A polynomial in variable x is an expression of the form:
$$p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$$
where \(a_0, a_1, \ldots, a_n\) are real coefficients and \(n\) is a non-negative integer (the degree if \(a_n \neq 0\)).
Rule: Combine like terms (same power of x).
Example: \((3x^3 - 2x + 5) + (x^3 + 4x^2 - 7)\)
Group by degree:
Result: \(4x^3 + 4x^2 - 2x - 2\)
Rule: Use the distributive property (every term times every term), then combine like terms.
Example: \((2x - 3)(x^2 + 4x - 1)\)
Distribute each term from the first factor:
Combine: \(2x^3 + 8x^2 - 3x^2 - 2x - 12x + 3 = 2x^3 + 5x^2 - 14x + 3\)
Memorize these — they appear constantly:
| Pattern | Formula |
|---|---|
| Square of a sum | \((a + b)^2 = a^2 + 2ab + b^2\) |
| Square of a difference | \((a - b)^2 = a^2 - 2ab + b^2\) |
| Difference of squares | \((a + b)(a - b) = a^2 - b^2\) |
| Cube of a sum | \((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\) |
| Cube of a difference | \((a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\) |
Example: Expand \((x + 2)^3\)
Using the formula with \(a = x\), \(b = 2\):
\((x + 2)^3 = x^3 + 3(x^2)(2) + 3(x)(4) + 8 = x^3 + 6x^2 + 12x + 8\)
Long Division: Used when dividing by polynomials of degree ≥ 2 or when you need both quotient and remainder.
Example: Divide \(x^3 - 2x^2 + 4\) by \(x - 2\)
$$\frac{x^3 - 2x^2 + 4}{x - 2}$$
Step 1: \(x^3 \div x = x^2\). Multiply: \(x^2(x-2) = x^3 - 2x^2\). Subtract: \(0\).
Step 2: Bring down: \(0x\). \(0 \div x = 0\). Multiply: \(0\). Subtract: \(0\).
Step 3: Bring down: \(4\). \(4 \div x = ?\) — can't divide evenly.
Result: \(x^2 + \frac{4}{x-2}\)
Synthetic Division: A shortcut when dividing by \((x - c)\).
Example: Divide \(x^3 - 2x^2 + 4\) by \(x - 2\) using synthetic division:
2 │ 1 -2 0 4
│ 2 0 0
─────────────────────
1 0 0 4
Quotient: \(x^2 + 0x + 0 = x^2\), Remainder: \(4\)
So: \(\frac{x^3 - 2x^2 + 4}{x - 2} = x^2 + \frac{4}{x-2}\)
Factoring is the reverse of multiplication — expressing a polynomial as a product of simpler polynomials. This skill is essential for simplifying expressions, solving equations, and computing limits.
Greatest Common Factor (GCF): Factor out the largest expression that divides every term.
Example: Factor \(6x^4 - 9x^3 + 15x^2\)
GCF of coefficients: \(\gcd(6, 9, 15) = 3\)
GCF of variables: \(x^2\) (lowest power)
Factor: \(3x^2(2x^2 - 3x + 5)\)
When to use: Four or more terms that can be grouped into pairs with common factors.
Example: Factor \(x^3 + 2x^2 - 3x - 6\)
Group: \((x^3 + 2x^2) + (-3x - 6)\)
Factor each group: \(x^2(x + 2) - 3(x + 2)\)
Factor out \((x + 2)\): \((x + 2)(x^2 - 3)\)
Case 1: \(a = 1\) (Monic trinomials)
Find two numbers that multiply to \(c\) and add to \(b\).
Example: Factor \(x^2 + 5x + 6\)
Need: product = 6, sum = 5
Numbers: 2 and 3
Factor: \((x + 2)(x + 3)\)
Case 2: \(a \neq 1\) (Non-monic trinomials)
AC Method: Find two numbers that multiply to \(ac\) and add to \(b\), then factor by grouping.
Example: Factor \(6x^2 + 11x - 10\)
Here \(a = 6\), \(b = 11\), \(c = -10\), so \(ac = -60\)
Need: product = -60, sum = 11
Numbers: 15 and -4
Rewrite middle term: \(6x^2 + 15x - 4x - 10\)
Group: \((6x^2 + 15x) + (-4x - 10)\)
Factor: \(3x(2x + 5) - 2(2x + 5)\)
Result: \((2x + 5)(3x - 2)\)
| Pattern | Formula |
|---|---|
| Difference of squares | \(a^2 - b^2 = (a + b)(a - b)\) |
| Perfect square trinomial | \(a^2 + 2ab + b^2 = (a + b)^2\) |
| Perfect square trinomial | \(a^2 - 2ab + b^2 = (a - b)^2\) |
| Sum of cubes | \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\) |
| Difference of cubes | \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\) |
Example: Factor \(x^3 - 8\)
Recognize: \(8 = 2^3\), so this is a difference of cubes.
Apply formula with \(a = x\), \(b = 2\):
\(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\)
Verification: \((x - 2)(x^2 + 2x + 4) = x^3 + 2x^2 + 4x - 2x^2 - 4x - 8 = x^3 - 8\) ✓
Example: Factor \(x^4 - 16\)
This is a difference of squares: \(x^4 - 16 = (x^2)^2 - 4^2 = (x^2 + 4)(x^2 - 4)\)
But \(x^2 - 4\) factors further: \(x^2 - 4 = (x + 2)(x - 2)\)
Final: \(x^4 - 16 = (x^2 + 4)(x + 2)(x - 2)\)
Note: \(x^2 + 4\) doesn't factor over ℝ.
Sometimes an expression has a repeated structure. Substitute to reveal a simpler pattern.
Example: Factor \(x^4 - 5x^2 + 4\)
Let \(u = x^2\). Then: \(u^2 - 5u + 4\)
Factor: \((u - 4)(u - 1)\)
Substitute back: \((x^2 - 4)(x^2 - 1)\)
Factor further: \((x + 2)(x - 2)(x + 1)(x - 1)\)
A rational expression is a ratio of two polynomials: \(\frac{P(x)}{Q(x)}\) where \(Q(x) \neq 0\).
The expression is undefined wherever the denominator equals zero.
Example: Find the domain of \(\frac{x + 3}{x^2 - 4}\)
Set denominator = 0: \(x^2 - 4 = 0 \Rightarrow x = \pm 2\)
Domain: All real numbers except \(x = 2\) and \(x = -2\), or \(\mathbb{R} \setminus \{-2, 2\}\)
Step 1: Factor numerator and denominator completely.
Step 2: Cancel common factors.
Example: Simplify \(\frac{x^2 - 9}{x^2 + 5x + 6}\)
Factor numerator: \(x^2 - 9 = (x + 3)(x - 3)\)
Factor denominator: \(x^2 + 5x + 6 = (x + 2)(x + 3)\)
Cancel \((x + 3)\): \(\frac{(x + 3)(x - 3)}{(x + 2)(x + 3)} = \frac{x - 3}{x + 2}\), valid for \(x \neq -3, -2\)
Multiplication: \(\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}\) (factor and cancel first)
Division: \(\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C}\)
Addition/Subtraction: Find LCD, convert, then combine numerators.
Example: Simplify \(\frac{2}{x - 1} + \frac{3}{x + 2}\)
LCD: \((x - 1)(x + 2)\)
Convert: \(\frac{2(x + 2)}{(x-1)(x+2)} + \frac{3(x - 1)}{(x-1)(x+2)}\)
Combine: \(\frac{2(x + 2) + 3(x - 1)}{(x-1)(x+2)} = \frac{2x + 4 + 3x - 3}{(x-1)(x+2)} = \frac{5x + 1}{(x-1)(x+2)}\)
A complex fraction has fractions in the numerator and/or denominator.
Strategy: Multiply numerator and denominator by the LCD of all inner fractions.
Example: Simplify \(\frac{\frac{1}{x} + \frac{1}{y}}{\frac{1}{x} - \frac{1}{y}}\)
LCD of inner fractions: \(xy\)
Multiply top and bottom by \(xy\):
$$\frac{\frac{1}{x} + \frac{1}{y}}{\frac{1}{x} - \frac{1}{y}} \cdot \frac{xy}{xy} = \frac{y + x}{y - x}$$
Goal: Eliminate radicals from denominators (standard form for final answers).
Multiply numerator and denominator by the radical.
Example: Rationalize \(\frac{5}{\sqrt{3}}\)
$$\frac{5}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3}$$
Multiply by the conjugate: \((a + b)(a - b) = a^2 - b^2\)
Example: Rationalize \(\frac{6}{2 + \sqrt{5}}\)
Conjugate of \(2 + \sqrt{5}\) is \(2 - \sqrt{5}\)
$$\frac{6}{2 + \sqrt{5}} \cdot \frac{2 - \sqrt{5}}{2 - \sqrt{5}} = \frac{6(2 - \sqrt{5})}{4 - 5} = \frac{6(2 - \sqrt{5})}{-1} = -6(2 - \sqrt{5}) = -12 + 6\sqrt{5}$$
Expand and simplify: \((x - 4)(2x^2 + 3x - 5)\)
Factor completely: \(x^3 - 4x^2 - x + 4\)
Simplify: \(\frac{x^2 - 4x + 3}{x^2 - 1}\)
Simplify the complex fraction: \(\frac{\frac{1}{x+h} - \frac{1}{x}}{h}\)
Factor completely: \(x^6 - 1\)
Rationalize the numerator and simplify: \(\frac{\sqrt{x} - \sqrt{a}}{x - a}\)
Find \(f'(x)\) for \(f(x) = \frac{1}{x^2}\) using the limit definition.
Part (a): Find and factor the characteristic polynomial for \(A = \begin{pmatrix} 5 & 2 \\ 3 & 4 \end{pmatrix}\).
Part (b): Verify that \(\lambda_1 + \lambda_2 = \text{tr}(A)\) and \(\lambda_1 \cdot \lambda_2 = \det(A)\)
Prove: \(a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + ab^{n-2} + b^{n-1})\)
Simplify: \(\frac{x^4 - 16}{x^3 + 8} \cdot \frac{x^2 - 2x + 4}{x^2 + 4} \div \frac{x - 2}{x + 2}\)
Expand \((x - 4)(2x^2 + 3x - 5)\)
Step 1: Distribute \(x\):
\(x(2x^2 + 3x - 5) = 2x^3 + 3x^2 - 5x\)
Step 2: Distribute \(-4\):
\(-4(2x^2 + 3x - 5) = -8x^2 - 12x + 20\)
Step 3: Combine like terms:
\(2x^3 + 3x^2 - 5x - 8x^2 - 12x + 20 = 2x^3 - 5x^2 - 17x + 20\)
Step 4: Factor out GCF (checking):
Wait, let me recalculate. The original problem states the answer should be \(4(2x^2 - 3x + 5)\).
Actually, looking at the source document more carefully, the example given shows:
\(2x^3 + 3x^2 - 5x - 8x^2 - 12x + 20\)
\(= 2x^3 + (3x^2 - 8x^2) + (-5x - 12x) + 20\)
\(= 2x^3 - 5x^2 - 17x + 20\)
Answer: \(2x^3 - 5x^2 - 17x + 20\)
Factor \(x^3 - 4x^2 - x + 4\)
Step 1: Try grouping:
\((x^3 - 4x^2) + (-x + 4)\)
Step 2: Factor each group:
\(x^2(x - 4) - 1(x - 4)\)
Step 3: Factor out \((x - 4)\):
\((x - 4)(x^2 - 1)\)
Step 4: Factor \(x^2 - 1\) as difference of squares:
\((x - 4)(x + 1)(x - 1)\)
Answer: \((x - 4)(x + 1)(x - 1)\)
Verification: \((x-4)(x+1) = x^2 - 3x - 4\), then \((x^2 - 3x - 4)(x - 1) = x^3 - x^2 - 3x^2 + 3x - 4x + 4 = x^3 - 4x^2 - x + 4\) ✓
Simplify \(\frac{x^2 - 4x + 3}{x^2 - 1}\)
Step 1: Factor numerator.
\(x^2 - 4x + 3\): Need two numbers with product 3 and sum -4. Numbers: -3 and -1.
\(x^2 - 4x + 3 = (x - 3)(x - 1)\)
Step 2: Factor denominator.
\(x^2 - 1 = (x + 1)(x - 1)\)
Step 3: Cancel common factors.
$$\frac{(x - 3)(x - 1)}{(x + 1)(x - 1)} = \frac{x - 3}{x + 1}$$
Answer: \(\frac{x - 3}{x + 1}\), for \(x \neq 1, -1\)
Simplify \(\frac{\frac{1}{x+h} - \frac{1}{x}}{h}\)
Step 1: Find common denominator for the numerator.
LCD of \(\frac{1}{x+h}\) and \(\frac{1}{x}\) is \(x(x+h)\).
$$\frac{1}{x+h} - \frac{1}{x} = \frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} = \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)}$$
Step 2: Divide by \(h\).
$$\frac{\frac{-h}{x(x+h)}}{h} = \frac{-h}{x(x+h)} \cdot \frac{1}{h} = \frac{-1}{x(x+h)}$$
Answer: \(\frac{-1}{x(x+h)}\)
Factor completely: \(x^6 - 1\)
Step 1: Use difference of squares.
\(x^6 - 1 = (x^3)^2 - 1^2 = (x^3 + 1)(x^3 - 1)\)
Step 2: Factor each cubic.
\(x^3 + 1\) is a sum of cubes: \(x^3 + 1 = (x + 1)(x^2 - x + 1)\)
\(x^3 - 1\) is a difference of cubes: \(x^3 - 1 = (x - 1)(x^2 + x + 1)\)
Step 3: Combine.
$$x^6 - 1 = (x + 1)(x^2 - x + 1)(x - 1)(x^2 + x + 1)$$
Step 4: Check if quadratic factors can be factored further.
For \(x^2 - x + 1\): Discriminant = \(1 - 4 = -3 < 0\). No real roots; doesn't factor over ℝ.
For \(x^2 + x + 1\): Discriminant = \(1 - 4 = -3 < 0\). No real roots; doesn't factor over ℝ.
Answer: \((x - 1)(x + 1)(x^2 - x + 1)(x^2 + x + 1)\)
Rationalize the numerator and simplify: \(\frac{\sqrt{x} - \sqrt{a}}{x - a}\)
Step 1: Multiply by the conjugate of the numerator.
$$\frac{\sqrt{x} - \sqrt{a}}{x - a} \cdot \frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}}$$
Step 2: Simplify the numerator using difference of squares.
$$(\sqrt{x})^2 - (\sqrt{a})^2 = x - a$$
Step 3: Write the result.
$$\frac{x - a}{(x - a)(\sqrt{x} + \sqrt{a})}$$
Step 4: Cancel \((x - a)\).
$$\frac{1}{\sqrt{x} + \sqrt{a}}$$
Answer: \(\frac{1}{\sqrt{x} + \sqrt{a}}\), for \(x \neq a\)
Find \(f'(x)\) for \(f(x) = \frac{1}{x^2}\) using the limit definition.
Step 1: Write the difference quotient.
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$$
Step 2: Find common denominator in the numerator.
$$\frac{1}{(x+h)^2} - \frac{1}{x^2} = \frac{x^2 - (x+h)^2}{x^2(x+h)^2}$$
Step 3: Expand \((x+h)^2\) and simplify the numerator.
$$x^2 - (x+h)^2 = x^2 - (x^2 + 2xh + h^2) = -2xh - h^2 = -h(2x + h)$$
Step 4: Substitute back.
$$\frac{\frac{-h(2x + h)}{x^2(x+h)^2}}{h} = \frac{-h(2x + h)}{h \cdot x^2(x+h)^2} = \frac{-(2x + h)}{x^2(x+h)^2}$$
Step 5: Take the limit as \(h \to 0\).
$$f'(x) = \lim_{h \to 0} \frac{-(2x + h)}{x^2(x+h)^2} = \frac{-(2x + 0)}{x^2 \cdot x^2} = \frac{-2x}{x^4} = \frac{-2}{x^3}$$
Answer: \(f'(x) = -\frac{2}{x^3}\)
Part (a): Find and factor the characteristic polynomial for \(A = \begin{pmatrix} 5 & 2 \\ 3 & 4 \end{pmatrix}\).
Using the formula \(p(\lambda) = \lambda^2 - (a+d)\lambda + (ad - bc)\):
$$p(\lambda) = \lambda^2 - 9\lambda + 14$$
Factor: Need two numbers with product 14 and sum 9. Numbers: 7 and 2.
$$p(\lambda) = (\lambda - 7)(\lambda - 2)$$
Eigenvalues: \(\lambda_1 = 7\), \(\lambda_2 = 2\)
Part (b): Verify the relationships.
\(\lambda_1 + \lambda_2 = 7 + 2 = 9 = \text{tr}(A)\) ✓
\(\lambda_1 \cdot \lambda_2 = 7 \cdot 2 = 14 = \det(A)\) ✓
Prove: \(a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + ab^{n-2} + b^{n-1})\)
Let \(S = a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + ab^{n-2} + b^{n-1}\)
This is a geometric series with first term \(a^{n-1}\), ratio \(\frac{b}{a}\), and \(n\) terms.
Compute \((a - b) \cdot S\):
$$a \cdot S = a^n + a^{n-1}b + a^{n-2}b^2 + \cdots + a^2b^{n-2} + ab^{n-1}$$
$$b \cdot S = a^{n-1}b + a^{n-2}b^2 + a^{n-3}b^3 + \cdots + ab^{n-1} + b^n$$
Subtract:
$$a \cdot S - b \cdot S = a^n - b^n$$
All middle terms cancel (telescoping sum).
Therefore: \((a - b) \cdot S = a^n - b^n\) ∎
This identity is used to analyze the error \(|x^n - x_0^n|\) when approximating a value \(x\) by \(x_0\):
$$|x^n - x_0^n| = |x - x_0| \cdot |x^{n-1} + x^{n-2}x_0 + \cdots + x_0^{n-1}|$$
If \(|x|, |x_0| \leq M\), then the sum has \(n\) terms each bounded by \(M^{n-1}\), so:
$$|x^n - x_0^n| \leq n M^{n-1} |x - x_0|$$
This shows that small errors in \(x\) can be amplified by a factor of \(nM^{n-1}\) when computing \(x^n\) — relevant to conditioning analysis.
Simplify: \(\frac{x^4 - 16}{x^3 + 8} \cdot \frac{x^2 - 2x + 4}{x^2 + 4} \div \frac{x - 2}{x + 2}\)
Step 1: Convert division to multiplication.
$$= \frac{x^4 - 16}{x^3 + 8} \cdot \frac{x^2 - 2x + 4}{x^2 + 4} \cdot \frac{x + 2}{x - 2}$$
Step 2: Factor all expressions.
Step 3: Substitute factored forms.
$$= \frac{(x^2 + 4)(x + 2)(x - 2)}{(x + 2)(x^2 - 2x + 4)} \cdot \frac{x^2 - 2x + 4}{x^2 + 4} \cdot \frac{x + 2}{x - 2}$$
Step 4: Cancel common factors.
Step 5: What remains.
$$= x + 2$$
Answer: \(x + 2\), for \(x \neq 2, -2\)
Lesson M1.3: Equations and Inequalities — solving polynomial, rational, radical, and absolute value equations; compound inequalities; sign analysis for rational inequalities.