More About Polynomials: Your Ultimate Study Guide!
Hey everyone! Welcome to your study notes for "More about Polynomials". Don't worry if the name sounds a bit intimidating. You've already met polynomials before, and this chapter is all about learning some cool new tricks to work with them.
We'll break down how to divide them (just like you did with numbers in primary school!), discover some amazing shortcuts called the Remainder and Factor Theorems, and learn how to handle fractions made of polynomials. These skills are super important because they are the building blocks for solving more complex problems in maths.
Let's take it one step at a time. You've got this!
1. Division of Polynomials: The "Long Division" Method
Remember long division with numbers? Like dividing 529 by 23? We can do the exact same thing with polynomials! It's a powerful tool to break down a big polynomial into smaller, more manageable parts.
First, what is the Division Algorithm?
This is just a formal way of saying what we already know from long division. When you divide one polynomial by another, you get a result and sometimes a leftover part.
The Rule: Dividend = (Divisor × Quotient) + Remainder
In polynomial language, if we divide a polynomial f(x) by another polynomial g(x), we get:
$$f(x) = g(x) \cdot Q(x) + R(x)$$
Where:
- f(x) is the Dividend (the polynomial being divided).
- g(x) is the Divisor (the polynomial you are dividing by).
- Q(x) is the Quotient (the main result of the division).
- R(x) is the Remainder (the leftover part).
An important rule: The degree of the remainder R(x) must be less than the degree of the divisor g(x).
Step-by-Step Guide to Long Division
Let's divide f(x) = 2x³ + 5x² - 4x - 5 by g(x) = x + 3.
Step 1: Arrange. Write it out in long division format. Make sure both polynomials are in descending order of power. If any terms are missing, add them with a zero coefficient (like 0x²).
Step 2: Divide. Divide the first term of the dividend (2x³) by the first term of the divisor (x).
$$2x³ ÷ x = 2x²$$. This is the first part of your quotient.
Step 3: Multiply. Multiply this result (2x²) by the entire divisor (x + 3).
$$2x² \cdot (x + 3) = 2x³ + 6x²$$. Write this underneath the dividend.
Step 4: Subtract. Subtract this new polynomial from the dividend. Be careful with negative signs!
$$(2x³ + 5x²) - (2x³ + 6x²) = -x²$$
Step 5: Bring Down. Bring down the next term from the dividend (-4x).
Step 6: Repeat! Now repeat the process with the new polynomial (-x² - 4x).
- Divide: $$-x² ÷ x = -x$$
- Multiply: $$-x \cdot (x + 3) = -x² - 3x$$
- Subtract: $$(-x² - 4x) - (-x² - 3x) = -x$$
- Bring Down: the next term, which is -5.
Repeat one last time with -x - 5.
- Divide: $$-x ÷ x = -1$$
- Multiply: $$-1 \cdot (x + 3) = -x - 3$$
- Subtract: $$(-x - 5) - (-x - 3) = -2$$
We can't divide any further because the degree of -2 (which is 0) is less than the degree of x+3 (which is 1). So, -2 is our remainder!
Result: The Quotient is 2x² - x - 1 and the Remainder is -2.
So, we can write: 2x³ + 5x² - 4x - 5 = (x + 3)(2x² - x - 1) - 2.
Key Takeaway for Section 1
Long division for polynomials follows the same "Divide, Multiply, Subtract, Bring Down" pattern as number division. It's a reliable method to find the quotient and remainder. Always remember to put in placeholders (like 0x) for missing terms!
2. The Remainder Theorem: A Clever Shortcut!
Doing long division can be... well, long! What if you only need to find the remainder? The Remainder Theorem is here to save the day. It's an amazing shortcut.
How does it work?
The theorem states:
When a polynomial f(x) is divided by a linear polynomial (x - a), the remainder is simply f(a).
Why? It's simple! From the division algorithm, we have f(x) = (x - a)Q(x) + R. If we plug in x = a, the first part becomes (a - a)Q(a) which is 0! So we are left with f(a) = R. Genius!
Let's Try It!
Let's use our previous example: Find the remainder when f(x) = 2x³ + 5x² - 4x - 5 is divided by x + 3.
- Find 'a'. Our divisor is x + 3. To match the form (x - a), we write it as x - (-3). So, a = -3.
- Calculate f(a). We need to find f(-3). Substitute x = -3 into the polynomial:
f(-3) = 2(-3)³ + 5(-3)² - 4(-3) - 5
f(-3) = 2(-27) + 5(9) + 12 - 5
f(-3) = -54 + 45 + 12 - 5
f(-3) = -2
The remainder is -2. Look at that! It's the same result we got from long division, but much faster.
Quick Review: What if the divisor is (ax - b)?
It's the same idea. Just find the value of x that makes the divisor zero.
For ax - b = 0, we have x = b/a.
So, the remainder when f(x) is divided by (ax - b) is f(b/a).
Key Takeaway for Section 2
To find the remainder when dividing f(x) by (x - a), just calculate f(a). No long division needed! This is one of the most useful tricks in this chapter.
3. The Factor Theorem: Finding the Building Blocks
The Factor Theorem is a special case of the Remainder Theorem. What happens if the remainder is 0? It means the divisor goes in perfectly! This makes it a factor.
Analogy: When you divide 12 by 3, the remainder is 0. This means 3 is a factor of 12. It's the exact same idea for polynomials.
The Theorem in a Nutshell
The theorem has two parts:
- If f(a) = 0, then (x - a) is a factor of the polynomial f(x).
- If (x - a) is a factor of f(x), then f(a) = 0.
How to Use the Factor Theorem
This theorem is fantastic for factorising higher-degree polynomials.
Example: Fully factorise f(x) = x³ + 2x² - 5x - 6.
Step 1: Find a possible factor. We need to find a number 'a' that makes f(a) = 0.
Pro Tip: Try the integer factors of the constant term (which is -6). The possible values for 'a' are ±1, ±2, ±3, ±6.
Let's test some:
- f(1) = (1)³ + 2(1)² - 5(1) - 6 = 1 + 2 - 5 - 6 = -8. Not zero.
- f(-1) = (-1)³ + 2(-1)² - 5(-1) - 6 = -1 + 2 + 5 - 6 = 0. Yes!
Since f(-1) = 0, we know that (x - (-1)), which is (x + 1), is a factor!
Step 2: Find the other factors. Now that we have one factor, we can use long division to find the rest. Divide the original polynomial by the factor we just found.
Divide (x³ + 2x² - 5x - 6) by (x + 1).
(If you do the long division, you'll get a quotient of x² + x - 6 and a remainder of 0).
So, f(x) = (x + 1)(x² + x - 6).
Step 3: Factorise the remaining quadratic. Now we just need to factorise x² + x - 6. This is easy!
x² + x - 6 = (x + 3)(x - 2)
Final Answer: f(x) = (x + 1)(x + 3)(x - 2). We've fully factorised it!
Key Takeaway for Section 3
The Factor Theorem is your key to breaking down big polynomials. If f(a) = 0, then (x - a) is a factor. Test factors of the constant term to find your first factor, then use long division.
4. H.C.F. & L.C.M. for Polynomials
You've found the Highest Common Factor (H.C.F.) and Lowest Common Multiple (L.C.M.) for numbers. Now, we'll do it for polynomials. This skill is essential for working with polynomial fractions.
Quick Review: H.C.F. & L.C.M. of Numbers
Let's find the H.C.F. and L.C.M. of 12 and 18.
1. Factorise: 12 = 2² × 3¹ and 18 = 2¹ × 3²
2. H.C.F: Take the common factors with the lowest powers. Common factors are 2 and 3. Lowest powers are 2¹ and 3¹. H.C.F. = 2 × 3 = 6.
3. L.C.M: Take all factors with the highest powers. All factors are 2 and 3. Highest powers are 2² and 3². L.C.M. = 2² × 3² = 36.
We use the exact same logic for polynomials!
Step-by-Step Guide for Polynomials
Example: Find the H.C.F. and L.C.M. of the polynomials:
P(x) = x² - 9
Q(x) = x² - x - 12
Step 1: Factorise everything completely! This is always the first step.
P(x) = (x - 3)(x + 3)
Q(x) = (x - 4)(x + 3)
Step 2: Find the H.C.F. Look for the factors they have in common.
Both have the factor (x + 3). That's it.
So, H.C.F. = (x + 3).
Step 3: Find the L.C.M. List all unique factors and take the highest power of each one.
The unique factors are (x - 3), (x + 3), and (x - 4). The highest power of each is 1.
So, L.C.M. = (x - 3)(x + 3)(x - 4).
Key Takeaway for Section 4
It's all about factorising first! Then remember these simple rules:
H.C.F. = Product of COMMON factors with their LOWEST powers.
L.C.M. = Product of ALL factors with their HIGHEST powers.
5. Working with Rational Functions (Polynomial Fractions)
A rational function is just a fancy name for a fraction with a polynomial on top and a polynomial on the bottom. Now we'll use our H.C.F. and L.C.M. skills to simplify, add, subtract, multiply, and divide them.
Simplifying, Multiplying, and Dividing
The golden rule for all of these is: FACTORISE FIRST, THEN CANCEL!
Example (Simplifying): Simplify $$(x² - 9) / (x² - x - 12)$$
$$ \frac{(x - 3)(x + 3)}{(x - 4)(x + 3)} = \frac{(x - 3)\cancel{(x + 3)}}{(x - 4)\cancel{(x + 3)}} = \frac{x - 3}{x - 4} $$
Example (Multiplying): Calculate $$ \frac{x+2}{x-5} \cdot \frac{x^2 - 25}{x^2 - 4} $$
$$ \frac{x+2}{x-5} \cdot \frac{(x-5)(x+5)}{(x-2)(x+2)} = \frac{\cancel{(x+2)}}{\cancel{(x-5)}} \cdot \frac{\cancel{(x-5)}(x+5)}{(x-2)\cancel{(x+2)}} = \frac{x+5}{x-2} $$
Example (Dividing): Calculate $$ \frac{x+1}{x-2} \div \frac{x^2+3x+2}{x^2-4} $$
Remember: Keep, Change, Flip!
$$ \frac{x+1}{x-2} \cdot \frac{x^2-4}{x^2+3x+2} = \frac{x+1}{x-2} \cdot \frac{(x-2)(x+2)}{(x+1)(x+2)} = 1 $$
Adding and Subtracting
Just like with number fractions, you need a common denominator. And what's the best common denominator? The L.C.M.!
Example: Calculate $$ \frac{3}{x-4} - \frac{2}{x+3} $$
Step 1: Find the L.C.M. of the denominators. The denominators are already factorised. The L.C.M. of (x-4) and (x+3) is simply (x-4)(x+3).
Step 2: Rewrite each fraction with the common denominator.
$$ \frac{3(x+3)}{(x-4)(x+3)} - \frac{2(x-4)}{(x-4)(x+3)} $$
Step 3: Combine the numerators. Be very careful with the subtraction sign!
$$ \frac{3(x+3) - 2(x-4)}{(x-4)(x+3)} = \frac{3x+9 - 2x+8}{(x-4)(x+3)} $$
Step 4: Simplify the numerator.
$$ \frac{x+17}{(x-4)(x+3)} $$
This is our final answer. Don't expand the denominator unless you're asked to.
Did you know?
The word "rational" in rational functions comes from the word "ratio", because it's a ratio of two polynomials. It has nothing to do with the numbers being "reasonable" or "logical"!
Key Takeaway for Section 5
Working with rational functions is just like working with regular fractions.
- For multiplication and division: Factorise and cancel.
- For addition and subtraction: Find the L.C.M. to get a common denominator, then combine.