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4. Polynomial Functions & Expressions
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United States of AmericaTennessee
Algebra 2

4.05 Polynomial equations and inequalities

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Concept summary

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 p_nx^n+p_{n-1}x^{n-1}+p_{n-2}x^{n-2}+\ldots+p_2x^2+p_1x+p_0=0 where n is a positive integer and p_n,p_{n-1},p_{n-2},\ldots,p_2,p_1,p_0 are real numbers.

We can find the roots of this polynomial equation by factoring. The maximum number of roots will be n for a polynomial of degree n.

Rational Roots Theorem

Given P\left(x\right)=p_nx^n+p_{n-1}x^{n-1}+p_{n-2}x^{n-2}+\ldots+p_2x^2+p_1x+p_0, where the coefficients are all integers. If P\left(x\right) has any rational roots then they must be of the form \dfrac{p}{q} where p is a factor of the constant term p_0 and q is a factor of the leading coefficient p_n.

When a polynomial expression is factored we can use the zero product property that if A\cdot B \cdot C \cdot D=0, then A=0, B=0, C=0, or D=0.

When solving a polynomial inequality, we can consider the sign (positive, negative, zero) of each factor individually to determine the sign of the product. For example, we can make a sign table for (x+3)(x-1)(x-5)>0:

x<-3x=-3-3<x<111<x<55x>5
(x+3)-0+++++
(x-1)---0+++
(x-5)-----0+
(x+3)(x-1)(x-5)-0+0-0+

So: (x+3)(x-1)(x-5)>0 is satisfied where (x+3)(x-1)(x-5) is positive, which is when x \in \left(-3,1\right) \cup \left(5, \infty\right).

Alternatively, we sketch a graph of the related polynomial function using the zeros and the leading coefficient to determine where it is positive, negative, or zero.

x
y

So: (x+3)(x-1)(x-5)>0 is satisfied where (x+3)(x-1)(x-5) is positive, which is when x \in \left(-3,1\right) \cup \left(5, \infty\right).

Worked examples

Example 1

Solve the equation \left(2x-9\right)\left(x^2-3x-10\right)=0.

Approach

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 factoring techniques.

Solution

The first factor is already linear, so we just need to try to write the quadratic factor as a product of linear factors.

To factor x^2-3x-10, we want to find two number that when multiplied give -10 and when added give -3.

The numbers -5 and 2 satisfy these conditions, so we can factor the quadratic as x^2-3x-10=(x-5)(x+2).

The equation \left(2x-9\right)\left(x^2-3x-10\right)=0 becomes (2x-9)(x-5)(x+2)=0

The Zero Product Property states that if A \cdot B \cdot C =0, then A =0, B=0, or C=0.

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

Reflection

We can check the answer by substituting each solution to \left(2x-9\right)\left(x^2-3x-10\right)=0.

  • \left(2\left(\dfrac{9}{2}\right)-9\right)\left(\left(\dfrac{9}{2}\right)^2-3\left(\dfrac{9}{2}\right)-10\right)=0 \cdot \left(\left(\dfrac{9}{2}\right)^2-3\left(\dfrac{9}{2}\right)-10\right)=0
  • \left(2\left(5\right)-9\right)\left(\left(5\right)^2-3\left(5\right)-10\right)=\left(2\left(5\right)-9\right)\cdot 0=0
  • \left(2\left(-2\right)^2-9\right)\left(\left(-2\right)^2-3\left(-2\right)-10\right)=\left(2\left(-2\right)-9\right) \cdot 0=0

This confirms that, x=\dfrac{9}{2},x=5,x=-2 are the solutions of \left(2x-9\right)\left(x^2-3x-10\right)=0.

Example 2

Solve the inequality 4x^3+9x>12x^2.

Approach

In order to use the Zero Product Property, the polynomial expression must be set equal to 0 so first we need to rearrange the inequality.

Solution

Subtract 12x^2 from both sides to get: 4x^3-12x^2+9x>0

We can proceed by factoring the left-hand side:

\displaystyle 4x^3-12x^2+9x\displaystyle >\displaystyle 0Given inequality in standard form
\displaystyle x\left(4x^2-12x+9\right)\displaystyle >\displaystyle 0Common factor of x
\displaystyle x\left((2x)^2-2(2x)(3)+3^2\right)\displaystyle >\displaystyle 0Rewrite the quadratic to show it is a perfect square trinomial
\displaystyle x(2x-3)^2\displaystyle >\displaystyle 0Factor the perfect square trinomial

Now we can use our knowledge of graphing polynomials to see where x(2x-3)^2>0.

x
y

To graph y=x(2x-3)^2:

  1. Find the x-intercepts:

    Solve x(2x-3)^2=0 to get x=0 with multiplicity 1 and 2x-3=0 \to x=\dfrac{3}{2} with multiplicity 2.

  2. Identify the end behavior: The leading coefficient is 4 which is positive, so it will go up and to the right.
  3. Use multiplicity: Since x=\frac{3}{2} has multiplicity 2 the graph will "touch" the axis without crossing.

From this graph we can see that the function values are positive when: x \in \left(0,\dfrac{3}{2}\right) \cup \left(\dfrac{3}{2}, \infty\right)

Reflection

We could also determine the values of x which satisfy this inequality using a table and see the same result:

x<0x=00<x<\dfrac{3}{2}x=\dfrac{3}{2}x>\dfrac{3}{2}
x-0+++
(2x-3)---0+
(2x-3)---0+
x(2x-3)^2-0+0+

Example 3

A toy car manufacturer produces x units per month. The monthly cost for the toy car is given by \\C(x) = 4 x^{2}+5x The monthly gross profit is given by G(x)=x^{3}+3x^2-8x+7

The cost C(x) and gross profit G(x) are both in dollars.

a

Form an expression for P(x), the net profit after producing and selling x units.

Approach

To find the expression for P(x), we first determine the relationship of the net profit to the gross profit and cost.

Solution

The net profit is obtained by subtracting the cost from the gross profit. That is, P(x)=G(x)-C(x) By substituting the given expressions we get the equation: P(x)=\left(x^{3}+3x^2-8x+7\right)-\left(4 x^{2}+5x\right)=x^3-x^2-13x+7

Therefore, P(x)=x^3-x^2-13x+7

b

Find the number of units that must be sold in order to make a profit of \$2962.

Approach

We first substitute the profit into the net profit equation, P\left(x\right), obtained in part (a). Then we solve the new equation for the positive real value of x since the number of units can never take negative values.

Solution

The net profit equation is given by P(x)=x^3-x^2-13x+7

To find the number of units that must be sold to earn a profit of \$2962, we set P(x)=2962 and solve for x. So we have x^3-x^2-13x+7=2962

We convert the equation into standard form by subtracting 2962 from both sides of the equation. x^3-x^2-13x-2955=0

Using the Rational Roots Theorem, we can identify potential roots of the equation in the form \dfrac{p}{q} where p is a factor of the constant term -2955 and q is a factor of the leading coefficient 1. Since q=1, we just need to look at p.

The possible values of p are \pm1,\pm3,\pm5,\pm15,\pm197,\pm591,\pm985, and \pm2955.

Recall the Factor Theorem states that, if P\left(c\right)=0 then \left(x-c\right) is a factor of P\left(x\right). So we can test each of the possible roots by substituting them into the equation for x. If the equation evaluates to 0 then we have found a root and therefore a factor of P\left(x\right).

Substituting 15 into the equation for x gives:

\displaystyle \left(15\right)^3-\left(15\right)^2-13\left(15\right)-2955\displaystyle =\displaystyle 0Substitution
\displaystyle 0\displaystyle =\displaystyle 0Evaluate

This shows that 15 is a root of the equation and therefore \left(x-15\right). Is a factor of the equation.

Using long or synthetic division, we can express the left-hand side of the equation as a product of two factors. That is, (x-15)\left(x^2+14x+197\right)=0

From here we can say that: x-15=0 \text{ or }x^2+14x+197=0

x-15=0 gives the solution x=15.

Since x^2+14x+197=0 is not factorable, we use quadratic formula to solve for the solution of x^2+14x+197=0.

\displaystyle x\displaystyle =\displaystyle \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}Quadratic formula
\displaystyle x\displaystyle =\displaystyle \dfrac{-14\pm\sqrt{\left(14\right)^2-4(1)(197)}}{2(1)}Substitute a=1,b=14 and c=197
\displaystyle x\displaystyle =\displaystyle -7\pm 2\sqrt{-37}Simplify

However, because -7+ 2\sqrt{-37} would not be a real solution, we can see that only the solution x=15 is viable since it is a positive real solution and x represents the number of units sold.

Therefore, 15 units of toy cars must be sold to make a profit of \$2962.

Outcomes

A2.N.Q.A.1

Use units as a way to understand real-world problems.*

A2.N.Q.A.1.B

Use appropriate quantities in formulas, converting units as necessary.

A2.A.APR.A.2

Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.

A2.A.CED.A.1

Create equations and inequalities in one variable and use them to solve problems in a real-world context.*

A2.A.CED.A.2

Create equations in two variables to represent relationships between quantities and use them to solve problems in a real-world context. Graph equations with two variables on coordinate axes with labels and scales, and use the graphs to make predictions.*

A2.F.IF.A.3

Understand geometric formulas as functions.*

A2.MP1

Make sense of problems and persevere in solving them.

A2.MP2

Reason abstractly and quantitatively.

A2.MP3

Construct viable arguments and critique the reasoning of others.

A2.MP4

Model with mathematics.

A2.MP6

Attend to precision.

A2.MP7

Look for and make use of structure.

A2.MP8

Look for and express regularity in repeated reasoning.

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