Grade 12 Mathematics in Papua New Guinea (PNG) is a crucial subject that plays a significant role in determining students' eligibility for higher education and future career opportunities. The questions often cover topics such as algebra, trigonometry, calculus, statistics, and geometry, requiring students to demonstrate problem-solving skills and a strong understanding of mathematical concepts. To help students prepare effectively, free online guides and resources have become increasingly available, offering step-by-step solutions, practice questions, and exam tips. Platforms like YouTube, educational blogs, and social media pages provide valuable content tailored to the PNG curriculum, including past exam papers and video tutorials explaining complex topics in simple terms. These free online resources not only boost students' confidence but also help them develop effective strategies to tackle tricky exam questions. Students are encouraged to make use of these platforms to improve their performance and increase their chances of excelling in Grade 12 Mathematics examinations.
Below is a Grade 12 Mathematics question that appeared in 2015 Advance Mathematics Paper. Please also watch the video at the end and subscribe to YouTube Channel.
Step 1: Apply the Factor Theorem
We are given that the polynomial:
\[P(x) = x^3 + ax^2 - bx - 10\]is divisible by \((x + 1)\) and \((x - 2)\).
Using the Factor Theorem:
\[P(-1) = 0\] \[P(2) = 0\]Step 2: Substitute \(x = -1\) into \(P(x)\)
Substituting \(-1\) into the polynomial:
\[(-1)^3 + a(-1)^2 - b(-1) - 10 = 0\]Simplifying:
\[-1 + a + b - 10 = 0\]Combine like terms:
\[a + b - 11 = 0 \quad \text{(Equation 1)}\]Step 3: Substitute \(x = 2\) into \(P(x)\)
Substituting \(2\) into the polynomial:
\[(2)^3 + a(2)^2 - b(2) - 10 = 0\]Simplifying:
\[8 + 4a - 2b - 10 = 0\]Combine like terms:
\[4a - 2b - 2 = 0\]Divide by 2:
\[2a - b - 1 = 0 \quad \text{(Equation 2)}\]Step 4: Solve the Equations
From Equation (1):
\[a + b = 11\]Substituting \(b = 11 - a\) into Equation (2):
\[2a - (11 - a) - 1 = 0\]Expanding and simplifying:
\[2a - 11 + a - 1 = 0\]Combine like terms:
\[3a - 12 = 0\]Solve for \(a\):
\[a = 4\]Substituting \(a = 4\) into Equation (1):
\[4 + b = 11\]Solve for \(b\):
\[b = 7\]