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3. Polynomial Expressions, Equations, & Functions
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Algebra 2

3.06 Solve polynomial equations

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Solving polynomial equations

A polynomial equation is an equation with polynomial expressions on both sides of the equation.

The standard form of a polynomial equation is given by a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_2x^2+a_1x+a_0=0 where n is a positive integer and a_n,\,a_{n-1},\,a_{n-2},\,\ldots,\,a_2,\,a_1,\,a_0 are constant coefficients.

There are various methods for solving polynomial equations, including:

  • factoring

  • quadratic formula

  • graphing

Let's compare these methods by finding the solutions of

-12x^{3}+72x=30x^{2}

Regardless of method, the polynomial should be rewritten in standard form:

0=12x^{3}+30x^{2}-72x

Solve by factoring

  • To solve a polynomial in factored form, we can set each factor equal to 0 and solve.

    \displaystyle 0\displaystyle =\displaystyle 6x(2x^{2}+5x-12)Take out a greatest common factor
    \displaystyle 0\displaystyle =\displaystyle 6x(2x-3)(x+4)Factor the quadratic completely

    6x=0 \qquad 2x-3=0 \qquad x+4=0

    x= \left\{-4 ,\, 0, \, \dfrac{3}{2}\right\}

    Not all polynomials are factorable, so an alternative method may be necessary.

Solve using the quadratic formula, x=\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}

  • A polynomial equation containing a quadratic expression can be solved using the quadratic formula. After factoring out the 6x that gave x=0 as a solution, we will use the quadratic formula for 2x^{2}+5x-12 with a=2,\,b=5,\, and c=-12.

    \displaystyle x\displaystyle =\displaystyle \dfrac{-5 \pm \sqrt{5^{2}-4(2)(-12)}}{2(2)}
    \displaystyle x\displaystyle =\displaystyle \dfrac{-5 \pm \sqrt{121}}{4}
    \displaystyle x\displaystyle =\displaystyle \dfrac{-5 - 11}{4},\,\dfrac{-5+11}{4}
    \displaystyle x\displaystyle =\displaystyle -4,\,\dfrac{3}{2}

    The solutions from both the quadratic formula and the greatest common factor give the same set of zeros, \left\{-4,\,0,\,\dfrac{3}{2}\right\}.

    This formula works for all quadratics, both factorable and non-factorable. However, this method cannot be used with higher-degree polynomials unless they can be factored down to a quadratic first.

Solve by graphing

  • The graph of a polynomial equation set equal to 0 will show the real solutions as x-intercepts. To graph the corresponding polynomial function of the given example, we will substitute f(x) or y for 0.f(x)=-12x^{3}-30x^{2}+72x

    -4
    -3
    -2
    -1
    1
    2
    3
    4
    x
    -180
    -160
    -140
    -120
    -100
    -80
    -60
    -40
    -20
    20
    y

    We see integer roots at -4 and 0, and can estimate a solution at 1.5, or \dfrac{3}{2}.

    This method can be used to find integer solutions or to estimate non-integer solutions, but it is not the best method for finding exact non-integer roots or imaginary solutions.

Examples

Example 1

Given the polynomial equation \left(4x^2-81\right)\left(x^2-3x-10\right)=0:

a

Determine the number and type of solutions.

Worked Solution
Create a strategy

Since the equation is already factored as two quadratics, the sign of each quadratic's discriminant, b^{2}-4ac, will tell us the number and type of solutions:

  • Positive: 2 real solutions

  • 0: 1 real solution

  • Negative: 2 complex, non-real solutions

Apply the idea

For the quadratic expression 4x^{2}-81, we will use values of a=4,\, b=0,\,and c=-81 to find the discriminant.

\displaystyle b^{2}-4ac\displaystyle =\displaystyle (0)^{2}-4(4)(-81)Substitute values of a,b, and c
\displaystyle =\displaystyle 0+1296
\displaystyle =\displaystyle 1296Simplify

The discriminant is positive, so 4x^{2}-81 has 2 real solutions.

For the quadratic expression x^{2}-3x-10, we will use values of a=1,\, b=-3,\,and c=-10 to find the discriminant.

\displaystyle b^{2}-4ac\displaystyle =\displaystyle (-3)^{2}-4(1)(-10)Substitute values of a,\,b,\, and c
\displaystyle =\displaystyle 9+40
\displaystyle =\displaystyle 49Simplify

The discriminant is positive, so x^{2}-3x-10 has 2 real solutions.

Combined, the number of real solutions to the polynomial is 4.

Reflect and check

We can check that the polynomial has 4 solutions using the fundamental theorem of algebra. This theorem says that the degree of the polynomial is equal to the number of its complex solutions.

If we were to multiply the given factors together, the degree with the highest term would come from multiplying 4x^2 in the first expression by x^2 in the second expression.4x^2\cdot x^2=4x^4This shows us that the degree of the polynomial is 4, so it has 4 complex solutions. We found 4 real solutions, and real numbers belong to the set of complex numbers, which means there are no imaginary or non-real solutions.

b

Solve the equation.

Worked Solution
Create a strategy

First, we determine whether the expression on the left-hand side of the equation can be expressed as a product of linear factors. If so, we can apply some factoring techniques.

Apply the idea

The equation contains a factored expression on the left-hand side and 0 on the right-hand side. The factors are both of degree 2. We need to express these quadratic factors as linear factors, if possible.

Observe that 4x^2-81=\left(2x\right)^2-\left(9\right)^2. This means that 4x^2-81 is a difference of two squares, so we have 4x^2-81=\left(2x-9\right)\left(2x+9\right).

\displaystyle \left(4x^2-81\right)\left(x^2-3x-10\right)\displaystyle =\displaystyle 0Original equation
\displaystyle \left(2x-9\right)\left(2x+9\right)\left(x^2-3x-10\right)\displaystyle =\displaystyle 0Difference of two squares identity
\displaystyle \left(2x-9\right)\left(2x+9\right)\left(x-5\right)\left(x+2\right)\displaystyle =\displaystyle 0Factor the quadratic

The zero product property states that if A \cdot B \cdot C \cdot D=0, then A =0, B=0, C=0 or D=0.

So we get the following equations:\begin{aligned}2x-9=0\\ 2x+9=0\\x-5=0 \\x+2=0\end{aligned} Solving each equation for x, we obtain the following solutions: x=\dfrac{9}{2},\,x=-\dfrac{9}{2},\,x=5,\,x=-2

Reflect and check

We can check the answer by using technology to graph y=\left(4x^2-81\right)\left(x^2-3x-10\right), then looking for the x-intercepts.

-5
-4
-3
-2
-1
1
2
3
4
5
x
-300
-150
150
300
450
600
750
900
y

The function is equal to zero at x=-\dfrac{9}{2},\,x=-2,\,x=\dfrac{9}{2},\,x=5 which confirms these are the solutions of \left(4x^2-81\right)\left(x^2-3x-10\right)=0.

Example 2

Given the equation 4x^5-9x^3=72-32x^2:

a

Determine the number and type of solutions.

Worked Solution
Create a strategy

The degree of a polynomial equation corresponds to the number of complex solutions, including repeated solutions.

After writing the equation in standard form, we can graph the corresponding polynomial function, identify the x-intercepts, then compare the intercepts and their multiplicities with the degree of the equation. This will help us determine the number of real and imaginary solutions.

Apply the idea

We can see that the equation 4x^5-9x^3=72-32x^2 is not in standard form because it is not equal to 0. We can convert the equation to standard form by moving the terms 72 and -32x^2 to the left-hand side of the equation using inverse operations.4x^5-9x^3+32x^2-72=0

Now, we can graph the corresponding polynomial function, f(x)=4x^{5}-9x^{3}+32x-72

-4
-3
-2
-1
1
2
3
4
x
-70
-60
-50
-40
-30
-20
-10
10
20
30
y

The graph shows 3 distinct x-intercepts, and the graph crosses the x-axis at each intercept. This means each intercept has a multiplicity of 1.

This tells us the polynomial equation has 3 real solutions.

Since the degree of the polynomial equation is 5, the polynomial must also have 2 complex, non-real solutions.

b

Find the roots of the polynomial equation and use the roots to confirm part (a).

Worked Solution
Create a strategy

Once in standard form, a polynomial equation can be solved in various ways, including factoring. Once factored, each factor will be set equal to zero and solved.

Apply the idea

In the previous part, we wrote the equation in standard form, giving 4x^5-9x^3+32x^2-72=0. Factoring and solving will give the exact real roots and allow us to solve for the imaginary solutions.

By factoring,

\displaystyle 4x^5-9x^3+32x^2-72\displaystyle =\displaystyle 0Standard form
\displaystyle (4x^5+32x^2)+(-9x^3-72)\displaystyle =\displaystyle 0Group terms based on common factors
\displaystyle 4x^2(x^3+8)-9(x^3+8)\displaystyle =\displaystyle 0Factor out the GCF: 4x^2 and -9
\displaystyle \left(4x^2-9\right)\left(x^3+8\right)\displaystyle =\displaystyle 0Factor out the common binomial
\displaystyle \left(2x-3\right)\left(2x+3\right)\left(x^3+8\right)\displaystyle =\displaystyle 0Difference of squares identity
\displaystyle \left(2x-3\right)\left(2x+3\right)\left(x+2\right)\left(x^2-2x+4\right)\displaystyle =\displaystyle 0Sum of cubes identity

Using the zero product property for the linear factors, we solve the following equations:

\begin{aligned}2x-3&=0,\,x=\dfrac{3}{2}\\ 2x+3&=0,\, x=-\dfrac{3}{2}\\x+2&=0,\,x=-2\end{aligned}

Since the expression x^2-2x+4 is not factorable, we will use the quadratic formula to solve:

\displaystyle x\displaystyle =\displaystyle \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}Quadratic formula
\displaystyle =\displaystyle \dfrac{-(-2)\pm\sqrt{\left(-2\right)^2-4(1)(4)}}{2(1)}Substitute a=1,\,b=-2 and c=4
\displaystyle =\displaystyle \dfrac{2\pm\sqrt{-12}}{2}Evaluate the operations
\displaystyle =\displaystyle \dfrac{2\pm2i\sqrt{3}}{2}Rewrite the radical
\displaystyle =\displaystyle 1\pm \sqrt{3}iSimplify

Therefore, the solutions are:x\in\left\{\dfrac{3}{2},\,-\dfrac{3}{2},\,-2,\,1+ \sqrt{3}i,\,1- \sqrt{3}i\right\}

These solutions match the predicted number of each type of solution in part (a) since there are 3 real solutions and 2 complex, non-real solutions.

Reflect and check

We can check the solutions by substituting the values and confirming equality. First checking the real solutions:

\displaystyle 4\left(\dfrac{3}{2}\right)^{5}-9\left(\dfrac{3}{2}\right)^{3}\displaystyle =\displaystyle 72-32\left(\dfrac{3}{2}\right)^{2}Substitute x=\dfrac{3}{2}
\displaystyle \dfrac{243}{8}-\dfrac{243}{8}\displaystyle =\displaystyle 72-72Evaluate the multiplication
\displaystyle 0\displaystyle =\displaystyle 0\checkmark
\displaystyle 4\left(-\dfrac{3}{2}\right)^{5}-9\left(-\dfrac{3}{2}\right)^{3}\displaystyle =\displaystyle 72-32\left(-\dfrac{3}{2}\right)^{2}Substitute x=-\dfrac{3}{2}
\displaystyle -\dfrac{243}{8}+\dfrac{243}{8}\displaystyle =\displaystyle 72-72Evaluate the multiplication
\displaystyle 0\displaystyle =\displaystyle 0\checkmark
\displaystyle 4(-2)^{5}-9(-2)^{3}\displaystyle =\displaystyle 72-32(-2)^{2}Substitute x=-2
\displaystyle -128+72\displaystyle =\displaystyle 72-128Evaluate the multiplication
\displaystyle -56\displaystyle =\displaystyle -56\checkmark

Checking the imaginary solutions:

\displaystyle 4(1+\sqrt{3}i)^{5}-9(1+\sqrt{3}i)^{3}\displaystyle =\displaystyle 72-32(1+\sqrt{3}i)^{2}Substitute x=1+\sqrt{3}i
\displaystyle 64-64\sqrt{3}i+72\displaystyle =\displaystyle 72+64-64\sqrt{3}iEvaluate the multiplication
\displaystyle 136-64\sqrt{3}i\displaystyle =\displaystyle 136-64\sqrt{3}i\checkmark
\displaystyle 4(1-\sqrt{3}i)^{5}-9(1-\sqrt{3}i)^{3}\displaystyle =\displaystyle 72-32(1-\sqrt{3}i)^{2}Substitute x=1-\sqrt{3}i
\displaystyle 64+64\sqrt{3}i+72\displaystyle =\displaystyle 72+64+64\sqrt{3}iEvaluate the multiplication
\displaystyle 136+64\sqrt{3}i\displaystyle =\displaystyle 136+64\sqrt{3}i\checkmark

All five values are valid solutions for the polynomial since each value balanced the two sides of the equation.

Example 3

Gabby is designing a marble basin that will house a set of planters in her garden. The basin's sides and bottom should be 1 foot thick. Its outer length should be twice the outer width and outer height.

She wants the basin to hold 36 cubic feet of water.

a

Gabby has determined that36=(2x-2)(x-2)(x-1) represents the dimensions of her basin. Determine all possible solutions for f(x).

Worked Solution
Create a strategy

Since we do not have zero on one side of the equation, we will expand the factors and subtract the 36 to the other side.

After the polynomial is expanded and the 36 is subtracted, the polynomial can be factored and its factors set equal to zero.

Apply the idea
\displaystyle 36\displaystyle =\displaystyle (2x-2)(x-2)(x-1)Original equation
\displaystyle 36\displaystyle =\displaystyle (2x^{2}-6x+4)(x-1)Multiply the first two binomials
\displaystyle 36\displaystyle =\displaystyle 2x^{3}-8x^{2}+10x-4Multiply the trinomial and binomial
\displaystyle 0\displaystyle =\displaystyle 2x^{3}-8x^{2}+10x-40Subtract 36
\displaystyle 0\displaystyle =\displaystyle 2\left(x^{3}-4x^{2}+5x-20\right)Factor the GCF
\displaystyle 0\displaystyle =\displaystyle 2\left[x^{2}\left(x-4\right)+5\left(x-4\right)\right]Factor by grouping
\displaystyle 0\displaystyle =\displaystyle 2(x^{2}+5)(x-4)Factor the GCF

Once factored, each factor will be set equal to zero and solved.

\displaystyle 0\displaystyle =\displaystyle x^{2}+5Write the first factor
\displaystyle -5\displaystyle =\displaystyle x^{2}Subtract 5 from both sides
\displaystyle \pm \sqrt{5}i\displaystyle =\displaystyle xSquare root both sides
\displaystyle 0\displaystyle =\displaystyle x-4Write the second factor
\displaystyle 4\displaystyle =\displaystyle xAdd 4 to both sides

The solutions of f(x) are x=-\sqrt{5}i, \, \sqrt{5}i, \, 4.

Reflect and check

We know that (x-4) has a real solution since it is a a linear expression. Since (x^{2}+5) is a quadratic, we could use the value of the discriminant, b^{2}-4ac, to determine what types of solutions it has. With a=1,\,b=0,\,and c=5, the discriminant is

(0)^{2}-4(1)(5)

-20

Since the discriminant of x^{2}+5 is negative, we know that f(x) has two complex, non-real solutions.

b

Determine which solution(s) would be best for Gabby's design, and describe your choice using the dimensions of the basin.

Worked Solution
Create a strategy

The value of x represents the thickness of the planter, so only real solutions should be considered. If there are multiple real solutions, they can be plugged in for x to see which values are the most realistic for the context.

Apply the idea

From the previous part, the solutions of f(x) are x=-\sqrt{5}i, \, \sqrt{5}i, \, 4. Only x=4 is a real solution.

Since x=4 is the only real solution and x represents the thickness, it is the only choice for Gabby's design. By substituting x=4 for each factor of (2x-2)(x-2)(x-1), we have dimensions of

2(4)-2=6 \qquad 4-2 = 2 \qquad 4-1=3

The dimensions of the basin are 6 feet by 2 feet by 3 feet.

Reflect and check

If our polynomial had multiple real solutions, we would evaluate the different real values in the context of the problem and see which gives the most realistic result.

Idea summary

The roots of a polynomial equation, p\left(x\right)=0, can be found using multiple methods after the equation has been written in standard form. Methods include

  • Factoring: If p(x) is factorable, then the zero product property can be used to find the solutions.

  • Quadratic formula: If the equation contains a factor ax^{2}+bx+c, the quadratic formula, x=\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}, can be used to find its solutions.

  • Graphing: The x-intercepts represent the real zeros of a polynomial. Graphing is best used to find integer solutions.

Outcomes

A2.EI.6

The student will represent, solve, and interpret the solution to a polynomial equation.

A2.EI.6c

Solve a polynomial equation over the set of complex numbers.

A2.EI.6b

Determine the number and type of solutions (real or imaginary) of a polynomial equation of degree three or higher.

A2.EI.6d

Verify possible solution(s) to polynomial equations of degree three or higher algebraically, graphically, and with technology to justify the reasonableness of answer(s). Explain the solution method and interpret solutions in context.

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