Solving Polynomials Using the Remainder Theorem

We are given the function:

f(x) = mx³ − 3x² + nx + 2

• When divided by (x + 3), the remainder is −3.
• When divided by (x − 2), the remainder is −4.

By the Remainder Theorem, if f(x) is divided by (x − a), the remainder is f(a).

Step 1: Substitute x = −3

f(−3) = m(−3)³ − 3(−3)² + n(−3) + 2

= −27m − 27 − 3n + 2

= −27m − 3n − 25

Given remainder −3:

−27m − 3n − 25 = −3

−27m − 3n = 22

27m + 3n = −22 (Equation 1)

Step 2: Substitute x = 2

f(2) = m(2)³ − 3(2)² + n(2) + 2

= 8m − 12 + 2n + 2

= 8m + 2n − 10

Given remainder −4:

8m + 2n − 10 = −4

8m + 2n = 6 (Equation 2)

Step 3: Solve the System of Equations

Multiply Equation 2 by 3 → 24m + 6n = 18

Multiply Equation... Continue reading "Solving Polynomial Remainder Theorem and Transformations" »

Customer Value: (Quality, Time, Flexibility, Customer Experience, Innovation) / Price

Sustainability Paradigms: Economic (Viable, Equitable), Environment (Viable, Bearable), Social (Bearable, Equitable)

- Efficiency (Optimization), Differentiator (Innovation), Driver (Motivation) | - Audits (Assess Sustainability Performance)

Table 7.3: Factors for Calculating Three-Sigma Limits for the X (Bar) Chart and R-Chart

*A sample is out of control if its value falls below the LCL or above the UCL*

Corporate Distribution Layers

Distributions are applied in the following order: Accumulated E&P + Current E&P, then basis, then the remainder is treated as capital gain.

• No Dividend: If there is no E&P, reduce current E&P first. Negative AEP still distributes E&P.
• Offset: Half of current E&P is lost to offset AEP before distribution.

Property Distributions

Recognize gains, but not losses.

• Dividend: FMV - Liabilities
• Basis: FMV
• Gains: FMV - Tax Basis

Distribution Table

The amount from CEP is calculated as: (Period distribution / Total distribution) * CEP. AEP is applied to the remainder, then to stock basis.

Stock Distributions and Redemptions

• Stock Distribution: Proportionate distributions are nontaxable and do not reduce E&

lim x->0 sinx/x = 1 | H.A.: compare degrees | V.A.: denom = 0 | Continuous if: f(a), lim, equal

DERIVATIVES: (x^n)'=nx^(n-1), (e^x)'=e^x, (a^x)'=a^x ln a, (lnx)'=1/x | (uv)'=u'v+uv', (u/v)'=(u'v-uv')/v^2 |chain: (f(g(x)))'=f'(g(x))g'(x)

TRIG DERIVATIVES: (sin)'=cos, (cos)'=-sin, (tan)'=sec^2 | (sec)'=sec·tan, (csc)'=-csc·cot, (cot)'=-csc^2

CRITICAL POINTS:f'=0 or DNE ⇒ crit pt | f'>0 inc | f'<0 dec | f''>0 conc up | f''<0 conc down | inflec = f'' signchange

INTEGRATION: ∫x^n dx = x^(n+1)/(n+1)+C | ∫e^x dx = e^x+C | ∫a^x dx = a^x/ln a+C | ∫1/x dx = ln|x|+C | ∫sin x dx

= -cos x+C | ∫cos x dx = sin x+C

FTC: Part 1: d/dx ∫_a^x f(t) dt = f(x) | Part 2: ∫_a^b f(x) dx = F(b)-F(a)

AREA & VOLUME: A = ∫_a^b (top - bot)... Continue reading "5 hr 20 min 20 sec corresponds to a longitude difference of" »

Gradient: \nabla f = \langle f_x,\ f_y,\ f_z\rangle Directional derivative (unit vector u): D_uf = \nabla f\cdot u Unit direction from A→B: u = \frac{B-A}{\|B-A\|} Tangent plane to z=f(x,y): z - z_0 = f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) Critical points: Solve f_x=0, f_y=0. D = f_{xx}f_{yy} - (f_{xy})^2 Min if D>0,f_{xx}>0. Max if D>0,f_{xx}<0. Saddle if D<0.

General double integral: Type I (y inner): \int_a^b \int_{g_1(x)}^{g_2(x)} f\, dy\, dx

Type II (x inner):

\int_c^d \int_{h_1(y)}^{h_2(y)} f\, dx\, dy

Polar:

x = r\cos\theta,\ y=r\sin\theta,\ dA=r\,dr\,d\theta

Circle: 0\le r\le R.

Line y=mx → \theta=\arctan m.

Region must be continuous in θ.

🔷

3. TRIPLE INTEGRALS

Cylindrical:

x=r\cos\theta,\ y=r\sin\theta,\ dV=r\,dz\,dr\,... Continue reading "Multivariable Calculus and Vector Analysis Formulas" »

1. Find the GCD of (143, 227)

The Euclidean Algorithm is used to find the Greatest Common Divisor (GCD) of two integers by repeatedly dividing the larger number by the smaller number until the remainder becomes 0.

• Step 1: 227 = 143 × 1 + 84
• Step 2: 143 = 84 × 1 + 59
• Step 3: 84 = 59 × 1 + 25
• Step 4: 59 = 25 × 2 + 9
• Step 5: 25 = 9 × 2 + 7
• Step 6: 9 = 7 × 1 + 2
• Step 7: 7 = 2 × 3 + 1
• Step 8: 2 = 1 × 2 + 0

Since the remainder becomes 0, the last non-zero remainder is 1.

Answer: GCD(143, 227) = 1

2. Find m and n such that gcd(a, b) = ma + nb for a = 56 and b = 72

Given:

a = 56, b = 72. We need to find integers m and n such that: gcd(56, 72) = 56m + 72n.

Step 1: Find the GCD using the Euclidean Algorithm

72 = 56 × 1 + 16
56 = 16 × 3 + 8
16 = 8 × 2 + 0
Hence,... Continue reading "Number Theory Problems: GCD, Congruences, and Fermat Numbers" »

Calculating Mean and Standard Deviation

The marks obtained in a statistics examination are normally distributed. If 15% of students scored ≥ 60 marks and 40% scored < 30 marks, find the mean (μ) and standard deviation (σ).

Given

• P(X ≥ 60) = 0.15
• P(X < 30) = 0.40

We need to find the values of μ and σ.

Step 1: Convert to Z-values

(i) P(X ≥ 60) = 0.15

Since P(X ≥ 60) = 0.15, then P(X < 60) = 0.85. From the standard normal table, P(Z < 1.04) = 0.85. Therefore:

(60 − μ) / σ = 1.04
μ = 60 − 1.04σ (Equation 1)

(ii) P(X < 30) = 0.40

From the standard normal table, P(Z < -0.25) = 0.40. Therefore:

(30 − μ) / σ = -0.25
μ = 30 + 0.25σ (Equation 2)

Step 3: Solve the Equations

Equating (1) and (2):

60 − 1.04σ = 30 + 0.25σ
30

Section A: R Basics and Data Types (Weeks 1-4)

Model Questions and Answers

Q: Create a vector v (3, NA, Inf, -Inf). Explain adding 5 to v.

A: The operation is element-wise. Missing values (NA) propagate, resulting in NA. Infinite values (Inf, -Inf) remain infinite.

v <- c(3, NA, Inf, -Inf)
print(v + 5)
# Output: [1]  8 NA Inf -Inf

Q: Given vector a, write code to count and replace NAs.

A: Assuming a <- c(10, 15, NA, 20).

• Count NAs: sum(is.na(a)) → 1.
• Replace NAs (e.g., with 0): a[is.na(a)] <- 0.

Q: Explain the difference between a[a > 12] and a[which(a > 12)].

A: Both select elements greater than 12 (15, 20). However:

• a[a > 12] → [1] 15 NA (Uses logical indexing; preserves the position of NA in the original vector as NA).
• a[which(

Problem 9: Z-Test for Two Means

A researcher wants to know the intelligence of students in a school. He selected two groups of students. In the first group, there are 150 students having a mean IQ of 75 with a S.D. of 15. In the second group, there are 250 students having a mean IQ of 70 with a S.D. of 20. Is there a significant difference between the means of the two groups?

Solution

Given

For Group I:

• Sample size, n₁ = 150
• Sample mean, x̄₁ = 75
• Standard deviation, s₁ = 15

For Group II:

• Sample size, n₂ = 250
• Sample mean, x̄₂ = 70
• Standard deviation, s₂ = 20

Level of significance: α = 0.05

Since both sample sizes are large (n₁, n₂ > 30), use the Z-test for the difference between two means.

Step 1: State the Hypotheses

• Null Hypothesis (H₀): μ₁ = μ₂