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2.1 — Remainder Theorem
Question 1
What is the remainder when \( f(x) = 2x^3 - 3x^2 + x - 5 \) is divided by \( (x-2) \)?
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Solution:
By the Remainder Theorem: \( f(2) = 2(8) - 3(4) + 2 - 5 = 16 - 12 + 2 - 5 = 1 \)
Question 2
If \( f(x) = x^3 + kx^2 - 3x + 2 \) divided by \( (x+1) \) gives remainder 7, find \( k \).
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Solution:
\( f(-1) = -1 + k + 3 + 2 = k + 4 = 7 \), so \( k = 3 \). Wait: \( f(-1) = (-1)^3 + k(-1)^2 -3(-1) + 2 = -1 + k + 3 + 2 = k + 4 = 7 \). So \( k = 3 \).
2.2 — Factor Theorem
Question 3
Which of the following is a factor of \( f(x) = x^3 - 7x + 6 \)?
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Solution:
Test \( f(1) = 1 - 7 + 6 = 0 \). Since \( f(1) = 0 \), by the Factor Theorem \( (x-1) \) is a factor.
Question 4
The factored form of \( 8x^3 - 27 \) is:
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Solution:
Difference of cubes: \( a^3 - b^3 = (a-b)(a^2+ab+b^2) \) where \( a=2x, b=3 \):
\( (2x-3)((2x)^2+(2x)(3)+3^2) = (2x-3)(4x^2+6x+9) \)
2.5 — Polynomial Inequalities
Question 5
The solution to \( (x-1)(x+2)(x-4) > 0 \) is:
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Solution:
Zeros at \( x = -2, 1, 4 \). Test intervals:
• \( x < -2 \): \( (-)(-)(-) = - \) ✗
• \( -2 < x < 1 \): \( (-)(+)(-) = + \) ✓
• \( 1 < x < 4 \): \( (+)(+)(-) = - \) ✗
• \( x > 4 \): \( (+)(+)(+) = + \) ✓
MHF4U — Advanced Functions, Grade 12 | Ontario Curriculum 2007 (Revised)