This is the only one where the College Board’s explanation was not satisfactory. This is in practice test #10, section 5, question 17 in the Blue Book.
If k and h are constants and x^2+kx+7 is equivalent to (x+1)(x+k), what is the value of k?
A) 0
B) 1
C) 7
D) 8
E) It cannot be determined from the information given.
I know the answer is D, but I do not really understand how to get there. Is there some quick way to do these types of problems? Thanks.
Imagine taking that trinomial and factoring it:
You would set up the parentheses, and put an x into each one.
Then you would look for 2 numbers that multiply to 7, but add up to k… and we already know that one of the numbers is 1. (Because the first factor is already (x+1) )
OK, so if the product is 7 and one term is 1, the other number is 7. The other factor is (x+7)
And 1 and 7 add up to 8.
@bjkmom @goldeagle81616 original problem probably said (x+1)(x+h), since x^2 + 8x + 7 is not equal to (x+1)(x+8)…
@MITer94 Yes you are right. My mistake. It was (x+1)(x+h) not (x+1)(x+k). Do you have a solution?
Since the two expressions are equivalent, that means they give the same value no matter what you use for x…
So try using x=0. You will find out something useful. Then try x=1. Combined with what you found when you set x=0, you will be on your way.
And btw, this is another one of those methods that will also work on the new SAT.
^Way back in the distant past when I was wet behind the ears in all things SAT, I would turn my nose at the “primitive” ways of solving math problems. As years went by, I have learned to appreciate the beauty of simplicity, and plugging in is often my weapon of choice when attacking the SAT/ACT forts.
(To be continued)
@goldeagle81616 any solutions are probably similar to what’s been stated above, but here is a slightly different one:
Instead of plugging x = 0 or 1, plug in x = -1 (since it is a root). Then (-1)^2 + (-1)k + 7 = 0 or k = 8.
Oops. Thanks for the catch.
So, if x=0:
h=7
Then, if x=1:
2+2h=k+8
k=2h-6
k=14-6
k=8
Right?
Jumping the gun. 
@goldeagle81616
It would be a little easier to plug in a newly found value of h in the equation.
A promised sequel to #5.
Sometimes plugging numbers in gets a bit more complicated:
Blue Book, p.527, q.8.
(x - 8)(x - k) = x^2 - 5kx + m
k and m are constants; equation is true for all values of x. What is the value of m?
(A) 8
(B) 16
© 24
(D) 32
(E) 40
@gcf101
(B) 16
I see what you mean. This strategy is quite effective. It seems to work as long as you choose 2 values of
x and go from there. Thanks for the advice and thanks as well to everyone else who contributed.
Totally welcome.
Once you have this method down, it’s worth it to notice what @gcf101 did to make things easier. Choosing an x value that is one of the root’s often cuts down on the amount of algebra you will have to do later. It isn’t essential, but it can help. I usually try x=0, x=1 and/or x= a root. So in that last question, x=0 and then x=8 are good choices.
Humble as I am, I have to give credit for plugging the root in the #1 question to @MITer94 (post #6). But I did use that method in #10 all on my own!
Old habits die hard. 
Can’t help but mention this powerful tool for cracking many quadratic equations based questions:
Simplified Vieta’s formula.
See how nicely both questions in this thread get unfolded now.