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Factorization

completing the square

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Algebra
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Mathematics
Elementary algebra
Algebra
Completing the square
Quadratic function
Factorization
Polynomial expansion
Partial fraction decomposition

Section 4.5: Vertex form with Completing the Square Completing the Square: Used to change quadratic functions in standard form to vertex form. f (x) = ax 2 + bx + c - f(x)=a(x?h) 2 +k ------------------------ Investigation This creates a Perfect Square 2 (b)2 Tririomiai, which factors To complete the square for an expression x + bx, add ------ Startwithx2 + bx Find (b)2 x2 +bx+() (x+ )2 / The Completed Square Factored form x2 +6x - x2 +4x 2 X - 8x _) - X - 2x oF W4CX 1rrrt! You Try?Complete the Square for the following. Then write the factored form. 1)x2 +14x 2)y2 +20y 2t\2S io) IOD (xiiJJ _ 0) 3)m2 -IOM L Omtn - 15 ) ;4 6 7 4)h2 -15h '4 P7 L_'1-' & ? **How do we go from standard form of mAquation to vertex form?

12

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Algebra
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Factorization
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Ruffini's rule

Nishat? ?Ahmed ? ?? ?? ?? ?? ?Test? ?Date:? ?9/19/17 TEST? ?1? ?REVIEW? ?SHEET - A? ?polynomial? ?written? ?in? ?standard? ?form? ?has? ?the? ?greatest? ?exponent? ?in? ?the? ?beginning? ?and? ?the constant? ?(or? ?the? ?smallest? ?exponent)? ?at? ?the? ?end.? ?Therefore,? ?the? ?exponents? ?are? ?organized? ?in descending? ?manner. - The? ?degree? ?of? ?a? ?polynomial? ?is? ?equivalent? ?to? ?the? ?value? ?of? ?the? ?greatest? ?exponent? ?in? ?the polynomial. - The? ?leading? ?coefficient? ?in? ?a? ?polynomial? ?is? ?the? ?coefficient? ?of? ?the? ?variable? ?with? ?the? ?greatest exponent. - The? ?constant? ?in? ?a? ?polynomial? ?does? ?not? ?have? ?a? ?variable,? ?it? ?simply? ?a? ?number.

Zsigmondy

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Calculus
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Fields of mathematics
Mathematics
Algebraic number theory
Number theory
Integer sequences
Zsigmondy's theorem
Factorization
Prime number
XTR
Fermat's Last Theorem
Coprime integers
Fermat's theorem on sums of two squares

The Zsigmondy Theorem PISOLVE 7/31/11 Abstract In this paper we will be discussing the uses and applications of Zsigmondy?s Theorem. As some of you may consider the theorem as ?large/ brutal? , it can actually be proved by elementary methods [1] - at the same time, it is applicable in so many Number Theoretic problems(All problems in this article are sourced in AoPS). We see no reasons not to use it. Intro Problem Let p > 3 be a prime. Show that every positive divisor of 2 p+1 3 is in the form 2kp+ 1. Solution: We show that all prime divisors are in this form, then the result readily follows.Let q| 2 p+1 3 . Then: q|22p ? 1 o2(q)|2p If o2(q) 6= 2p, we have 3 cases: o2(q) = 1 Then q|1, clearly impossible. o2(q) = 2

Steps for Factoring Polynomials

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1. Pull out greatest common factor or GCF [This is a very important step that is often skipped] Ex: or 2. Look at number of terms a. 2 Terms: If there are two terms (binomial), then check to see if it is i. A difference of squares: Ex: ii. A difference of cubes Ex: iii. A sum of cubes Ex: If there are two terms and the expression is none of the above, it is prime! b. 3 Terms: If there are three terms (trinomial), then factor into two binomials thinking of the FOIL method in reverse. Find factors of c that add up to b. Start by identifying the factors of c. The sign before c determines whether signs are the same or different in the 2 binomials that are produced, .

Factoring Polynomials of Degree 3

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Factoring

Factoring Polynomials of Degree 3 Factoring a 3 - b 3 An expression of the form a 3 - b 3 is called a difference of cubes. The factored form of a 3 - b 3 is (a - b)(a 2 + ab + b 2) : (a - b)(a 2 + ab + b 2) = a 3 - a 2 b + a 2 b - ab 2 + ab 2 - b 3 = a 3 - b 3 For example, the factored form of 27x 3 - 8 ( a = 3x, b = 2 ) is (3x - 2)(9x 2 + 6x + 4) . Similarly, the factored form of 125x 3 -27y 3 ( a = 5x, b = 3y ) is (5x - 3y)(25x 2 +15xy + 9y 2) . To factor a difference of cubes, find a and b and plug them into (a - b)(a 2 + ab + b 2) . Factoring a 3 + b 3 An expression of the form a 3 + b 3 is called a sum of cubes. The factored form of a 3 + b 3 is (a + b)(a 2 - ab + b 2) : (a + b)(a 2 - ab + b 2) = a 3 + a 2 b - a 2 b - ab 2 + ab 2 + b 3 = a 3 - b 3 .
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