Question

factor \\(x^3 - 27\\).

a) \\((x + 3)(x^2 + 3x + 9)\\)
b) \\((x + 3)(x^2 - 3x + 9)\\)
c) \\((x - 3)(x^2 - 3x + 9)\\)
d) \\((x - 3)(x^2 + 3x + 9)\\)

question 15 (5 points)
use synthetic division to find the quotient if polynomial \\(p(x) = x^3 - 7x + 6\\) is divided by \\(x - 2\\).

a) \\(x^2 + 2x + 3\\)
b) \\(x^2 + 2x - 3\\)
c) \\(x^2 - 2x - 3\\)
d) \\(x^2 - 2x + 3\\)

question 16 (5 points)
consider the function \\(f(x) = x^4 - x^3 + 2x^2 - 2x\\). how many real roots does it have?

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

Explanation:

Factor the difference of cubes

Using the Difference of Squares Factoring knowledge point as a foundation, we apply the difference of cubes formula:

$$ a^3 - b^3 = (a - b)(a^2 + ab + b^2) $$

Here, \(a = x\) and \(b = 3\) since \(3^3 = 27\):

$$ x^3 - 27 = (x - 3)(x^2 + 3x + 9) $$

Perform synthetic division

Using the Synthetic Division knowledge point

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The quotient is \(x^2 + 2x - 3\).

Factor the quartic function

To find the real roots of \(f(x) = x^4 - x^3 + 2x^2 - 2x\), we first factor by grouping:

$$ f(x) = x^3(x - 1) + 2x(x - 1) = (x^3 + 2x)(x - 1) $$

Factoring out \(x\) from the first term:

$$ f(x) = x(x^2 + 2)(x - 1) $$

Determine the number of real roots

We set each factor to zero to find the roots:

1. \(x = 0\) (real root)
2. \(x - 1 = 0 \implies x = 1\) (real root)
3. \(x^2 + 2 = 0 \implies x^2 = -2\) (no real roots, only complex roots \(x = \pm i\sqrt{2}\))

Thus, there are exactly 2 real roots.

Answer:

Question 14

• (A) \((x + 3)(x^2 + 3x + 9)\)
• (B) \((x + 3)(x^2 - 3x + 9)\)
• (C) \((x - 3)(x^2 - 3x + 9)\)
• (D) \((x - 3)(x^2 + 3x + 9)\) (Correct answer)

Question 15

• (A) \(x^2 + 2x + 3\)
• (B) \(x^2 + 2x - 3\) (Correct answer)
• (C) \(x^2 - 2x - 3\)
• (D) \(x^2 - 2x + 3\)

Question 16

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