Question

final review assignment
this assignment includes multiple choice and short answer questions. for multiple choice questions, select the best answer. each is worth 1 mark. marks assigned to short answer questions are indicated for each question. be sure to show all necessary work.

1. the non - permissible value(s) for the rational expression \\(\frac{2x^2 - x - 1}{4x^2 - 3x - 1}\\) is/are

a. \\(-\frac{1}{4}\\)
b. \\(-\frac{1}{2}\\)
c. 1
d. all of the above
e. both a and c

1. the simplified form of the rational expression \\(\frac{3a^3 + 3a^2 - 36a}{a^3 + 2a^2 - 15a}\\) is

a. \\(\frac{3(a + 4)}{a + 5}, a\
eq0, 3, - 5\\)
b. \\(\frac{3(a + 4)}{a + 5}, a\
eq - 5\\)
c. \\(\frac{3a(a + 4)}{a + 5}, a\
eq0, 3, - 5\\)
d. \\(\frac{3a(a + 4)}{a + 5}, a\
eq - 5\\)

Explanation:

Question 1

Step 1: Find denominator's roots

To find non - permissible values, set the denominator \(4x^{2}-3x - 1=0\).
Factor the quadratic: \(4x^{2}-3x - 1=(4x + 1)(x - 1)\).
Set each factor equal to zero:

• For \(4x+1 = 0\), we have \(4x=-1\), so \(x =-\frac{1}{4}\).
• For \(x - 1=0\), we have \(x = 1\).

Step 2: Analyze options

Option A is \(-\frac{1}{4}\) (a non - permissible value), Option C is \(1\) (a non - permissible value), Option B: Solve \(x=-\frac{1}{2}\), substitute into \(4x^{2}-3x - 1\): \(4\times(-\frac{1}{2})^{2}-3\times(-\frac{1}{2})-1=4\times\frac{1}{4}+\frac{3}{2}-1 = 1+\frac{3}{2}-1=\frac{3}{2}
eq0\), so \(x =-\frac{1}{2}\) is not a non - permissible value. So both A and C are non - permissible values.

Step 1: Factor numerator and denominator

• Factor numerator \(3a^{3}+3a^{2}-36a\):

First, factor out \(3a\): \(3a(a^{2}+a - 12)\).
Then factor \(a^{2}+a - 12=(a + 4)(a - 3)\). So numerator is \(3a(a + 4)(a - 3)\).

• Factor denominator \(a^{3}+2a^{2}-15a\):

First, factor out \(a\): \(a(a^{2}+2a - 15)\).
Then factor \(a^{2}+2a - 15=(a + 5)(a - 3)\). So denominator is \(a(a + 5)(a - 3)\).

Step 2: Simplify the rational expression

The rational expression \(\frac{3a(a + 4)(a - 3)}{a(a + 5)(a - 3)}\).
Cancel out common factors \(a\) (where \(a
eq0\)) and \((a - 3)\) (where \(a
eq3\)):
We get \(\frac{3(a + 4)}{a + 5}\).
Now, find non - permissible values:

• From original denominator \(a^{3}+2a^{2}-15a=a(a + 5)(a - 3)\), set equal to zero: \(a = 0\), \(a=-5\), \(a = 3\). So \(a

eq0,3,-5\).

Step 3: Analyze options

Option A: \(\frac{3(a + 4)}{a + 5},a
eq0,3,-5\) matches our simplified form and non - permissible values. Option B: Missing \(a
eq0,3\), Option C: Incorrect factoring (extra \(a\) in numerator after canceling), Option D: Incorrect factoring (extra \(a\) in numerator after canceling) and missing non - permissible values.

Answer:

E. Both A and C

Question 2