Answer:
16
x
^4 − 96
x^
3 + 216
x
^2 − 216
x + 81
Step-by-step explanation:
Use the binomial expansion theorem to find each term. ∑
∑The binomial theorem states ( a + b ) n = n ∑ k = 0 n C k ⋅ ( a n − k b k ) . 4 ∑ k = 0 4 ! ( 4 − k ) ! k ! ⋅ ( 2 x ) 4 − k ⋅ ( − 3 ) k
Expand the summation. 4 ! ( 4 − 0 ) ! 0 ! ⋅ ( 2 x ) 4 − 0 ⋅ ( − 3 ) 0 + 4 ! ( 4 − 1 ) ! 1 ! ⋅ ( 2 x ) 4 − 1 ⋅ ( − 3 ) + 4 ! ( 4 − 2 ) ! 2 ! ⋅ ( 2 x ) 4 − 2 ⋅ ( − 3 ) 2 + 4 ! ( 4 − 3 ) ! 3 ! ⋅ ( 2 x ) 4 − 3 ⋅ ( − 3 ) 3 + 4 ! ( 4 − 4 ) ! 4 ! ⋅ ( 2 x ) 4 − 4 ⋅ ( − 3 ) 4
Simplify the exponents for each term of the expansion. 1 ⋅ ( 2 x ) 4 ⋅ ( − 3 ) 0 + 4 ⋅ ( 2 x ) 3 ⋅ ( − 3 ) + 6 ⋅ ( 2 x ) 2 ⋅ ( − 3 ) 2 + 4 ⋅ ( 2 x ) ⋅ ( − 3 ) 3 + 1 ⋅ ( 2 x ) 0 ⋅ ( − 3 ) 4
Simplify each term.
= 16
x
^4 − 96
x^
3 + 216
x
^2 − 216
x + 81
Answer: B
Step-by-step explanation: There are two ways to solve this question. The faster way is to multiply each side of the given equation by ax−2 (so you can get rid of the fraction). When you multiply each side by ax−2, you should have:
24x2+25x−47=(−8x−3)(ax−2)−53
You should then multiply (−8x−3) and (ax−2) using FOIL.
24x2+25x−47=−8ax2−3ax+16x+6−53
Then, reduce on the right side of the equation
24x2+25x−47=−8ax2−3ax+16x−47
Since the coefficients of the x2-term have to be equal on both sides of the equation, −8a=24, or a=−3.
The other option which is longer and more tedious is to attempt to plug in all of the answer choices for a and see which answer choice makes both sides of the equation equal. Again, this is the longer option, and I do not recommend it for the actual SAT as it will waste too much time.
The final answer is B.
Long answer:
The domain is all real numbers except -6 and 0. Hence, domain D: {x ∈ ℝ| x ≠ –6, 0}
Hence the range is all real numbers except -3 and 3. Hence, range R: (–∞, –3) ∪ (3, ∞)
Given the following functions:
First we need to get the composite function f(g(x))
Get the domain
The domain the values of x for which the function exists. The function cannot exists at when x = -6 and x = 0
Hence the domain is all real numbers except -6 and 0. Hence, domain D: {x ∈ ℝ| x ≠ –6, 0}
The range is the value of y for which the function exists. The function cannot exists at when x = -6 and x = 0
Hence the range is all real numbers except -3 and 3. Hence, range R: (–∞, –3) ∪ (3, ∞)