f(x)=x^2+4(x-3)
Which of the following is an equivalent form of the expression above that displays the zeroes of the function as constants or coefficients in the expression?
An even faster way to do the above problem is to consider that the x-intercept form is the factored form. So you can just copy the expression and use the factor command - menu-algebra-factor
CAS solution
factor(x^2+4(x-3),x) Enter
returns (x-2)(x+6)
Official Practice Test 4 Section 4 Question 28
f(x)=(x+6)(x-4)
Which of the following is an equivalent form of the function f above in which the minimum value of f appears as a constant or coefficient?
A. x^2-24
B. x^2+2x-24
C. (x-1)^2-21
D. (x+1)^2-25
CAS Solution
This question asks for the vertex form, so use the Complete the Square command. Remember to put the multiplication sign between the factors.
completesquare((x+6)(x-4),x)
returns (x+1)^2-25
Alternately, recognize that only choices C and D are in vertex form, and test these by subtracting them from the original. Remember to put parentheses around the second expression to avoid sign mistakes. Since you have two similar answer choices, test the second one first.
D: (x+6)(x-4)-((x+1)^2-25) Enter
returns 0
@Plotinus I feel like a few of these examples cross the line between “CAS calculator hacks that instantly kill the problem” to “solutions using the CAS, but most other solutions are faster.”
For example, PSAT S4 #23 can be solved algebraically pretty quickly, but another fast method (that has been promoted much here) is to plug in numbers. For example, let a=3 and b=8, then a^4 = 81 and all you need to do is see which of the answers is equal to 81 when b = 8 (which is D, (b+1)^2).
Another possible pitfall with the CAS is you have to really know what you’re doing. For example, I highly doubt the following works:
solve(b = a^2 + 1, a) +/- sqrt(b-1) a^4
[/QUOTE]
since the scope of the variable “a” is only within the “solve” equation. So typing a^4 will only give you a^4. I don’t have a CAS to confirm, but I’m fairly sure this is true (of course, you can remedy it by defining a := sqrt(b-1), but why?).
I didn’t even know a “complete the square” method existed on a TI graphing calculator. How many of your students know that it exists?
Here is an even faster (and cheap) way to use the calculator:
min((x+6)(x-4),x) //I forget the exact command name, whatever minimizes the value of the function -25 And lo and behold, choice D contains the number -25 in it.
[/QUOTE]
Yes, I agree, but some of the examples are just to illustrate methods than can be applied to similar, much harder problems. There are not that many official practice problems out there, so nobody knows exactly what is going to turn up on the real tests. I developed most of these methods for old SAT problems for which some of the methods may have had more obvious advantages. They may make sense again for the new SAT depending upon what turns up.
I don’t think you have reproduced my solution here accurately. This was the solution I posted:
CAS Solution
solve(b=a^2-1,a) Enter
returns a=-sqrt(b+1) and b≥-1 or a=sqrt(b+1) and b≥-1
sqrt(b+1)^4 Enter
returns (b+1)^2
In the second step, I manually raise the solution obtained in the first step to the power of 4. I didn’t use the variable a in the second step. Of course the variable a would not be defined outside the solve equation unless you use “Define”.
There is another way to do this problem in one step, but I am not revealing all my trade secrets here. These are just a few crumbs… 
Some operations are very straightforward, but others are a little more complicated. A student has to have time to work with the calculator and practice with it both with untimed homework and under simulated test conditions to know how to do things easily and to know the best way to use it for him or her as an individual.
Some students even went to the test with BOTH the TI-84’s they used for school AND their TI-Nspire CX CAS’s. They used the 84 for most operations and the Nspire for a few things. We are talking about squeezing some extra points with CAS here, not plugging the whole test into the calculator. Just think of all those people who are going to miss NMSF because of 1 question. What if it was CAS hackable …?
How is this faster? Does it take less time to do
fmin((x+6)(x-4),x) than to do competesquare((x+6)(x-4),x)?
Further, fmin and fmax return the critical X-value, not the y-value, The student will still have to plug the x-value back into the equation to determine the y-value. You need BOTH the x and the y value to make sure that you are picking the correct answer choice. This is a completesquare not an fmin problem.
I don’t know the answer to that question because in the past I used Complete the Square constantly for Math Level 2, but not at all for regular SAT math. The regular SAT students may have seen it because the Nspire has menus, and when you open the algebra menu there is a list of clickable options: Solve, Factor, Expand, Zeroes, Compete the Square, etc.
However, it looks like Complete the Square is going to be great for the new SAT. I guarantee you all my SAT students will know about Complete the Square very shortly.
That, and a lot of other things many people don’t know exist on the calculator…
Here is an example of how a CAS method that seems dumb for an easy problem can turn out to be very smart when a harder version of the same problem turns up.
The method is to find an equivalent expression by subtracting the answer choices from the original and seeing which choice gives a difference=0.
CB rated “medium” problem:
Which of the following is equivalent to 3(x+5)-6?
A. 3x-3
B. 3x-1
C. 3x+9
D. 15x-6
I think all my students can do 3(x+5)-6=3x+15-6=3x+9 with a pencil faster and more accurately than they can copy the equation into their calculators and subtract the answer choices to see which one gives a difference=0.
So you might conclude the CAS approach is dumb.
But what about this problem that occurred later on the same test?
CB rated “hard” problem
(2x-2)^2 - (2x-2)
Which of the following is an equivalent expression?
A. 2x-2
B. 2x^2-6x+6
C. 4x^2-10x+2
D. (2x-3)(2x-2)
A lot of pencil students who got this problem wrong or who took a long time to get it right would solve it faster and more accurately with the method that was dumb for the easy version: subtract the answer choices, starting from the most complicated looking one.
D: (2x-2)^2 - (2x-2)-((2x-3)(2x-2)) Enter
returns 0
So maybe the dumb method isn’t so dumb?
What about picking numbers and backsolving?
Picking numbers and backsolving work very well for some problems and for some people, but sometimes CAS might be better.
Even on Math Level 1, there are questions like:
2x^2+2y^2+12x-24y+60=0
What are the center and radius of this circle?
CAS Solution
completesquare(2x^2+2y^2+12x-24y+60=0,x,y)
returns 2(x-3)^2+2(y-6)^2=30
At this point you have to divide by 2 and take the square root to get the radius, so a little work is required, but no heavy lifting.
Also
What are the points of intersection of a circle with center at the origin and radius 2, and a circle with a center (2,0) and radius 2?
CAS Solution
solve (x^2+y^2=4
(x-2)^2+y^2=4, x,y)
returns x=1 and y=sqrt(3) or x=1 and y=-sqrt(3)
You can do a lot of great things with non-linear systems. We have already seen the intersection of a parabola and a line. Who knows what else might be in store. These problems are going to be in the “kill immediately with CAS” category. 
I originally developed the “x=y iff x-y=0” CAS hack for the following problem:
For which of the following functions is f(x)=f(1-x)?
A. f(x)=1-x
B, f(x)=1-x^2
C. f(x)= x^2-(1-x)^2
D. f(x)=x^2(1-x)^2
E. f(x)=x/(1-x)
CAS Solution: (Test the answer choices from most complicated to least complicated.)
C: Define f(x)= x^2-(1-x)^2 Enter
returns Done
f(x)-f(1-x) Enter
returns 4x-2
D: Define f(x)=x^2(1-x)^2 Enter (copy the C Define line and change the subtraction sign to a multiplication sign)
returns Done
f(x)-f(1-x) Enter (copy from the earlier line entry)
returns 0
I admit, that hack was a lot more fun…
@Plotinus I know, and my goal wasn’t to reproduce your solution. What I was saying was, if an unsuspecting student that didn’t really know what he was doing tried to compute a^4 that way.
If you know that the complete the square method exists, then perhaps not.
All my Math Level 2 students knew about Complete the Square. It’s great for CAS-killing many Math Level 2 conics problems. It looks like some version of those problems might be imported to the rSAT. That’s one reason I said that there are still CAS-hackable problems on the new SAT, and maybe even more of them.
I just got a copy of the CB PSAT/NMSQT Practice Test #2.
Section 4, question 29 (Grid-in) CB Difficulty Rating: h (hard)
y=x^2-4x+3
y=x-1
If (x,y) is a solution to the system of equations above, what is one possible value of the product of x and y?
CAS Solution:
solve[y=x^2-4x+3
y=x-1, (x,y)] Enter
Returns
x=1 and y=0 or x=4 and y=3
Answers: 0 or 12
Section 4, question 27 (last question in mc section) CB Difficulty Rating: h (hard)
x= 1/3 y
154-4y=10x
The equations of two lines are shown above. If the lines are graphed in the xy-plane, which of the following ordered pairs represents the point at which the lines would intersect?
A) (1,3)
B) (3,9)
C) (5,15)
D) (7,21)
CAS Solution:
solve[x= 1/3 y
154-4y=10x, (x,y)] Enter
Returns
x=7 and y=21
Answer: D
Section 4, question 26 (second to last question of mc section) CB Difficulty Rating: h (hard)
If function f is defined by f(x)=3x^2-5x+4, what is f(x-4)?
A) f(x-4)=3x^2-4x
B) f(x-4)=3x^2-5x+72
C) f(x-4)=3x^2-29x+52
D) f(x)=3x^2-29x+72
CAS Solution
Define f(x)=3x^2-5x+4 Enter
f(x-4) Enter
Returns
3x^2-29x+72
Answer: D
Section 4, question 10 CB Difficulty Rating: m (medium)
4x - 1/2 x - 7 = 7(1/2 x - 7)
Which of the following describes the solution to the equation above?
A) x=0
B) x=10 1/2
C) The equation has infinitely many solutions.
D) The equation has no solutions.
CAS Solution:
solve(4x - 1/2 x - 7 = 7(1/2 x - 7),x) Enter
Returns
False
Answer: D
Section 4, question 9 CB Difficulty Rating: m (medium)
(x^2y-3y^2+5xy^2) - (-x^2y + 3xy^2 -3y^2)
Which of the following is equivalent to the expression above?
A) 2x^2y+2xy^2
B)8xy^2-6y^2
C)2x^2y+8xy^2-6y^2
D) x^4y^2-9xy^4-15xy^2
CAS Solution
(x^2y-3y^2+5xy^2) - (-x^2y + 3xy^2 -3y^2) Enter
Returns
2x^2y+2xy^2
Answer: A
OMG, that calculator seems too OP
Ban It!
Or move these five questions to the non-calculator section.
SAT Official Practice Test #5 Section 4 Question 29
x^2 + 20x + y^2 + 16y = −20
The equation above defines a circle in the xy-plane. What are the coordinates of the center of the circle?
A) (−20,−16)
B)(−10, −8)
C) (10, 8)
D) (20,16)
CAS Solution
completesquare(x^2 + 20x + y^2 + 16y = −20) Enter
Returns
(x+10)^2 + (y+8)^2=144
At this point, the student has to know the center form of the equation of the circle and to reverse the signs of the constants to find the center. If you do not know this, you can graph the equation to find the center.
Answer: B
SAT Official Practice Test #5 Section 4 Question 30
y=x^2-a
In the equation above, a is a positive constant and the graph of the equation in the xy‐plane is a parabola. Which of the following is an equivalent form of the equation?
A) y=(x+a)(x-a)
B) y=(x+ sqrt a)(x− sqrt a)
C) y=(x+a/2)(x-a/2)
D) y=(x+a)^2
CAS Solution
The most brainless way to approach this equivalent expression problem is to subtract each answer choice from the given equation and see which choice yields a difference of 0. Use the copy and edit functions to enter the expressions more quickly.
A: (x^2-a)-(x+a)(x-a) Enter
Returns
a^2-1
B: (x+a)(x-a)-(x+ sqrt a)(x− sqrt a) Enter
Returns
0
Answer: B
SAT Official Practice Test #6 Section 4 Question 34 (Grid-In)
In the xy-plane, the graph of y = 3x^2 − 14x intersects the graph of y=x at the points (0, 0) and
(a,a). What is the value of a ?
CAS Solution:
solve(y = 3x^2 − 14x
y=x, (x,y)) Enter
Returns
x=0 and y=0 or x=5 and y=5
Answer: 5