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Solving Polynomial Equations

Polynomials are usually written in standard form. However, depending on what they describe and what information is needed, it's sometimes useful to write them in factored form. By rewriting polynomials as products of their factors, it's possible to solve polynomial equations.
Concept

Polynomial Equation

A polynomial equation is an equation that contains a polynomial expression. An example is x3x=7x2+3. x^3 - x = 7x^2 + 3. By rearranging the equation so that one side is 0,0, it's possible to identify the degree of the equation. The equation above can be rewritten as x37x2x3=0. x^3 - 7x^2 - x - 3 = 0.

The maximum number of solutions of an equation is given by the degree. Since the highest exponent is 3,3, this equation has a maximum of 33 solutions. In order to solve some polynomial equations, algebraic methods, such as the Quadratic Formula and the Zero Product Property, can be used. Alternatively, a graphic solution works for any polynomial equation and numerical methods may also be used.
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Exercise

Solve the equation by factoring. 2x3+3x218x27=0 2x^3+3x^2-18x-27=0

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Solution
To begin, we must determine which type of factoring we can use to solve the given polynomial equation. Notice that there is no common factor amongst all four terms. (2xxx)+(3xx)(233x)(333) (2 \cdot x \cdot x \cdot x) + (3 \cdot x \cdot x) - (2 \cdot 3 \cdot 3 \cdot x) - (3 \cdot 3 \cdot 3) The first three terms all have an xx in common, while the last three terms have a 33 in common. Thus, we must factor by grouping the first two terms and the last two terms. In the first pair, the GCF is x2,x^2, and in the second it is -9.\text{-} 9. Let's factor out the GCFs separately.
2x3+3x218x27=02x^3+3x^2-18x-27=0
x2(2x+3)18x27=0x^2(2x+3)-18x-27=0
x2(2x+3)9(2x+3)=0x^2(2x+3)-9(2x+3)=0
Now, both terms have the factor 2x+3,2x+3, which can be factored out. x2(2x+3)9(2x+3)=(x29)(2x+3)=0 x^2(2x+3)-9(2x+3)=\left(x^2-9\right)(2x+3)=0 Using the Zero Product Property, we can separate this equation into two new equations, that can be solved individually.
(x29)(2x+3)=0\left(x^2-9\right)(2x+3)=0
x29=0(I)2x+3=0(II)\begin{array}{lc}x^2-9=0 & \text{(I)}\\ 2x+3=0 & \text{(II)}\end{array}
x2=92x+3=0\begin{array}{l}x^2=9 \\ 2x+3=0 \end{array}
x=±92x+3=0\begin{array}{l}x=\pm\sqrt{9} \\ 2x+3=0 \end{array}
x=±32x+3=0\begin{array}{l}x=\pm3 \\ 2x+3=0 \end{array}
x=±32x=-3\begin{array}{l}x=\pm3 \\ 2x=\text{-}3 \end{array}
x=±3x=-1.5\begin{array}{l}x=\pm3 \\ x=\text{-}1.5 \end{array}
Thus, the original equation has three solutions: x=-3,x=\text{-}3, x=-1.5,x=\text{-}1.5, and x=3.x=3.
Method

Solving a Polynomial Equation Graphically

Similar to other equations, polynomial equations can be solved graphically. Consider the following equation. x32x22x+1=2x23x5 x^3-2x^2-2x+1=2x^2-3x-5 Usually, all terms are gathered to one side to create an equivalent equation. x34x2+x+6=0 x^3-4x^2+x+6=0 The polynomial expression on the left-hand side can be viewed as the function y=x34x2+x+6.y=x^3-4x^2+x+6. The solutions to the original equation are the zeros of the function. Graphing the function allows the zeros to be seen easily.

They are x=-1,x=\text{-}1, x=2,x=2, and x=3.x=3. Thus, the solutions to the original equation are x=-1,x=\text{-}1, x=2,x=2, and x=3.x=3. By substituting the solutions into the equation, it's possible to determine if the solutions are exact or approximate.
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Exercise

Sketch the following polynomial function using zeros and end behavior. y=x3+26x2+105x y=x^3+26x^2+105x

Show Solution
Solution
Let's start by solving the equation to find the zeros of the function. x3+26x2+105x=0 x^3+26x^2+105x=0 Since all terms contain x,x, we can factor xx out and then use the Zero Product Property.
x3+26x2+105x=0x^3+26x^2+105x=0
x(x2+26x+105)=0x\left(x^2+26x+105\right)=0
x=0x2+26x+105=0\begin{array}{l}x=0 \\ x^2+26x+105=0 \end{array}
We get x=0,x=0, and a quadratic equation. This means one solution is x=0.x=0. We can find the others by solving the quadratic equation.
x2+26x+105=0x^2+26x+105=0
x=-26±2624110521x=\dfrac{\text{-}{\color{#009600}{26}}\pm\sqrt{{\color{#009600}{26}}^2-4\cdot{\color{#0000FF}{1}}\cdot{\color{#FF0000}{105}}}}{2\cdot{\color{#0000FF}{1}}}
x=-26±6764202x=\dfrac{\text{-}26\pm\sqrt{676-420}}{2}
x=-26±2562x=\dfrac{\text{-}26\pm\sqrt{256}}{2}
x=-26±162x=\dfrac{\text{-}26\pm 16}{2}
x=-42/2x=-10/2\begin{array}{l}x=\text{-}42/2 \\ x=\text{-}10/2 \end{array}
x1=-21x2=-5\begin{array}{l}x_1=\text{-}21 \\ x_2=\text{-}5 \end{array}
The equation has the solutions x=0,x=0, x=-21x=\text{-}21 and x=-5.x=\text{-}5. These are the function's zeros. To graph the function, we can plot the zeros in a coordinate plane.

The coefficient in the x3x^3-term is positive, meaning that when xx approaches infinity, the graph extends upward.

Since the function is a third-degree polynomial — odd — the left end extends in the opposite direction: downward.

Lastly, let's sketch the rest of the graph. We don't know exactly how it looks, but it intersects the xx-axis at x=-21,x=\text{-}21, x=-1x=\text{-}1 and x=0.x=0.

Note that other sketches are possible, as long as the zeros and end behavior are the same.

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