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Introduction to Polynomials and the Factor Theorem
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Today, we will explore the factorisation of polynomials. Let's begin by discussing what a polynomial is. Can anyone tell me what defines a polynomial?
A polynomial is an algebraic expression made up of variables and coefficients.
Correct! Now, has anyone heard of the Factor Theorem?
I think it relates to finding roots of polynomials.
Exactly! The Factor Theorem states that if p(a) = 0 for a polynomial p(x), then (x - a) is a factor of p(x). Now, letβs see how this works with an example.
What happens if the polynomial is a constant?
Good question! Non-zero constant polynomials do not have any zeros. Any idea why?
Because replacing x with any number keeps the polynomial the same!
Exactly! In summary, a polynomialβs factors reveal its roots, which we can find using the Factor Theorem.
Applying the Factor Theorem
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Letβs take a polynomial, say p(x) = x^3 + 3x^2 + 5x + 6. How do we find if x + 2 is a factor?
We can substitute -2 into p(x)!
Yes! Let's calculate it. p(-2) equals what?
That would be -8 + 12 - 10 + 6, which equals 0!
Right! So x + 2 is indeed a factor. Let's summarize this: if p(a) = 0, then (x - a) is a factor.
Can every polynomial always be factored this way?
Great inquiry! It works with polynomials of degree higher than 1, but not all polynomials can be factored into linear factors over the reals.
So, we might have to deal with irreducible factors?
Precisely! Understanding these factors aids significantly in polynomial equations.
Splitting the Middle Term
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Now we'll focus on a valuable technique for quadratics, called βsplitting the middle termβ. For the polynomial 6xΒ² + 17x + 5, how can we find two numbers that help in factorisation?
We need two numbers that add to 17 and multiply to 30, right?
Absolutely! What pairs could work for you?
I think 2 and 15 work since 2 + 15 = 17 and 2 Γ 15 = 30.
Great job! So we rewrite the polynomial as 6xΒ² + 2x + 15x + 5. Now letβs factor by grouping. What do we get?
It factors into (3x + 1)(2x + 5)!
Exactly right! Remember, this is a powerful method for factorising quadratic polynomials.
Working with Cubic Polynomials
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Finally, letβs factorise a cubic polynomial, like xΒ³ - 23xΒ² + 142x - 120. What's our approach?
We can start by finding a root!
Exactly! What could we check? We want values that divide -120.
We can try 1, -1, 2, etc.
Yes! Testing p(1) = 0, we see x - 1 is a factor. Now how do we factor out x - 1?
We can use synthetic or polynomial long division!
Fantastic! After dividing, what is our result?
Itβs (x - 1)(xΒ² - 22x + 120). Now we can factor xΒ² - 22x + 120 as well!
Exactly! So the complete factorisation is (x - 1)(x - 10)(x - 12). Well done!
Introduction & Overview
Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.
Quick Overview
Standard
In this section, students learn about the Factor Theorem, which provides a connection between the roots of polynomials and their factors. Various methods for factorising polynomials, notably by splitting the middle term, are also discussed with practical examples to reinforce the concepts.
Detailed
Factorisation of Polynomials
In this section, we delve into the factorisation of polynomials, crucial for simplifying expressions and solving polynomial equations.
Key Concepts Covered:
- Factor Theorem: A polynomial of degree n > 1 has a factor x - a if p(a) = 0, making a a zero of the polynomial. Conversely, if x - a is a factor, then p(a) = 0.
- Example Demonstrations: The section features examples showing the application of the Factor Theorem in identifying factors of polynomials.
- Method of Splitting the Middle Term: For quadratic polynomials of the form axΒ² + bx + c, the middle term can be written as the sum of two numbers whose product gives ac. This method allows us to express the polynomial as the product of two binomials.
Through various examples, such as factorising the polynomial 6xΒ² + 17x + 5 by splitting the middle term into suitable components, students see practical applications of these concepts. Additionally, students learn how to use the Factor Theorem to verify factors of cubic polynomials, encouraging an understanding of polynomial behavior by examining its roots.
Audio Book
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Introduction to Factor Theorem
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Let us now look at the situation of Example 10 above more closely. It tells us that since the remainder, \( q\left( \frac{1}{2} \right) = 0 \), \( (2t + 1) \) is a factor of \( q(t) \), i.e., \( q(t) = (2t + 1) g(t) \) for some polynomial \( g(t) \). This is a particular case of the following theorem.
Factor Theorem: If \( p(x) \) is a polynomial of degree \( n > 1 \) and \( a \) is any real number, then (i) \( x - a \) is a factor of \( p(x) \), if \( p(a) = 0 \), and (ii) \( p(a) = 0 \), if \( x - a \) is a factor of \( p(x) \).
Detailed Explanation
The Factor Theorem helps us understand the relationship between a polynomial and its factors. If replacing \( x \) with a number \( a \) in the polynomial gives a result of \( 0 \), then \( (x - a) \) is a factor of that polynomial. This is a crucial tool for polynomial factorization, as it allows us to determine possible factors by checking various values of \( a \).
Examples & Analogies
Imagine you're baking a cake. If you find out that a specific ingredient (like sugar) is missing (which represents the polynomial equaling zero), you can conclude that you canβt bake the cake (the polynomial doesnβt exist without that factor). Hence, if the polynomial equals zero at some point, it indicates a missing factor, just like missing an essential ingredient.
Applications of Factor Theorem
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Example 6: Examine whether \( x + 2 \) is a factor of \( x^3 + 3x^2 + 5x + 6 \) and of \( 2x + 4 \).
Solution: The zero of \( x + 2 \) is \( -2 \). Let \( p(x) = x^3 + 3x^2 + 5x + 6 \) and \( s(x) = 2x + 4 \)
Then, \( p(-2) = (-2)^3 + 3(-2)^2 + 5(-2) + 6 = -8 + 12 - 10 + 6 = 0 \)
So, by the Factor Theorem, \( x + 2 \) is a factor of \( x^3 + 3x^2 + 5x + 6 \).
Again, \( s(-2) = 2(-2) + 4 = 0 \)
So, \( x + 2 \) is a factor of \( 2x + 4 \).
Detailed Explanation
This example applies the Factor Theorem to test whether \( x + 2 \) is a factor of two polynomials. By substituting \( -2 \) into the polynomials \( p(x) \) and \( s(x) \) and checking if the result is \( 0 \), we confirm that the factorization holds for both polynomials, demonstrating the practical utility of the Factor Theorem.
Examples & Analogies
Think of it like checking if a key opens a lock. If it does, you know that the key (factor) fits perfectly to unlock that specific lock (polynomial). Here, substituting \( -2 \) was like trying the key, and if it turns out to unlock (results in zero), we confirm that the factor indeed works.
Finding Unknown Values
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Example 7: Find the value of \( k \), if \( x - 1 \) is a factor of \( 4x^3 + 3x^2 - 4x + k \).
Solution: As \( x - 1 \) is a factor of \( p(x) = 4x^3 + 3x^2 - 4x + k \), \( p(1) = 0 \)
Now, \( p(1) = 4(1)^3 + 3(1)^2 - 4(1) + k = 4 + 3 - 4 + k = 0 \)
This gives \( k = -3 \).
Detailed Explanation
In this example, we are tasked with finding an unknown value, \( k \), that allows a polynomial to have a specific factor. By using the normal condition that a polynomial equals zero when substituting the root, we solve for \( k \) to ensure the factor, \( x - 1 \), holds true in that polynomial.
Examples & Analogies
Consider it akin to mixing different liquids to achieve a specific compound. If you know that for a reaction to be valid (the polynomial to hold true), certain amounts of ingredients must be used (the specific value of \( k \)), thatβs what you're determining in this process.
Quadratic Polynomial Factorisation
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We will now use the Factor Theorem to factorise some polynomials of degree 2 and 3. You are already familiar with the factorisation of a quadratic polynomial like \( x^2 + lx + m \). You had factorised it by splitting the middle term \( lx \) as \( ax + bx \) so that \( ab = m \). Then \( x^2 + lx + m = (x + a)(x + b) \).
Detailed Explanation
This section clarifies how to factor quadratic polynomials using a method known as splitting the middle term. By identifying two numbers that add up to the coefficient of the middle term and multiply to give the constant term, we can break the polynomial into two linear factors. This is essential for solving quadratic equations and simplifying expressions.
Examples & Analogies
Think of splitting the middle term like breaking a rope at specific points to create equal segments. Each piece (factor) represents part of the whole equation, allowing us to manage it better and find solutions more easily, much like simplifying a task into smaller steps.
Example of Quadratic Factorisation
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Example 8: Factorise \( 6x^2 + 17x + 5 \) by splitting the middle term, and by using the Factor Theorem.
Solution 1: (By splitting method) If we can find two numbers \( p \) and \( q \) such that \( p + q = 17 \) and \( pq = 6 \times 5 = 30 \), then we can get the factors.
So, let us look for the pairs of factors of 30. Some are 1 and 30, 2 and 15, 3 and 10, 5 and 6. Of these pairs, 2 and 15 will give us \( p + q = 17 \).
Therefore, \( 6x^2 + 17x + 5 = 6x^2 + (2 + 15)x + 5 = 6x^2 + 2x + 15x + 5 = 2x(3x + 1) + 5(3x + 1) = (3x + 1)(2x + 5) \).
Detailed Explanation
This example walks through the process of factoring a quadratic polynomial by splitting the middle term. By recognizing that we can separate the middle term into two components that add up to the same value, we can find factors that simplify the polynomial into a product of two binomials, facilitating further calculations.
Examples & Analogies
Imagine you have a large pizza (the quadratic), and you want to share it equally with friends. By dividing it into portions (factors) that are manageable (like splitting the middle term), you ensure everyone gets a fair share (the simplified form), making it easier to handle.
Cubic Polynomial Factorisation
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Now, let us consider factorising cubic polynomials. Here, the splitting method will not be appropriate to start with. We need to find at least one factor first, as you will see in the following example.
Example 10: Factorise \( x^3 - 23x^2 + 142x - 120 \). Solution: Let \( p(x) = x^3 - 23x^2 + 142x - 120 \) We shall now look for all the factors of \( -120 \).
Detailed Explanation
When dealing with cubic polynomials, the process begins by identifying at least one factor. This is different from quadratic polynomials because cubic polynomials can often have multiple roots. By systematically testing possible values, we can find a factor that simplifies our polynomial, leading to further factorization.
Examples & Analogies
Think of it as troubleshooting an engine that has multiple issues (cubic polynomial). Before you can fix everything, you first need to identify at least one problem (factor) to start resolving the issue, making it easier to refine and improve the overall function.
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
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Factor Theorem: A polynomial of degree n > 1 has a factor x - a if p(a) = 0, making a a zero of the polynomial. Conversely, if x - a is a factor, then p(a) = 0.
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Example Demonstrations: The section features examples showing the application of the Factor Theorem in identifying factors of polynomials.
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Method of Splitting the Middle Term: For quadratic polynomials of the form axΒ² + bx + c, the middle term can be written as the sum of two numbers whose product gives ac. This method allows us to express the polynomial as the product of two binomials.
-
Through various examples, such as factorising the polynomial
6xΒ² + 17x + 5by splitting the middle term into suitable components, students see practical applications of these concepts. Additionally, students learn how to use the Factor Theorem to verify factors of cubic polynomials, encouraging an understanding of polynomial behavior by examining its roots.
Examples & Real-Life Applications
See how the concepts apply in real-world scenarios to understand their practical implications.
Examples
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Example 1: Factorise 6xΒ² + 17x + 5 by splitting the middle term into 2x + 15x, resulting in factors (3x + 1)(2x + 5).
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Example 2: Verify if x - 1 is a factor of xΒ³ - 23xΒ² + 142x - 120 by finding p(1) = 0.
Memory Aids
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
π΅ Rhymes Time
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To find a factor and do it right, plug in the root, watch it ignite!
π Fascinating Stories
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Once upon a time, polynomials were shy. They hoped to be split into factors, oh my! One day, a hero found their roots, and together they danced in factor groups.
π§ Other Memory Gems
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FOR FACTORS: Find Out Rational Roots For Factorisation!
π― Super Acronyms
FLT (Factor, Linear, Theorem)
- Remember what FLT stands for in factoring!
Flash Cards
Review key concepts with flashcards.
Glossary of Terms
Review the Definitions for terms.
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Term: Polynomial
Definition:
An algebraic expression that consists of variables and coefficients, combined using addition, subtraction, multiplication, and non-negative integer exponents.
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Term: Factor Theorem
Definition:
A theorem stating that a polynomial p(x) has a factor (x - a) if p(a) = 0.
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Term: Middle Term Splitting
Definition:
A method of factorising quadratic polynomials by rewriting the middle term as a sum of two terms.
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Term: Coefficient
Definition:
A numerical factor in a term of a polynomial.
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Term: Root
Definition:
A value of x that makes the polynomial equal to zero.
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Term: Quadratic Polynomial
Definition:
A polynomial of degree 2, typically expressed in the form axΒ² + bx + c.