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use pascals triangle to find the coefficients of the expanded polynomia…

Question

use pascals triangle to find the coefficients of the expanded polynomial \\((a + b)^3\\).

a) 1 3 3 1
b) 1 2 1
c) 1 4 6 4 1
d) 1 1

question 16 (5 points)
find all real and non-real roots of the function \\(f(x) = x^2 + 49\\).

a) \\(x = i + 7, i - 7\\)
b) \\(x = -49i, 49i\\)
c) \\(x = -7, 7\\)
d) \\(x = -7i, 7i\\)

question 17 (5 points)
what are the real solutions to the equation \\(4x^3 - x^2 - 4x + 1 = 0\\)?

a) \\(x = -1, 1\\)
b) \\(x = -1, 1/4, 1\\)
c) \\(x = -1, -1/4, 1\\)
d) \\(x = -1, 0, 1/4\\)

Explanation:

Expand using Pascal's Triangle

Using the Pascals Triangle Expansion knowledge point

$$ LATEXBLOCK0 $$

Find roots of the quadratic function

Using the Complex Quadratic Solutions knowledge point

$$ LATEXBLOCK1 $$

Solve the cubic equation

To find the real solutions of \(4x^3 - x^2 - 4x + 1 = 0\), we can group the terms to factor the polynomial:

$$ (4x^3 - 4x) - (x^2 - 1) = 0 $$

Factor out common terms from each group:

$$ 4x(x^2 - 1) - 1(x^2 - 1) = 0 $$

Factor out the common binomial \((x^2 - 1)\):

$$ (4x - 1)(x^2 - 1) = 0 $$

Set each factor to zero to find the roots:

$$ 4x - 1 = 0 \implies x = \frac{1}{4} $$
$$ x^2 - 1 = 0 \implies x^2 = 1 \implies x = \pm 1 $$

Thus, the real solutions are \(x = -1, \frac{1}{4}, 1\).

Answer:

Question 1

  • A) 1 3 3 1 (Correct answer)
  • B) 1 2 1
  • C) 1 4 6 4 1
  • D) 1 1

Question 16

  • A) \(x = i + 7, i - 7\)
  • B) \(x = -49i, 49i\)
  • C) \(x = -7, 7\)
  • D) \(x = -7i, 7i\) (Correct answer)

Question 17

  • A) \(x = -1, 1\)
  • B) \(x = -1, \frac{1}{4}, 1\) (Correct answer)
  • C) \(x = -1, -\frac{1}{4}, 1\)
  • D) \(x = -1, 0, \frac{1}{4}\)