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How to Find Real Zeros of Any Polynomial Function 

Mr H Tutoring
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By finding the potential zeros and checking, we can find one of the real zeros. (If it exists) Depending on the degree of the function, we can use the same method to find the 2nd zero or factor the function to find the rest of the zeros.
The lesson also includes a quick lesson on using the synthetic division to check if a number is a zero of the function.

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28 сен 2024

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Комментарии : 15   
@KartikHina
@KartikHina 14 дней назад
Make a series on calculs please sir
@jacobbelury3417
@jacobbelury3417 День назад
This was in algebra 2 for like 1 week barely a day lmao
@charlesrobinson3977
@charlesrobinson3977 2 дня назад
You say this method works for "any polynomial function", but don't you mean "any polynomial function with integral coefficient"?
@Mathgoddesssupports
@Mathgoddesssupports 14 дней назад
❓🙋‍♂️❓Why in the synthetic division do you multiply and then *add*? I see that it works but don’t understand why it does since division generally involves multiplying and then subtracting. TIA for anyone’s clarification.
@Pankaj_Kumar_2008
@Pankaj_Kumar_2008 14 дней назад
The answer is very simple my friend. Cause in the first case when you divide a polynomial, you use a divisor which is a factor of the polynomial and then multiply and subtract. However when it comes to the Synthetic Division, you do actually multiply it by the inverse of that factor, for that reason you add them. Let me elaborate, 1. For Long Division, we take the divisor as (x-a). 2. For Synthetic Division, we take x=a. Clearly both of them are additive inverse of each other, so that's why we add at one whereas subtract to the other to have them in Equilibrium. It went too long I guess 😅😅😅
@carultch
@carultch 14 дней назад
What synthetic division does, is replace operations such as subtracting a negative, with adding a positive. It recognizes the self-cancellation of the negative signs, and replaces it with the more intuitive operation of simple addition, and simple multiplication. It makes it so you don't need to guess terms, but sets them up in a more straight-forward method of calculating them
@steveschmidt5156
@steveschmidt5156 14 дней назад
Outstanding. Thank you.
@davidnewell3232
@davidnewell3232 14 дней назад
Mind your "p"s and "q"s.
@NameFirst-jv9gj
@NameFirst-jv9gj 14 дней назад
😂😂😂
@anestismoutafidis4575
@anestismoutafidis4575 5 часов назад
2•1^3 +11•1^2- 7•1 -6 =0 First zero-number=1 (2x-7/x)•(x^2+5,5x)+32,5 => (2x-7/x) 2•1,87 -7/1,87=3,74 -3,74=0 Second zero-number: 1,87 (x^2+5,5x) (-5,5)^2 +5,5 • (-5,5)=0 Third zero-number= -5,5 Zero numbers{ -5,5; 1; 1,87}ℝ
@gregnixon1296
@gregnixon1296 13 дней назад
My school system kicked the rational root theorem to the curb last year. Sad days.
@josephshaff5194
@josephshaff5194 13 дней назад
👍
@geremymuccleswood307
@geremymuccleswood307 14 дней назад
imagine p is 100 and q is 16…
@carultch
@carultch 14 дней назад
In that case, your candidates for roots would be: 1/16, 1/8, 1/4, 1/2, 5/16, 5/8, 5/4, 5/2, 1, 2, 4, 5, 10, 20, and 100 And the negatives of all of the above. Usually, you try to use other clues as well, such as Descartes' rule of signs, and the sum and product of the roots, which can be directly determined from the b-term, the final constant term, the a-term, and whether the degree of the polynomial is odd or even. This allows you to narrow down your search for possible roots. Polynomial roots in general, will add to -b/a, and will multiply to k/a for even-ordered polynomials, and multiply to -k/a for odd-ordered polynomials.
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