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Topic: **Polynomial functions homework****Question:**
I do not know how to put the legendre polynomial equation (can be founded on wikipedia) onto MATLAB. I understand that they have legendre(x,n) but we have to create a function ourselves for homework.

June 17, 2019 / By Brook

Here are some helpful information: http://www.mathworks.in/matlabcentral/newsreader/view_thread/311250

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Hi, On every type of factoring problem ALWAYS look first for a GCF, a greatest common factor. There won't always be one, but if there is, you should divide it out first. A GCF will carry along through the problem and be the first thing in an answer with a GCF. Now we are going to look at problems with only 2 terms. Even on these, always check first for a GCF. Then the problems could be the difference of perfect squares, the difference of perfect cubes, or the sum of perfect cubes. First we will look at the difference of perfect squares. Suppose you had 4x² - 81 This is the difference of perfect squares. 4x² is really (2x) squared. 81 is 9². So both factors start with 2x and both end with 9. One binomial has a "-" and the other has a "+". The factors are (2x - 9)(2x + 9). Suppose you had 16a² - 49 This is the difference of perfect squares. 16a² is really (4a) squared. 49 is 7². So both factors start with 4a and both end with 7. One binomial has a "-" and the other has a "+". The order does not matter. The factors are (4a + 7)(4a - 7). Suppose you had 18x² - 50 18 and 50 are not perfect squares because no number times itself equals either 18 or 50. The problem is that we did not look first for a GCF! This time both terms are divisible by 2. If we divide that out first, we get: 2(9x² - 25) Now we have a difference of perfect squares! Keep the GCF of 2 out front. The parentheses has a difference of perfect squares. 9x² is really (3x) squared. 25 is 5². So both factors start with 3x and both end with 5. One binomial has a "-" and the other has a "+". The factors are 2(3x - 5)(3x + 5). Suppose you had 16x^4 - 81 This is the difference of perfect squares. 16x^4 is really (4x²) squared. 81 is 9². So both factors start with 4x² and both end with 9. One binomial has a "-" and the other has a "+". The order does not matter. (4x² + 9)(4x² - 9) This is correct so far, but it is not done this time. Notice that with x² in terms with 4 and 9, that they are perfect squares again, so the factor that is the DIFFERENCE of perfect squares can factor again. (The sum of perfect squares can not factor.) So (4x² - 9) factors again into: (4x² + 9)(4x² - 9) (4x² + 9)(2x - 3)(2x + 3) <== This is the answer in factored form. Now we're going to look at the sum and difference of perfect cubes. Again always look first for a GCF. If there is one, factor it out. Perfect cube problems will always factor into a binomial times a trinomial, like this: x³ + 125 = (x + 5)(x² - 5x + 25) The general rule for these is: 1) Factor out a GCF if there is one. 2) Make 2 parentheses for a binomial and a trinomial. 3) The sign from the problem goes in the binomial. The opposite sign from what you just used goes in the trinomial, and is always followed by a "+" as the second sign in the trinomial. This is the only step where what you do is different on the sum of cubes from on the difference of cubes. 4) Figure out what you would cube to get the first term of the problem and put that expression as the first term of your binomial. Then figure out what you would cube to get the second term of the problem and put that expression as the second term of your binomial. 5) Your trinomial already has its signs, so don't worry about the signs. To find terms of the trinomial, write the first term of your binomial 3 times followed by the second term of your binomial 3 times out to the side. Multiply the first 2 things together to get the trinomial's first term. Multiply the next 2 things together to get the trinomial's middle term. Multiply the last 2 things together to get the trinomial's third term. So for, x³ + 125, there is no GCF. Parentheses would be ( + )( - + ) To get x³, you'd cube x. To get 125, you'd cube 5. So your binomial becomes (x + 5)( - + ) Now write x x x 5 5 5 x*x = x² x*5 = 5x 5*5 = 25 So your trinomial becomes (x + 5)(x² - 5x + 25) Here's another example: So for 8x³ + 27y³, there is no GCF Parentheses would be ( + )( - + ) To get 8x³, you'd cube 2x. To get 27y³, you'd cube 3y. So your binomial becomes (2x + 3y)( - + ) Now write 2x 2x 2x 3y 3y 3y 2x*2x = 4x² 2x*3y = 6xy 3y*3y = 9y² So your trinomial becomes (2x + 3y)(4x² - 6xy + 9y²) I' m going to explain factoring trinomials the way that I teach it, which is probably different from what you've done. Try to follow the steps. This method will work on any trinomial. 3y² - 13y - 10 Look for a GCF None this time. If there was one, factor it out. Then temporarily start both parentheses with the first number and variable. (3y.......)(3y..........) First sign goes in first parentheses. (3y..-....)(3y..........) Product of signs goes in 2nd parentheses. (3y..-....)(3y...+.....) <== plus is because neg x neg = positive Now multiply your first and third numbers together. Ignore their signs - you've already done them. 3 x 10 = 30 So, out to the side list pairs of factors of 30. ± 30 ------ 1, 30 2,

Make me your contact, and look at some of my answers. I write very thorough answers, and they have examples already, which will save you time of writing or finding an example. or, I can copy and paste one or two... Question: Would anyone help me please? quadratic formula. x^2-3x-3= -5 Answer: Sure. The first thing you want to do is add 5 to both sides to zero it out. x^2-3x-3= -5 add 5: x^2-3x+2 Ok, let's see what we know so far. The x^2 doesn't have a coefficient, a number in front of it, so we know it'll be (x )(x ). The last number, positive 2, we get by multiplying, and since the middle term is a negative, we know we have to multiply two negatives to get the positive 2, so now we know we have (x - )(x - ). Now we need two numbers that when multiplied give us positive 2, and when added/subtracted give us negative 3. Those numbers must be -2 and -1, those are the only factors 2 has. So, the factors of your equation are: (x-2)(x-1). Question: Factor the polynomial x2 + 2x – 15? Answer: Ok, let' break it down to what we know so far. There isn't a coefficient, or a number in front of the x^2, so we know that we have (x )(x ). The signs are + and -, and we know that we multiply to the the last number. The last number is -15, and in order to get a negative number by multiplying, we have to multiply a negative and a positive. So now, we know we have (x+ )(x- ). Now we need to find two numbers that when multiplied give us -15, and when added/subtracted give us a positive 2. I think the numbers are 5 and 3, but where do we put the 5 and where do we put the 3? Since the middle factor, 2x, is positive, we want the larger factor of 15 to go with the + sign, so your polynomial, x^2+2x-15 is: (x+5)(x-3). wpf. wpf.

well if you have a trinomial like x squared + 10x + 24 you know that X times X is x squared so start of with this: (x ) (x ) since everything is positive in the trinomial, your factored answer will be (x + ?) (x + ?) Now what you have to do is figure out which two number will equal 24 when multiplied together and will also add up to 10 when you add them together. If you think for a minute, you will realize the the two numbers are 6 and 4. so your answer now is (x + 6) (x + 4) The trinomial is now factored. email me if you have any other questions. hope this helped. emai: [email protected]

Try the rational roots test. The possible rational roots of a polynomial are all the factors of the constant term positive and negative divided by all the factors of the leading coefficient positive and negative. For example: 2x^3+x^3-6x^2+12 The middle two terms are irrelevant. The factors of 12 are: +/- 1,2,3,4,6,12 (okay maybe should've chosen a simpler term) The factors of 2 are: +/- 1,2 Divide every factor of 12 into each factor of two to get: +/- 1,2,3,4,6,12 and +/- 1/2,1,3/2,2,3,6 +/- 1/2,1,3/2,3,4,2,6,12 are the possible rational roots. Then you can use synthetic division to see if the rational roots work. If none of them work, then the roots are either irrational or complex.

Let's say that you have two vectors, x and y, which contains all of the coordinates x = [1 2 3 4 5 7 9] y = [3 5 1 8 0 7 11] you can calculate your formulas very easily... I assume that a0, a1 and N are known scalar constants. sum_y = a0*N + a1 * sum(x) sum_xy = a0 * sum(x) + a1*sum(x.^2) I don't know what you want to do with sum_y and sum_xy or what they represent, but that's the way to calculate them according to your formulas. Try it with your own numbers.

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i dont get your question.... why u want to use a loop in a linear rgression? its usually solution=A/b; where A is the Jacobian matrix of the system...

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