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If (x – 2k) is a factor of f (x), which of the following must be true? f (2k) = 0 f (-2k) = 0 A root of f (x) is x =
Find an answer to your question ✅ “If (x – 2k) is a factor of f (x), which of the following must be true? f (2k) = 0 f (-2k) = 0 A root of f (x) is x = – 2k. A y intercept of …” in 📘 Mathematics if you’re in doubt about the correctness of the answers or there’s no answer, then try to use the smart search and find answers to the similar questions.
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If (x – 2k) is a factor of f (x) , which of the following must be true? – If a root is r1 then a factor of the polynomial must be x-r1. A y intercept" in Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.Scoring well on GMAT Reading Comprehension is a critical factor for nailing a perfect score. Learn to implement the FIRE strategy with Scoreleap to buttress your Verbal section on GMAT. Notice that we are asked to find which of the options MUST be true, not COULD be true.At what point(s) are the following true? (Select all that apply). dy/dx and d… Transcribed Image Text from this Question. The graph of a function y = f(x) is shown. At what point(s) are the following true?
If x 0 and x/|x| < x, which of the following must be true? | Forum – The common factor of the given expressions has to be determined. The common factor is (x – 3). Approved by eNotes Editorial Team.altavistard altavistard. Answer: If (x+k) is a factor of f(x), then -k is a root (or solution) of f(x). Step-by-step explanationy=f(x)=(x-2k). if x=2k , then. Find the equation of the straight linePassing through A (1, 3) and is parallelW perpendicular to the straight linepassing through B(a.s) and c (4.6).
Solved: The Graph Of A Function Y = F(x) Is Shown. | Chegg.com – What must be a factor of $f(x)$? The question is from a SAT practice test. I'm not entirely sure what the question is asking. At first I thought that the $x Then I thought, that the $x$ table values must be multiplied by the equations to get $f(x)$, for example, in the equation $x-3$ multiplied by $0$ if $x$ is…B.)Neither x = -8 nor x = 8 is a root of f(x). Find the width of the rectangle.? 8 answers.5. Which of the following must be true? L 0, then Bill. ionexistent. For all real numbers x and y, let f be a function such that f(x+ y) = f (x) + f (y) + 2xy and The graph of y =f'(x) is shown. Which of the following statements about the function f(x) are true? I. f(x) is decreasing for all x between a and c.
Partial Fraction Decomposition – Example 5 – .
❖ Synthetic Division – A Shortcut for Long Division! ❖ – In this video we're gonna do an example of what's called synthetic division Synthetic division is when you're basically dividing a polynomial by a first degree polynomial so x to the first power it may or may not have a number in there if it was just x you would just break things up individually that'd be easy anyway So suppose we wanna divide, again, x cubed minus 2x squared plus 3x minus 4 we're gonna divide that by (x-2) What you do in this case is this Whatever we're dividing by…
So notice there's a minus 2 — it's x minus 2 we actually take the opposite sign and use positive 2 and that's what you stick outside And then notice I have 1 x cubed minus 2x squared plus 3x minus 4 Notice there's no terms missing. It starts from the third degree second degree first degree and then we just have a number All we do is write the coefficients So there's a 1 a negative 2 in front of the x squared so a positive 3 in front of the x and there's a negative 4 just hanging out And with synthetic division what you do is the first term you just drop it right down there's nothing to do And then the idea is we multiply So I'm gonna take positive 2 times 1, and then stick that number in the next spot So it says postive 2 times 1 will give us positive 2 And then what we do is we add those numbers together in this column So negative 2 plus 2 is gonna give me zero And then we repeat this process Now we take positive 2, multiply by zero and then we stick it in the next column So it says positive 2 times zero is just gonna give us zero So if we add those together we get positive 3 And then again I take positive 2 times positive 3 I'm just gonna erase my little red dots here Take positive 2 times positive 3, that'll give me positive 6 We add the column together and we get the number 2 This last number is gonna be your remainder If I could spell correctly here And what it says is, in summary, it says We can actually — So the numbers in the bottom are what are important 'Kay so it says we can actually rewrite x cubed minus 2x squared plus 3x minus 4 divided by (x-2) So I'll just keep my numbers at the bottom It says we can actually write that as — Okay so the highest power is x cubed and to make things one degree less so this is gonna go with my x squared term this is gonna go with my x term this is gonna be my constant and again the other part is my remainder It says we can actually write x cubed minus 2x squared plus 3x minus 4 divided by (x-2) as 1 x squared plus zero x So usually zero x we'll leave out so I'll erase it So 1 x squared plus 3 and then my remainder is 2 and we put that over what we were originally dividing by which is (x-2) And now we have broken up our original fraction into something a little bit more simple So this would be your solution after you did the synthetic division So a useful little trick One other thing I wanna point out about synthetic division Notice that if we look at the thing on top If I call that f(x) So if I say f(x) equals x cubed minus 2x squared plus 3x minus 4 Suppose I wanted to evaluate this — Suppose I want to plug in the number 2 Well it turns out if you plug in the number 2 If you do the synthetic division with (x-2), if you do that synthetic division it says whatever the remainder is that is actually gonna be the solution when you plug that value in So just a coincidence here that I plug 2 in and got 2 out But notice if I plug 2 in I'll get 2 cubed minus 2 times 2 squared plus 3 times 2 minus 4 So this is 8 minus 8 plus 6 minus 4 Well the 8 and the 8 cancel out positive 6 minus 4 is, hey, positive 2 Okay, so synthetic division is actually also a way to evaluate polynomials That remainder, again, is gonna be your solution And sometimes if you have a real tedious problem, using synthetic division can be useful For example if you're plugging things into a calculator Typically I don't use it, really hardly ever but that's just a personal preference of mine so Alright, I hope this example makes some sense I think I'm gonna do one other as well and just kinda reinforce this stuff If you have any questions just send me an email .
Linear Algebra Example Problems – Augmented Matrix Of A Consistent Linear System – So we've been looking at different examples
of solving systems of linear equations.
This video is similar in that we're going
to be looking at a system of linear equations, but it's a little bit different because the
system of equations actually has a variable in it. So here are the equations we're going to look
at. So the first one is x1 plus 3×2 equals 4,
and the second equation is 2×1 plus ax2 equals a negative 3. So right here, we have this variable a that
is a variable or an unknown– it's not an unknown. It's a variable system the system of equations. And what we're going to do is we're going
to figure out for which values of a does this system of equations have "a" solution. So for what value of a is the system consistent? So that's the question that we're going to
try to solve here. So the way we're going to do this is we're
in still form an augmented matrix. So here are the coefficients of x1's and x2's. And then the right side of the equation–
on the equal side, 4, negative 3, they go right here. This is our standard augmented matrix. This looks a little different from the previous
videos though because the number right here for the x2 factor is not a number, but it's
this variable a. We can still manipulate these equations and
do the same type of operations that we're used to doing though. We can still manipulate this into a form that
we can try to solve the equations for. So let's go ahead and do that. Let's go ahead and let the second equation
equal itself minus 2 times the first. So I'm going to try to get a 0 right here
just like I always do when working with systems of equations in this augmented matrix. If I do that, then I end up with a 0 there,
and I have a minus 6 here, and then I'll have a negative 11 right here. So if I look at this equation now if we write
out what this means, what do I have? This equation says that 0x1 plus the quantity
a minus 6 times x2 is equal to a negative 11. Well, 0 times x1 is just 0, so this just simplifies
to the quantity a minus 6 times x2 is equal to a negative 11. So the way I like to think about this is what
would happen if this factor a minus 6 was equal to zero? So if this term right here was 0 in its totality,
I'd have 0 times x2 equals negative 11 or 0 equals a negative 11. So I'd end up with a contradiction. So I can't have a minus 6 equals 0. Another way to think of it is if I was to
solve for x2 by dividing this side of the equation by a minus 6 and dividing the other
side of the equation by a minus 6, I'd end up with this expression right here for x2.
x2 is equal to a negative 11 divided by the quantity a minus 6. Well, obviously, I can only do that division
as long as a minus 6 is not 0. So same condition pops up, I don't want a
minus 6 to be equal to 0. In the first case, I would have an inconsistent
equation, 0 equals negative 11. In the other case, I've violated a fundamental
law of divided by a 0. Either case, that's a bad thing. Both of those conditions tell me that as long
as a is not 6, I'm OK. As long as a isn't 6, so factor won't be a
0 and I won't be dividing by 0 here. So this system will actually have a solution
for all a, so this is the symbol for all, except when a is equal to 6. When a equals 6, things fall apart. I have a contradiction. But for any other value, I am OK. So we've answered the question. This system of equations is consistent. It has a solution for any value of a except
for the value a equals six 6. So just to wrap things up, as a specific example,
if I happened to have a equals 5, I can actually go through and solve for the values of x1
and x2, and I would get x1 as a negative 29 and x2 is equal to 11. This would be a unique solution to the system
of equations. Or if for instance, a was equal to negative
4. If I solve for x1 and x2, I would get the
unique solution x1 equals 0.7, and x2 equals 1.1. I could keep doing this for forever. I could take– there's an infinite number
of a's I could pick, and for each value of a I pick, there is a unique solution as long
as I don't pick a equals 6. So that's this problem. It's really similar to what we've done before
in terms of solving systems of equations, but it was a little bit different because
we had a variable that was a coefficient of one of our unknowns. In this case, it was the coefficient of the
x2 unknown in the second equation. And we found out that we can solve this system
of equations for all values of a except for one value and that value was a equal to 6. .