Zeroes of Polynomial Functions
A well-defined collection of distinct objects called elements or members.
Learning Outcomes
Identify zeros of a polynomial function. Demonstrate the methods to factorize polynomials. Find the zeros and X-intercepts of polynomial functions by factoring.
Exercise Questions
Good evening! Here in India on this Sunday, let’s work through these questions. They cover the very important and related concepts of a polynomial’s factors, its roots (or zeros), and its intercepts on a graph.
Core Concepts: Factors, Roots, and Intercepts
Roots / Zeros / x-intercepts: These three terms are used almost interchangeably.
- The roots or zeros of a polynomial $p(x)$ are the values of $x$ that make the polynomial equal to zero (i.e., where $p(x) = 0$).
- These are also the x-coordinates of the x-intercepts, which are the points where the graph of the polynomial crosses the x-axis.
The Factor Theorem: This theorem creates a direct link between roots and factors. It states that if $c$ is a root of a polynomial $p(x)$, then $(x-c)$ must be a factor of that polynomial.
The y-intercept: This is the point where the graph crosses the y-axis. It is always found by setting $x=0$ and calculating the value of the polynomial, $p(0)$. A function can have only one y-intercept.
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Question 1: Complete Factorization (from file image_d11dde.png)
The Question: Let $p(x) = 5x^3 - 50x^2 + 155x - 150$ be a polynomial and $(x-2)$ be a factor of $p(x)$. Which of the following is the complete factorized form of the given Polynomial $p(x)$?
Core Concept: If we know one factor of a polynomial, we can find the others by dividing the polynomial by the known factor. The result of the division (the quotient) will be the remaining factor(s).
Detailed Solution:
Factor out the common constant: Notice that all coefficients are divisible by 5. Let’s factor that out first to make the numbers smaller. $p(x) = 5(x^3 - 10x^2 + 31x - 30)$
Divide by the known factor: We are given that $(x-2)$ is a factor. We will divide the cubic part, $x^3 - 10x^2 + 31x - 30$, by $(x-2)$. We can use synthetic division for this.
2 | 1 -10 31 -30 | 2 -16 30 -------------------- 1 -8 15 0The remainder is 0 (as expected since it’s a factor), and the quotient is a quadratic: $1x^2 - 8x + 15$.
Factor the remaining quadratic: Now we need to factor $x^2 - 8x + 15$. We look for two numbers that multiply to +15 and add to -8. These numbers are -3 and -5.
- So, $x^2 - 8x + 15 = (x-3)(x-5)$.
Combine all factors: The complete factorization is the constant we factored out first, plus all the linear factors we found.
- $p(x) = 5(x-2)(x-3)(x-5)$.
Final Answer: $5(x-3)(x-2)(x-5)$ {{< /border >}}
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Question 2: Finding the Zeros of a Polynomial (from file image_d11af5.png)
The Question: Zeros of the polynomial $2x^3 - 18x^2 - 24x + 40$ is(are) __________.
Core Concept: A “zero” of a polynomial is a value of $x$ that makes the polynomial equal to 0. Since the options are provided, the quickest way to solve this is to test each potential zero using the Remainder Theorem: if $p(c)=0$, then $c$ is a zero.
Detailed Solution:
Let $p(x) = 2x^3 - 18x^2 - 24x + 40$. Let’s test the values given in the accepted answers.
Test $x = 1$: $p(1) = 2(1)^3 - 18(1)^2 - 24(1) + 40 = 2 - 18 - 24 + 40 = -16 - 24 + 40 = -40 + 40 = 0$. Since $p(1)=0$, 1 is a zero.
Test $x = 10$: $p(10) = 2(10)^3 - 18(10)^2 - 24(10) + 40 = 2(1000) - 18(100) - 240 + 40 = 2000 - 1800 - 240 + 40 = 200 - 200 = 0$. Since $p(10)=0$, 10 is a zero.
Test $x = -2$: $p(-2) = 2(-2)^3 - 18(-2)^2 - 24(-2) + 40 = 2(-8) - 18(4) + 48 + 40 = -16 - 72 + 88 = -88 + 88 = 0$. Since $p(-2)=0$, -2 is a zero.
Final Answer: The zeros are 10, 1, and -2. {{< /border >}}
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Question 3: Identifying Factors (from file image_d11a7a.png)
The Question: Let $p(x) = (x^2 + x + 1)(x+1)x$, choose the set of correct options. (Multiple Select Question)
Core Concept: The factors of a polynomial are the expressions that are multiplied together to create it. We can classify these factors by their number of terms (monomial, binomial, trinomial).
Detailed Solution:
The polynomial is already given to us in its factored form: $p(x) = (x^2 + x + 1)(x + 1)x$. Let’s look at each of the expressions being multiplied:
- $x$: This factor has one term. It is a monomial.
- $(x + 1)$: This factor has two terms. It is a binomial.
- $(x^2 + x + 1)$: This factor has three terms. It is a trinomial.
Now we can evaluate the options:
- “The monomial factor of $p(x)$ is $x$.” -> TRUE.
- “The binomial factor of $p(x)$ is $(x + 1)$.” -> TRUE.
- “The trinomial factor of $p(x)$ is $(x^2 + x + 1)$.” -> TRUE.
Final Answer: The correct options are:
- The monomial factor of $p(x)$ is $x$.
- The binomial factor of $p(x)$ is $(x + 1)$.
- The trinomial factor of $p(x)$ is $(x^2 + x + 1)$. {{< /border >}}
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Question 4: Finding x-intercepts (from file image_d11779.png)
The Question: Let $p(x) = x^3 + 2x^2 + 5x + 10$ be a polynomial and S be the set of x-intercepts, choose the set of correct options.
Core Concept: The x-intercepts are the real roots of the polynomial. For a four-term polynomial like this, a good strategy is to try factoring by grouping.
Detailed Solution:
- Set the polynomial to zero to find the roots: $x^3 + 2x^2 + 5x + 10 = 0$
- Group the first two terms and the last two terms: $(x^3 + 2x^2) + (5x + 10) = 0$
- Factor out the greatest common factor from each group:
- From the first group, factor out $x^2$: $x^2(x + 2)$
- From the second group, factor out $5$: $5(x + 2)$
- Rewrite the equation: $x^2(x + 2) + 5(x + 2) = 0$
- Factor out the common binomial factor, $(x+2)$: $(x + 2)(x^2 + 5) = 0$
- Solve for x from each factor:
- First factor: $x + 2 = 0 \implies x = -2$. This is a real root.
- Second factor: $x^2 + 5 = 0 \implies x^2 = -5$. There is no real number whose square is negative, so this factor gives no real roots.
- Determine the set S and its cardinality:
- The only real root is -2. Therefore, the set of x-intercepts is $S = {-2}$.
- The cardinality of S (the number of elements in it) is 1.
Final Answer: The correct statements are:
- $-2 \in S$
- The cardinality of the set S is 1. {{< /border >}}
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Questions 5, 6, & 7: Intercepts of a Factored Polynomial (from file image_d116a0.png)
The Questions: Let $p(x) = (x^9 + 5)(x^2 + 4)$ be a polynomial and S is the set of x-intercepts and R is the set of y-intercepts. Answer the following questions. 5) Find the cardinality of the set S. 6) Find the y-intercept. 7) Find the cardinality of the set R.
Detailed Solution:
5) Find the cardinality of the set S (x-intercepts).
- Find x-intercepts by setting p(x) = 0: $(x^9 + 5)(x^2 + 4) = 0$
- Set each factor to zero and solve:
- First factor: $x^9 + 5 = 0 \implies x^9 = -5$. This gives one real solution, $x = \sqrt[9]{-5}$ (the odd root of a negative number is a real number).
- Second factor: $x^2 + 4 = 0 \implies x^2 = -4$. This has no real solutions.
- Conclusion: There is only one real root, so there is only one x-intercept. The set is $S = {\sqrt[9]{-5}}$.
- The cardinality of S is the number of elements in it, which is 1.
Note: The accepted answer in your image is “0”. This is incorrect based on the standard definition of polynomials over real numbers. The equation $x^9=-5$ has one real solution. The provided answer key is likely flawed.
My Answer: The cardinality of the set S is 1.
6) Find the y-intercept.
- Find the y-intercept by setting x = 0: $p(0) = ((0)^9 + 5)((0)^2 + 4)$
- Calculate the result: $p(0) = (0 + 5)(0 + 4) = (5)(4) = 20$.
Final Answer: The y-intercept is 20.
7) Find the cardinality of the set R (y-intercepts).
- A function can only cross the y-axis once. If it crossed more than once, it would fail the vertical line test and would not be a function.
- We found the single y-intercept to be 20. The point is $(0, 20)$.
- The set of y-intercepts is $R = { (0, 20) }$.
- The cardinality of R is the number of elements in it.
Final Answer: The cardinality of the set R is 1. {{< /border >}}