Solving Polynomial Equations

Solving Polynomial Equations is a fundamental aspect of algebra that involves finding values of the variable that make the equation true.

Understanding Polynomial Equations

A polynomial equation is any expression of the form $ a_n x^n + a_{n-1} x^{n-1} + \ldots + a_2 x^2 + a_1 x + a_0 = 0 $.

Solving Linear Polynomials

Linear polynomials of the form $ax + b = 0$ can be solved using $ x = -\cfrac{b}{a} $.

Solving Quadratic Polynomials

For a quadratic polynomial $ax^2 + bx + c = 0$, use the quadratic formula: $x = \cfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Solving Higher Degree Polynomials

For cubic or quartic equations, such as $ax^3 + bx^2 + cx + d = 0$, use factoring, synthetic division, or special formulas specific to these polynomials.

Steps to Solve Polynomial Equations

Standard Form: Ensure the polynomial is written in the form $a_n x^n + a_{n-1} x^{n-1} + \ldots + a_2 x^2 + a_1 x + a_0 = 0$
Factor Out GCD: Factor the greatest common divisor from the polynomial if possible
Identify Possible Roots: Use techniques like the Rational Root Theorem for potential roots $\pm\frac{p}{q}$, where $ p $ is a factor of the constant term and $ q $ is a factor of the leading coefficient
Test and Divide: Use synthetic division to divide the polynomial by $(x - r)$ where $r$ is a root, to reduce the polynomial degree
Repeat: Continue the process with the reduced polynomial
Check All Roots: Substitute the roots back into the original equation to verify they are correct
Consider Complex Roots: Non-real roots can appear in conjugate pairs if the polynomial has real coefficients

Example

Solve $15x^2 + 3x - 8 = 5x - 7$

First, we can express the equation in Standard Form. We can do so by moving all terms onto the same side and setting the equation equal to $0$:

$15x^2 + 3x - 8 - 5x + 7 = 0$

Next, we can collect like terms:

$15x^2 - 2x - 1 = 0$

Then, we can factor the expression. In this instance, we will use the quadratic equation.

Since we can identify that $\textcolor{red}{a = 15}$, $\textcolor{green}{b = -2}$, and $\textcolor{blue}{c = -1}$, we can plug these values into the quadratic formula to solve for the solutions:

$x = \cfrac{-\textcolor{green}{b} \pm \sqrt{\textcolor{green}{b}^2 - 4\textcolor{red}{a}\textcolor{blue}{c}}}{2\textcolor{red}{a}}$

$x= \cfrac{-(\textcolor{green}{-2}) \pm \sqrt{(\textcolor{green}{-2})^2 - 4(\textcolor{red}{15})(\textcolor{blue}{-1})}}{2(\textcolor{red}{15})}$

$x = \cfrac{2 \pm \sqrt{4 + 60}}{30}$

$x = \cfrac{2 \pm \sqrt{64}}{30}$

$x = \cfrac{2 \pm 8}{30}$

After, we can use $\pm$, to determine the $2$ roots either by adding the 2 values of subtracting the $2$ values in the numerator.

We can determine the first root by adding the values in the numerator:

$x_1 = \cfrac{2 + 8}{30}$

$x_1 = \cfrac{10}{30}$

$x_1 = \cfrac{1}{3}$

We can determine the first root by subtracting the values in the numerator:

$x_2 = \cfrac{2-8}{30}$

$x_2 = \cfrac{-6}{30}$

$x_2 = -\cfrac{1}{5}$

Therefore, we can determine the solutions are $\boldsymbol{\cfrac{1}{3}}$ and $\boldsymbol{-\cfrac{1}{5}}$.

Solve the following equations, then sketch the function that corresponds to the one with all the terms moved to the left hand side

$6x^3 + 49x^2 + 8x - 12 = 2x + 4$

Show Answer

First, we can rewrite the equation in Standard Form by moving all terms to one side and setting it to $0$:

$6x^3 + 49x^2 + 8x - 12 - 2x - 4 = 0$

Next, we can simplify the equation by collecting like terms:

$6x^3 + 49x^2 + 6x - 16 = 0$

Next, we can factor the equation. To find the factors of this equation, we can use the rational root theorem (or $\cfrac{p}{q}$) where $p$ represents the factor of the trailing constant while $q$ represents the factor of the leading coefficient.

The factors of the leading coefficient, $6$, are $1, 2, 3, 6$ while the factors of the constant term, $-16$, are $1, 2, 4, 8, 16$.

If we substitute these values into the current expression, we get $P\left(\cfrac{1}{2}\right) = 0$

$f\left(\cfrac{1}{2}\right) = 6\left(\cfrac{1}{2}\right)^3 + 49\left(\cfrac{1}{2}\right)^2 + 6\left(\cfrac{1}{2}\right) - 16$

$f\left(\cfrac{1}{2}\right) = 6\left(\cfrac{1}{8}\right) + 49\left(\cfrac{1}{4}\right) + 3 - 16$

$f\left(\cfrac{1}{2}\right) = {3}{4} + \cfrac{49}{4}-\cfrac{52}{4}$

$f\left(\cfrac{1}{2}\right) = 0$

We can determine that $(2x - 1)$ is a factor of the polynomial. Now dividing $(2x - 1)$ by the polynomial, we have:

$$ \require{enclose} \begin{array}{r} 3x^2+26x+16 \\[-3pt] 2x-1 \enclose{longdiv}{6x^3 + 49x^2 + 6x - 16} \\[-3pt] \underline{-6x^3+3x^2}\phantom{2} \\[-3pt] 52x^2+6x-16 \\[-3pt] \underline{-52x^2+26x}\phantom{2} \\[-3pt] 32x-16 \\[-3pt] \underline{-32x+16}\phantom{2} \\[-3pt] 0 \\[-3pt] \end{array} $$

We can determine that the polynomial factored is:

$(2x -1)(3x^2+26x+16)$

We can factor the expression even further and determine the solutions:

$(2x -1)(x+8)(3x+2)$

$x_1 = \cfrac{1}{2}$

$x_2 = -8$

$x_3 = \cfrac{1}{3}$

Therefore, we can determine the solutions of this function as $\boldsymbol{-8}, \boldsymbol{\cfrac{1}{2}},$ and $\boldsymbol{\cfrac{1}{3}}$.

We can sketch a graph of this function as:

Graph of the cubic polynomial equation, 6x³+49x²+6x-16=0.

$113x - 30 = 8x^3 - 30x^2$

Show Answer

First, we can rewrite the equation in Standard Form by moving all terms to one side and setting it to $0$:

$-8x^3 + 30x^2 + 113x - 30 = 0$

The factors of the leading coefficient, $-8$, are $1, 2, 4, 8$ while the factors of the constant term, $30$, are $1, 2, 3, 5, 6, 10, 15, 30$.

If we substitute these values into the current expression, we get $P\left(\cfrac{1}{4}\right) = 0$

$f\left(\cfrac{1}{4}\right) = -8\left(\cfrac{1}{4}\right)^3 + 30\left(\cfrac{1}{4}\right)^2 + 113\left(\cfrac{1}{4}\right) - 30$

$f\left(\cfrac{1}{4}\right) = -8(\cfrac{1}{64}) + 30(\cfrac{1}{16}) + \cfrac{113}{4} - 30$

$f\left(\cfrac{1}{4}\right) = -\cfrac{1}{8} + \cfrac{15}{8} + \cfrac{226}{8} - \cfrac{240}{8}$

$f\left(\cfrac{1}{4}\right) = 0$

We can determine that $(4x - 1)$ is a factor of the polynomial. Now dividing $(4x - 1)$ by $f(x)$, we have:

$$ \require{enclose} \begin{array}{r} -2x^2+7x +30 \\[-3pt] 4x-1 \enclose{longdiv}{-8x^3 + 30x^2 + 113x - 30} \\[-3pt] \underline{8x^3-2x^2}\phantom{2} \\[-3pt] 28x^2+113x-30 \\[-3pt] \underline{-28x^2+7x}\phantom{2} \\[-3pt] 120x-30 \\[-3pt] \underline{-120x -30}\phantom{2} \\[-3pt] 0 \\[-3pt] \end{array} $$

We can determine that $f(x)$ factored is:

$(4x -1)(-2x^2+7x +30)$

We can factor the expression even further and determine the solutions:

$(4x -1)(x-6)(-2x-5)$

$x_1 = \cfrac{1}{4}$

$x_2 = 6$

$x_3 = -\cfrac{5}{2}$

Therefore, we can determine the solutions are $\boldsymbol{-\cfrac{5}{2}}, \boldsymbol{\cfrac{1}{4}},$ and $\boldsymbol{6}$.

We can sketch a graph of this expression as such:

Graph of the cubic polynomial equation, -8x³+30x²+113x-30=0.