MAT 1015 Sections

3.2.1: Algebraic Analysis of Polynomial Functions

Learning Objectives

  • Evaluate function values
  • Describe the meaning of assigning a function a value
  • Perform addition and subtraction of polynomials
  • Perform multiplication of polynomials

Function Values

As we have seen in previous chapters, to evaluate a function, we substitute a value for the independent variable in the function equation and simplify the arithmetic expression. This value is the function value corresponding to the value of the independent variable. For example, to evaluate the value of the function [latex]f(x)=x^3-2x^2+5x-14[/latex] when [latex]x = 1[/latex], we substitute [latex]x[/latex] with the value 1 and simplify:

 [latex]\begin{aligned} f(x)&=x^3-2x^2+5x-14\\f(1) &= (1)^3 - 2(1)^2 + 5(1) - 14 \\f(1)&=1-2+5-14 \\f(1)&= -10\end{aligned}[/latex]

Therefore, the function value when [latex]x = 1[/latex] is –10, or, [latex]f(1) = -10[/latex].

Example 1

For [latex]g(x)=3x^3-x^2+2x-5[/latex], determine the function value,

1. when [latex]x=2[/latex]

2. when [latex]x=-1[/latex]

3. [latex]g(0)[/latex]

4. [latex]g\left (-\frac{1}{2}\right )[/latex]

Solution

Substitute the [latex]x[/latex]-value into the function and simplify.

1.

[latex]\begin{aligned}g(x)&=3x^3-x^2+2x-5\\g(2)&=3(2)^3-(2)^2+2(2)-5\\&=24-4+4-5\\&=19\end{aligned}[/latex]

2.

[latex]\begin{aligned}g(x)&=3x^3-x^2+2x-5\\g(-1)&=3(-1)^3-(-1)^2+2(-1)-5\\&=-3-1-2-5\\&=-11\end{aligned}[/latex]

3.

[latex]\begin{aligned}g(x)&=3x^3-x^2+2x-5\\g(0)&=3(0)^3-(0)^2+2(0)-5\\&=0-0+0-5\\&=-5\end{aligned}[/latex]

4.

[latex]\begin{aligned}g(x)&=3x^3-x^2+2x-5\\g\left (-\frac{1}{2}\right )&=3\left (-\frac{1}{2}\right )^3-\left (-\frac{1}{2}\right )^2+2\left (-\frac{1}{2}\right )-5\\&=-\dfrac{3}{8}-\dfrac{1}{4}-1-5\\&=\dfrac{-3-2-8-40}{8}\\&=-\dfrac{53}{8}\end{aligned}[/latex]

Try It 1

For [latex]f(x)=-x^3-x^2+3x+7[/latex], determine the function value,

1. when [latex]x=2[/latex]

2. when [latex]x=-1[/latex]

3. [latex]f(0)[/latex]

4. [latex]f(-2)[/latex]

The Meaning of Assigning a Function a Value

Assigning a polynomial function a value (e.g., [latex]f(x) = 2[/latex]) has several meanings depending on perspective. First, it means finding the value(s) in the domain that is (are) mapped onto this function value in the range. For example, figure 1 shows the mapping of [latex]x[/latex]-values to [latex]f(x)=x^2[/latex]. If [latex]f(x)=a[/latex], we need to find the value in the domain that results in a value of [latex]a[/latex] for [latex]f(x)[/latex].

mapping of x^2

Figure 1.The meaning of assigning a function a value: mapping

Second, since a function is composed of ordered pairs [latex](x, y)[/latex] (or points [latex](x, y)[/latex] on the coordinate plane), it may be understood as finding the [latex]x[/latex]-coordinate given the value of the [latex]y[/latex]-coordinate on the coordinate plane. For example, figure 2 shows the graph of a polynomial [latex]y=f(x)[/latex]. If [latex]f(x)=1[/latex] then [latex]y=1[/latex] and the corresponding points on the graph when [latex]y=1[/latex] are (–3, 1) and (–1, 1). Consequently, when [latex]f(x)=1[/latex], [latex]x=–1,\;–3[/latex].

(x,y) points on graph

FIgure 2. The meaning of assigning a function a value: coordinate points on a graph

Third, assigning a function a value, [latex]f(x) = 2[/latex] for example, means finding the intersection point between the graph of the function and the horizontal line [latex]y = 2[/latex]. Figure 3 shows the function [latex]f(x)=x^2-4x+5[/latex] and the horizontal line [latex]y=2[/latex]. The parabola and the line intersect at the points (1, 2) and (3, 2). This shows that when [latex]f(x)=2[/latex], the corresponding [latex]x[/latex]-values are 1 and 3.

Graph of parabola intersecting with a horizontal line

Figure 3. The meaning of assigning a function a value: intersection of graph with horizontal line

Example 2

Use Desmos to determine the value(s) of [latex]x[/latex] when [latex]f(x)=-1[/latex] for [latex]f(x)=x^2+6x+7[/latex].

Solution

To determine when [latex]x^2+6x+7=-1[/latex] we look for the intersection point(s) between the graphs [latex]y=x^2+6x+7[/latex] and [latex]y=-1[/latex].

Using Desmos:

graph of parabola and a line

From the graph, the intersection points of the parabola and the line are (–4, –1) and (–2, –1).

So, [latex]f(x)=-1[/latex] when [latex]x=-4,\;-2[/latex].

Try It 2

Use Desmos to determine the value(s) of [latex]x[/latex] when [latex]f(x)=1[/latex] for [latex]f(x)=-x^2-4x-2[/latex].

Addition and Subtraction

We can add and subtract polynomial functions by combining like terms. For example, [latex]5{x}^{2}[/latex] and [latex]-2{x}^{2}[/latex] are like terms and can be added to get [latex]3{x}^{2}[/latex], but [latex]3x[/latex] and [latex]3{x}^{2}[/latex] are not like terms, since they have different exponents, therefore, they cannot be added.

Example 3

Find the sum of the polynomials, [latex]f(x)=12{x}^{2}+9x - 21[/latex] and [latex]g(x)=4{x}^{3}+8{x}^{2}-5x+20[/latex]

Solution

[latex]\begin{aligned}f(x)+g(x)&=\left(12{x}^{2}+9x-21\right)+\left(4{x}^{3}+8{x}^{2}-5x+20\right)\\&=4{x}^{3}+\left(12{x}^{2}+8{x}^{2}\right)+\left(9x - 5x\right)+\left(-21+20\right) \hfill & \text{Combine like terms}.\hfill \\ &=4{x}^{3}+20{x}^{2}+4x - 1\hfill & \text{Simplify}.\hfill \end{aligned}[/latex]

When we subtract polynomials, we need to pay attention to the sign of the terms we are combining. In the following example we will show how to distribute the negative sign to each term of a polynomial that is being subtracted from another.

Example 4

Find the difference: [latex]f(x)-g(x)[/latex] when [latex]7{x}^{4}-{x}^{2}+6x+1[/latex] and [latex]g(x)=5{x}^{3}-2{x}^{2}+3x+2[/latex]

Solution

[latex]\begin{aligned}f(x)-g(x)&=\left(7{x}^{4}-{x}^{2}+6x+1\right)-\left(5{x}^{3}-2{x}^{2}+3x+2\right)\\&=7{x}^4-{x}^2+6x+1-5{x}^3+2{x}^{2}-3x-2\text{ }\hfill & \text{Distribute –1 to all terms being subtracted}.\hfill \\ &=7{x}^{4}-5{x}^{3}+\left(-{x}^{2}+2{x}^{2}\right)+\left(6x - 3x\right)+\left(1 - 2\right)\text{ }\hfill & \text{Combine like terms}.\hfill \\ &=7{x}^{4}-5{x}^{3}+{x}^{2}+3x - 1\hfill & \text{Simplify}.\hfill \end{aligned}[/latex]

Note that finding the difference between two polynomials is the same as adding the opposite of the second polynomial to the first.

Try It 3

1. Find [latex]f(x)+g(x)[/latex] when [latex]f(x)=3x^2-4x+2[/latex] and [latex]g(x)=-3x^2-2x+1[/latex]

2. Find [latex]f(x)-g(x)[/latex] when [latex]f(x)=5x^2-3x+2[/latex] and [latex]g(x)=3x^2-3x-7[/latex]

3. Find [latex]f(x)+g(x)-h(x)[/latex] when [latex]f(x)=3x^3-4x^2+4[/latex], [latex]g(x)=2x^2-4x+7[/latex] and [latex]h(x)=2x^3-8x^2-5x+3[/latex]

Multiplication

Monomials

The most basic polynomial functions to multiply are monomials, i.e. polynomials with just one term. We use the properties of exponents to achieve this. For example, [latex]x \times x=x^2[/latex]. This is because there are two [latex]x[/latex]s multiplied together, and we use the exponent 2 to specify the number of times the same base (e.g., [latex]x[/latex]) is multiplied by itself. If we have [latex]y \times y \times y \times y[/latex], we may write it as [latex]y^4[/latex]; [latex]y[/latex] multiplied by itself 4 times. If we have [latex]x \times x^2 \times x^5[/latex] we can expand each factor to get [latex]x \cdot (x \cdot x) \cdot (x \cdot x \cdot x \cdot x \cdot x)[/latex]. Therefore, there are a total of eight [latex]x[/latex]s that are multiplied together: [latex]x \times x^2 \times x^5=x^8[/latex]. Notice that we can add the exponents: [latex]1+2+5=8[/latex]. This is an example of an exponential property called the multiplicative property of exponents.

Multiplicative property of exponents

For any [latex]x, m,[/latex] and [latex]n\in\mathbb{R}[/latex],

[latex]x^m\cdot x^n=x^{m+n}[/latex]

In other words, to multiply exponential terms with the same base, we keep the common base and add the exponents.

Example 5

Simplify:

1. [latex]x^5\cdot x^7[/latex]

2. [latex]y^4\cdot y^3\cdot y^0[/latex]

3. [latex]a^5\cdot a^2\cdot a^8\cdot a^4[/latex]

4. [latex]x^4\cdot y^2\cdot x^5[/latex]

Solution

1. [latex]x^5\cdot x^7=x^{5+7}=x^{12}[/latex]   Same base, so keep the base and add the exponents.

2. [latex]y^4\cdot y^3\cdot y^0=y^{4+3+0}=y^7[/latex]   Same base, so keep the base and add the exponents. Recall that [latex]y^0=1[/latex] for all non-zero [latex]y[/latex]-values. [latex]0^0[/latex] is not defined.

3. [latex]a^5\cdot a^2\cdot a^8\cdot a^4=a^{5+2+8+4}=a^{19}[/latex]    Same base, so keep the base and add the exponents.

4. [latex]x^4\cdot y^2\cdot x^5=x^{4+5}y^2=x^9y^2[/latex]     Watch the bases. This cannot be simplified further since [latex]x[/latex] and [latex]y[/latex] are different bases.

If the exponential terms have a coefficient other than 1, we multiply the coefficients and multiply the variables.

Example 6

Simplify:

1. [latex]3x^3\cdot 4x^4[/latex]

2. [latex]-5y^2\left(3y^5\right)\left(2y^4\right)[/latex]

3. [latex]7x^4\cdot3y^7\cdot7x^2\cdot\left(-3y^2\right)[/latex]

Solution

1. [latex]3x^3\cdot 4x^4=(3\cdot4)\left(x^3\cdot x^4\right)=12x^7[/latex]

2. [latex]-5y^2\left(3y^5\right)\left(2y^4\right)=(-5\cdot 3\cdot 2)\left(y^2 \cdot y^5 \cdot y^4 \right) = -30y^{11}[/latex]

3.  [latex]7x^4\cdot 3y^7 \cdot 7x^2 \cdot \left(-3y^2\right)=\left(7\cdot 3\cdot 7\cdot(-3) \right)\left(x^{4+2}y^{7+2}\right)=-441x^6y^9[/latex]

Try It 4

Simplify:

1. [latex]x^5\cdot x^9[/latex]

2. [latex]4y^6\cdot5y^3[/latex]

3. [latex]-3x^3\cdot5x^6\cdot\dfrac{4}{9}x[/latex]

4. [latex]-6y^5\left(5x^2\right)\left(y^8\right)(8x)[/latex]

Monomial times a Polynomial

To multiply polynomial functions, we need a way to multiply over addition. The distributive property allows us to multiply each term in the first polynomial by each term in the second polynomial. We then simplify by combining like terms.

We have used the distributive property in the past to multiply a number into an algebraic expression. For example, [latex]\color{blue}{2}(x+7)=\color{blue}{2}\cdot x+\color{blue}{2}\cdot 7=2x+14[/latex]. We say that we distribute the [latex]2[/latex] to each term in the expression.

We can also apply the distributive property to terms. For example, [latex]\color{blue}{3x}(2x-4)=\color{blue}{3x}\cdot 2x-\color{blue}{3x}\cdot 4=6x^2-12x[/latex].

the distributive property of multiplication over addition

[latex]a(b+c)=ab+ac[/latex]

where [latex]a,\;b,\;[/latex]and [latex]c[/latex] are terms.

The following video shows examples of using the distributive property to find the product of monomials (single terms) and polynomials.

Example 7

Find [latex]f(x)\cdot g(x)[/latex] when [latex]f(x)=-3x^2[/latex] and [latex]g(x)=4x^2-3x+2[/latex].

Solution

[latex]\begin{aligned}f(x)\cdot g(x)&=-3x^2 \left (4x^2-3x+2 \right )\\ &=-3x^2(4x^2)-3x^2(-3x)-3x^2(2)\\&=-12x^4+9x^3-6x^2\end{aligned}[/latex]

Try It 5

Find [latex]f(x)\cdot g(x)[/latex] when [latex]f(x)=4x^3[/latex] and [latex]g(x)=-3x^2+4x-5[/latex].

We now expand the distributive property to include polynomials. When multiplying polynomial functions, the distributive property allows us to multiply each term of the first polynomial function by each term of the second. We then add the products together and combine like terms to simplify.

distributive property and polynomials

[latex]\begin{array}{c}\left (a_nx^n+...+a_1x+a_0\right )\left (b_mx^m+...+b_1x+b_0\right )\\=a_nx^n\left (b_mx^m+...+b_1x+b_0\right )+a_{n-1}x^{n-1}\left (b_mx^m+...+b_1x+b_0\right )+...+a_0\left (b_mx^m+...+b_1x+b_0\right )\end{array}[/latex]

Every term in the first polynomial gets multiplied by every term in the second polynomial.

Example 8

Multiply the polynomials:  [latex]f(x)=3x+2[/latex] and [latex]g(x)=3x^2+4x+2[/latex]

Solution

Every term in the binomial gets multiplied onto every term in the trinomial:

[latex]\begin{aligned}f(x)\cdot g(x)&=(\color{blue}{3x}+\color{red}{2})(3x^2+4x+2)\\&=\color{blue}{3x}(3x^2+4x+2)+\color{red}{2}(3x^2+4x+2)\\&=\color{blue}{3x}(3x^2)+\color{blue}{3x}\color{black}{(4x)+}\color{blue}{3x}(2)+\color{red}{2}(3x^2)+\color{red}{2}(4x)+\color{red}{2}(2)\\&=9x^3+12x^2+6x+6x^2+8x+4\\&=9x^3+(12x^2+6x^2)+(6x+8x)+4\\&=9x^3+18x^2+14x+4\end{aligned}[/latex]

No matter how many terms are in the polynomials, we always multiply using the distributive property where every term in the first polynomial is multiplied onto every term in the second polynomial.

Example 9

Multiply the polynomial functions:  [latex]f(x)=4x-3[/latex] and [latex]g(x)=5x-6[/latex]

Solution

Every term in the first binomial gets multiplied onto every term in the second binomial:

[latex]\begin{aligned}f(x)\cdot g(x)&=(4x-3)(5x-6)\\&=\color{blue}{4x}(5x-6)\color{red}{-3}(5x-6)\\&=\color{blue}{4x}(5x)+\color{blue}{4x}(-6)\color{red}{-3}(5x)\color{red}{-3}(-6)\\&=20x^2-24x-15x+18\\&=20x^2-39x+18\end{aligned}[/latex]

Try It 6

Multiply the polynomial functions:  [latex]f(x)=7x-4[/latex] and [latex]g(x)=5x+3[/latex]

As the number of terms in the polynomials increases, the more important it becomes to organize the terms after we distribute. One way to help organize terms, is to use vertical multiplication.

Example 10

Multiply the functions: [latex]f(x)=3x+6[/latex] and [latex]g(x)=5x^{2}+3x+10[/latex]

Solution

Set up the problem in a vertical form, and begin by multiplying [latex]3x+6[/latex] by [latex]+10[/latex]. Place the products underneath, as shown. Now multiply [latex]3x+6[/latex] by 10, and place the two terms beneath the like terms.

[latex]\begin{array}{r}3x+\,\,\,6\,\\\underline{\times\,\,\,\,\,\,5x^{2}+\,\,3x+10}\\+30x+60\,\end{array}[/latex]

Now multiply [latex]3x+6[/latex] by [latex]+3x[/latex]. Notice that [latex]\left(6\right)\left(3x\right)=18x[/latex]; since this term is like [latex]30x[/latex], place it directly beneath it.

[latex]\begin{array}{r}3x\,\,\,\,\,\,+\,\,\,6\,\,\\\underline{\times\,\,\,\,\,\,5x^{2}\,\,\,\,\,\,+3x\,\,\,\,\,\,+10}\\+30x\,\,\,\,\,+60\,\,\\+9x^{2}\,\,\,+18x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Finally, multiply [latex]3x+6[/latex] by [latex]5x^{2}[/latex]. Notice that [latex]30x^{2}[/latex] is placed underneath [latex]9x^{2}[/latex].

[latex]\begin{array}{r}3x\,\,\,\,\,\,+\,\,\,6\,\,\\\underline{\times\,\,\,\,\,\,5x^{2}\,\,\,\,\,\,+3x\,\,\,\,\,\,+10}\\+30x\,\,\,\,\,+60\,\,\\+9x^{2}\,\,\,+18x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+15x^{3}+30x^{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}[/latex]

Now add like terms.

[latex]\begin{array}{r}3x\,\,\,\,\,\,+\,\,\,6\,\,\\\underline{\times\,\,\,\,\,\,5x^{2}\,\,\,\,\,\,+3x\,\,\,\,\,\,+10}\\+30x\,\,\,\,\,+60\,\,\\+9x^{2}\,\,\,+18x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+15x^{3}\,\,\,\,\,\,+30x^{2}\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\+15x^{3}\,\,\,\,\,\,+39x^{2}\,\,\,\,+48x\,\,\,\,\,+60\end{array}[/latex]

The answer is [latex]f(x)\cdot g(x)=15x^{3}+39x^{2}+48x+60[/latex].

Try It 7

FInd the product of [latex]f(x)=3x+2[/latex] and [latex]g(x)=5x^2-3x+1[/latex].

Another way to organize terms is do a hybrid version of the horizontal and vertical methods by writing like terms beneath one another as we distribute horizontally. This collects like terms in the same column so they are easily recognized and combined.

Example 11

Find the product of [latex]f(x)=2x+1[/latex] and [latex]g(x)=3{x}^{2}-x+4[/latex].

Solution

[latex]\begin{array}{ll}f(x)\cdot g(x)&=\left(2x+1\right)\left(3{x}^{2}-x+4\right)& \\ &= \color{blue}{2x}\left(3{x}^{2}-x+4\right)\color{red}{+1}\left(3{x}^{2}-x+4\right)&\text{Distribute }\color{blue}{2x}\text{ and }\color{red}{+1} \\&= 6{x}^{3}-2{x}^{2}+8x\\&\;\;\;\;\;\;\;\;\;\;\,+3{x}^{2}-\;\,x+4&\text{Write like terms under one another}\\&= 6{x}^{3}+{x}^{2}+7x+4&\text{Combine like terms}\end{array}[/latex]

Try It 8

Find the product of [latex]f(x)=4x-1[/latex] and [latex]g(x)=2{x}^{2}-x+3[/latex].

Whichever method we choose to use, we will always end up with the same answer.

The following video shows more examples of multiplying polynomials.

We can multiply polynomial functions using the distributive property no matter how many terms are in the polynomials.

Example 12

Find the product of [latex]f(x)=x^2+4x-1[/latex] and [latex]g(x)=2{x}^{2}-3x+5[/latex].

Solution

[latex]\begin{array}{l}f(x)\cdot g(x)&=\left (\color{red}{x^2}\color{blue}{+4x}\color{green}{-1}\right )\left (2x^2-3x+5\right )\\&=\color{red}{x^2}\left (2x^2-3x+5\right )\color{blue}{+4x}\left (2x^2-3x+5\right )\color{green}{-1}\left (2x^2-3x+5\right )\\&=2x^4-3x^3+5x^2 \\&\;\;\;\;\;\;\;\;\;\;\; +8x^3-12x^2+20x\\&\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-2x^2\,+\;\,3x-5}\\&=2x^4+5x^3-9x^2+23x-5\end{array}[/latex]

Try It 9

Find the product of [latex]f(x)=3x^2+x-2[/latex] and [latex]g(x)=x^2-5x+4[/latex].

 


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