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Salman Khan Google Chrome Homework - lcon Sudlbrpnyehbnsmokapchonewoid-5578685825 mathxl.c Section BO1 (Spring 202 171 Calculus & Analytic Geometry Math Section...

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Salman Khan Google Chrome Homework - lcon Sudlbrpnyehbnsmokapchonewoid-5578685825 mathxl.c Section BO1 (Spring 202 171 Calculus & Analytic Geometry Math Section 4.9 Enhancec Homework: Assignment 4 of 34 (4 complete) Score: 0 of 4.9.9 what is the value of C? If F(x) = 322 _ 9x + C ad F( - 1)= 9x + C and F(-1)= -15,then €-D] If F(x) = 3x2 _ Simplify your answer:)KeeShow

Salman Khan Google Chrome Homework - lcon Sudlbrpnyehbnsmokapchonewoid-5578685825 mathxl.c Section BO1 (Spring 202 171 Calculus & Analytic Geometry Math Section 4.9 Enhancec Homework: Assignment 4 of 34 (4 complete) Score: 0 of 4.9.9 what is the value of C? If F(x) = 322 _ 9x + C ad F( - 1)= 9x + C and F(-1)= -15,then €-D] If F(x) = 3x2 _ Simplify your answer:) Kee Show



Answers

In Exercises 47 - 54, write the function in the form $ f(x) = (x - k)q(x) + r $ for the given value of $ k $, demonstrate that $ f(k) = r $.

$ f(x) = -4x^3 + 6x^2 + 12x + 4 $, $ k = 1 -\sqrt{3} $

The goal in this problem is to take our function f of X and write it in this form. So that means that we need to divide by X minus K where K is 1/5. So we're dividing by X minus 1/5. So that means we're going to put 1/5 outside the box for synthetic division and the coefficients of F of X in a row. We have 10 x cubed negative 22 X squared, negative three x and four. Then we bring down the first number. We multiply it by 1/5 and that gives us two. We write that in the next face and add the Colin and we get negative. 20. We multiply that by 1/5 and we get negative for We write that in the next space and we have a column and we get negative seven. We multiply that by 1/5 and we get negative 7/5. We write that in the next face and add the column. Think of four as 25th. So 25th plus negative 7/5 would be 13 5th Okay, so that last number is the remainder. The number before that is the constant. The number before that is the coefficient on the X term and then every before that is a coefficient on the X squared term. So that means our function F of X can be written as X minus 1/5 times the quotient 10 X squared, minus 20 x minus seven, plus the remainder 13 5th The other goal for this problem is to show that 13 5th are remainder is also the function value. In other words, we want to show that half of 1/5 is 13 5th and the way we can do that is we can calculate f of 1/5 using regular substitution into the original function f of X. So that would mean we have 10 times 1/5 cube minus 22 times 1/5 squared, minus three times 1/5 plus four just using the original function that we were given. Now the wise thing to do would be to use your calculator with all these fractions, you can work it out yourself if you'd like to use your calculator, and you will find that you get 13 5th So that shows that the remainder is indeed the function value

Our goal in this problem is to be able to rewrite our function f of X in this form as a product of a divisor and a quotient, and then plus the remainder. So the value of K we're using is two plus square root, too. So are divisor. Is X minus that? So we're going to put two plus square root too, outside the box for synthetic division, and we're going to put our coefficients from F of X in a row. Negative three x cubed eight x squared 10 x and negative eight. And now we're going to do synthetic division. So we bring down the negative three and we multiply it by the number outside the box. And that's going to give us negative six minus three square root, too. And we're going to add the column and we have two minus three square root, too. Then we multiply that by the number outside the box. So we have two minus three square root to, and we're multiplying it by two plus square root, too. So let's do that off to the side. Using the oil method. The first is for the outside is to square root to the inside is negative. Six square root two and the last is negative six. And so when we combine like terms, we get negative two minus four square root, too. Now we're going to add the column, and we have eight minus four square root, too. Now we're going to multiply that by the number outside the box. So again, off to the side eight minus four square root to multiplied by two plus square root, too. So, using the foil technique, the first is 16. The outside is eight square root to the inside is negative. Eight square root two on the last is negative eight. So we can cancel the outside and inside and out the other terms, and we get eight. So we write that in the next face. Add the column and we have zero, so zero is our remainder. The next number is the constant. The next number is the coefficient on the X term, and the next number is a coefficient on the X squared term. So that means we're going to be able to write our polynomial as f of X equals X minus K, which is going to be X minus two minus square root, too multiplied by the quotient, which is negative. Three X squared plus two minus three square root, two times X plus eight minus four square root, too, with the remainder of zero. Okay. The other thing we want to show is that the function value is equal to the remainder. And what that means is we want to show that F of two plus square root to is actually zero. That is why coordinate is actually just equal to the remainder. So what we need to do in order to to verify that is to take our original function F of X and a substitute two plus square root you in it and show that we get zero so F of two plus square root, too. Substituting that into our function, we would have negative three times two plus square root, two cubed plus eight times two plus square root, two squared, plus 10 times two plus square root too minus eight. Okay, we're going to simplify this, so there's a couple things we want to do off to the side to be able to simplify. We went to Cuba to plus square root to and we went to square two plus square root, too. So let's start with squaring it two plus square root two times itself. When you use the foil technique, you get four class four square root two plus two, so you get six plus four square root, too. Okay, we excuse me. We also want to Cuba. So what if we multiply it by one more two plus square root, too, and then using the oil technique on that, we get 12 for the for the first plus six square root to for the outside, plus eight square root to for the inside, plus eight for the last. And when we combine like terms, we have 20 plus 14 square root, too. All right, so now we're simplifying our F of two plus square root to, and we're hoping to show that it equals zero. So now let's distribute our negative three, and we have negative 60 minus 42 square root, too. Plus, now let's distribute are 8 48 plus 32 square root, too. Now it's distribute the 10 plus 20 plus 10 square root to minus eight. Now let's combine the like terms negative. 42 square root two plus 32 square root, two plus 10 square root to those all add to zero negative. 60 Class 48 Last 20 minus eight. Those all add to zero as well, so we just get zero. So we have shown that the function value is equal to the remainder.

The goal for this problem is to take our function f of X and write it in this form. So that means we need to divide by X minus K and in this case, K's negative square root five. So, using synthetic division, we would have negative square root five outside the box and then we would have the coefficients of our Poland polynomial in a row. So we have one x cubed two x squared, negative five x and negative four. So we bring down the first number one. We multiply it by the negative square root five and we get negative square root. Five. We write that in the next space and we had the column and we have two minus square root five. Then we multiply that by negative square root five and we get negative. Two square root five plus five and we had the column and we get negative. Two square root five. Then we multiply that by negative square root five and we get two times 5 10 and we write that in the next base. Add the Colin and we get six. So the last number is the remainder. The number before that is the constant. The number before that is the coefficient on the X term and the never before that is the coefficient on the X squared term. So that means we can write our function f of X as our divisor. X minus negative square root fi, which is X plus square root fight. Times are quotient one X squared a plus to minus square root five X minus two square root fight, plus our remainder six. The other thing we want to do in this problem is show that the remainder six is equal to the function value. In other words, we want to show that F of negative square root five is actually six, using our original function f of X, so let's go ahead and find F of negative square root. Five. That would be negative. Square root. Five cubed plus two times negative square root. Five squared minus five times negative square root five minus four. Let's simplify that and make sure we get six negative square root. Five. Cubed would be negative. Five square root five. Two times negative square root. Five Squared would be two times five, and that would be 10. Then we have minus five times negative square root five. So that would be plus square root. Clyde find, Square replied. And then we have minus four so we can cancel the minus five square root five and the plus five square root fied. And all we have left is 10 minus four. We just six. So you see, the remainder is equal to the function value.

The goal of this problem is to take our function f of X and right in this core. And so to do that, we need to use synthetic division and divide by X minus k que is negative 2/3. So we're going to put negative 2/3 outside the box and the coefficients of our polynomial f of X in a row. So we have 15 x to the fourth 10 x cubed negative six x squared, zero x We need zero placeholder and 14. Okay, Now for the synthetic division, bring down the 15 multiply it by negative 2/3 and we get negative 10. Right that in the next space. And add the Colin when we get zero, Multiply that by negative 2/3 and we get zero right that in the next space and had the column and we have negative six. Multiply that by nature 2/3 and we get four right that in the next space and how the column and we have four. Multiply that by the negative 2/3 and we get negative 8/3. Add the column and we end up with 34 3rd Okay, That last number is the remainder. The number before that is the constant. The number before that is the coefficient on the X term than ever before. That is a coefficient on the X squared term, and then ever before that is the coefficient on the X cubed turn. So that means our polynomial f of X can be written as X minus negative 2/3 which would be X plus 2/3 times 15 x cubed, minus six X plus four plus remainder 34 3rd Okay. The other thing we want to show for this problem is that the remainder 34 3rd is the function value. In other words, we want to show that f of negative 2/3 is 34 3rd And to show that what we can do is find f of negative 2/3 using direct substitution with the original function, so f of negative 2/3 would be 15 times negative, 2/3 to the fourth power, plus 10 times negative 2/3 to the third power minus six times negative, 2/3 squared plus 14. Okay, so a wise thing to do would be to put that in your calculator. I'm gonna see if I can tough it out and do this in my head. So I want to raise negative 2/3 to the fourth power, and that would be 16 81st. And I want to raise negative 2/3 to the third power, and that would be negative. Eight. 27th. I'm gonna multiply that by the 10. I want to multiply negative 2/3 by itself. And that would be four nights were gonna multiply that by the negative six and then at the 14. Okay, at this point, it would be wise to go ahead and let the calculator help with this type that all in changed a fraction, and you do get 34 3rd So that shows that the function value is equal to the remainder.


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