^2 – 11x – 12.\)
\(
\begin{array}{r|rrrr}
-4 & 1 & 2 & -11 & -12 \\
& & -4 & 8 & 12 \\
\hline
& 1 & -2 & -3 & 0 \\
\end{array}
\)
\(\text{This yields a quotient of } x^2 – 2x – 3.\)
\(\,\,\,\,\, x^2 – 2x – 3 = (x + 1)(x – 3)\)
\(\,\,\,\,\, x = -4, -1, 3\)
\(\textbf{7)}\) \(f(x)=x^3-6x^2+11x-6\) The zeroes are \(x=1, 2, 3\)
\(\text{We use synthetic division with }x = 1 \text{ and the coefficients of }x^3 – 6x^2 + 11x – 6.\)
\(\,\,\,
\begin{array}{r|rrrr}
1 & 1 & -6 & 11 & -6 \\
& & 1 & -5 & 6 \\
\hline
& 1 & -5 & 6 & 0 \\
\end{array}
\)
\(\text{This yields a quotient of }x^2 – 5x + 6.\)
\(\,\,\,\,\,x^2 – 5x + 6 = (x – 2)(x – 3)\)
\(\,\,\,\,\,x = 1, 2, 3\)
\(\text{We use synthetic division with }x = 1 \text{ and the coefficients of }x^3 – 6x^2 + 11x – 6.\)
\(\,\,\,
\begin{array}{r|rrrr}
1 & 1 & -6 & 11 & -6 \\
& & 1 & -5 & 6 \\
\hline
& 1 & -5 & 6 & 0 \\
\end{array}
\)
\(\text{This yields a quotient of }x^2 – 5x + 6.\)
\(\,\,\,\,\,x^2 – 5x + 6 = (x – 2)(x – 3)\)
\(\,\,\,\,\,x = 1, 2, 3\)
\(\textbf{8)}\) \(f(x)=2x^3+x^2-5x+2\) The zeroes are \(x=-2, 1, \displaystyle \frac{1}{2}\)
\(\text{We use synthetic division with } x = -2 \text{ and the coefficients of } 2x^3 + x^2 – 5x + 2.\)
\(
\begin{array}{r|rrrr}
-2 & 2 & 1 & -5 & 2 \\
& & -4 & 6 & -2 \\
\hline
& 2 & -3 & 1 & 0 \\
\end{array}
\)
\(\text{This yields a quotient of } 2x^2 – 3x + 1.\)
\(\,\,\,\,\, 2x^2 – 3x + 1 = (x – 1)\left(2x – 1\right)\)
\(\,\,\,\,\, x = -2, 1, \frac{1}{2}\)
\(\text{We use synthetic division with } x = -2 \text{ and the coefficients of } 2x^3 + x^2 – 5x + 2.\)
\(
\begin{array}{r|rrrr}
-2 & 2 & 1 & -5 & 2 \\
& & -4 & 6 & -2 \\
\hline
& 2 & -3 & 1 & 0 \\
\end{array}
\)
\(\text{This yields a quotient of } 2x^2 – 3x + 1.\)
\(\,\,\,\,\, 2x^2 – 3x + 1 = (x – 1)\left(2x – 1\right)\)
\(\,\,\,\,\, x = -2, 1, \frac{1}{2}\)
Express each in linear factored form
\(\textbf{9)}\) Express \(f(x)=x^3+2x^2-11x-12\) in linear factored form. \(f(x)=(x+4)(x+1)(x-3)\)
\(\textbf{10)}\) Express \(f(x)=x^3-6x^2+11x-6\) in linear factored form. \(f(x)=(x-1)(x-2)(x-3)\)
\(\textbf{11)}\) Express \(f(x)=2x^3+x^2-5x+2\) in linear factored form. \(f(x)=(2x-1)(x+2)(x-1)\)
See Related Pages\(\)
\(\bullet\text{ Adding and Subtracting Polynomials}\)
\(\,\,\,\,\,\,\,\,(4d+7)−(2d−5)…\)
\(\bullet\text{ Multiplying Polynomials}\)
\(\,\,\,\,\,\,\,\,(x+2)(x^2+3x−5)…\)
\(\bullet\text{ Dividing Polynomials}\)
\(\,\,\,\,\,\,\,\,(x^3-8)÷(x-2)…\)
\(\bullet\text{ Dividing Polynomials (Synthetic Division)}\)
\(\,\,\,\,\,\,\,\,(x^3-8)÷(x-2)…\)
\(\bullet\text{ Synthetic Substitution}\)
\(\,\,\,\,\,\,\,\,f(x)=4x^4−3x^2+8x−2…\)
\(\bullet\text{ End Behavior}\)
\(\,\,\,\,\,\,\,\, \text{As } x\rightarrow \infty, \quad f(x)\rightarrow \infty \)
\(\,\,\,\,\,\,\,\, \text{As } x\rightarrow -\infty, \quad f(x)\rightarrow \infty… \)
\(\bullet\text{ Completing the Square}\)
\(\,\,\,\,\,\,\,\,x^2+10x−24=0…\)
\(\bullet\text{ Quadratic Formula and the Discriminant}\)
\(\,\,\,\,\,\,\,\,x=-b \pm \displaystyle\frac{\sqrt{b^2-4ac}}{2a}…\)
\(\bullet\text{ Complex Numbers}\)
\(\,\,\,\,\,\,\,\,i=\sqrt{-1}…\)
\(\bullet\text{ Multiplicity of Roots}\)
\(\,\,\,\,\,\,\,\,\)
\(…\)
\(\bullet\text{ Rational Zero Theorem}\)
\(\,\,\,\,\,\,\,\, \pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12…\)
\(\bullet\text{ Descartes Rule of Signs}\)
\(\,\)
\(\bullet\text{ Roots and Zeroes}\)
\(\,\,\,\,\,\,\,\,\text{Solve for }x. 3x^2+4x=0…\)
\(\bullet\text{ Linear Factored Form}\)
\(\,\,\,\,\,\,\,\,f(x)=(x+4)(x+1)(x−3)…\)
\(\bullet\text{ Polynomial Inequalities}\)
\(\,\,\,\,\,\,\,\,x^3-4x^2-4x+16 \gt 0…\)
In Summary
The Rational Zero Theorem states that if a polynomial (with rational coefficients) has a rational zero, then that zero can be expressed in the form of a fraction p/q, where p is a factor of the constant term of the polynomial and q is a factor of the leading coefficient of the polynomial. It is usually used when the degree of a polynomial is greater than 2 and traditional methods of factoring won’t work. It is commonly used in conjunction with polynomial division or synthetic division.
The Rational Zero Theorem is typically introduced in an algebra 2 or pre-calculus course along with the chapter on polynomials.