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- to be able to provide our viewers of the information they needed regarding factoring in Intermediate Algebra subject in friendly approach.
- to change negative perspective of mathematics into positive.
- to develop the mathematical skills of the viewers and motivate them.
- help teachers in discussing different factoring methods by the use of media
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CONCLUSION AND GENERALIZATION
Before factoring any polynomial, write the polynomial in descending order of one of the variables. Then note how many terms there are, and proceed by using one or more of the following techniques.
1. ALWAYS Factor out the Greatest Common Factor (GCF) first. Look for this in every problem. This includes factoring out the negative sign if it proceeds the leading term.
Example: − x2 + 6x − 3 = −1(x2 − 6x + 3)
Example: 4x3 y5 − 8x 2y 3 = 4x 2 y 3(xy2 − 2) where 4x 2 y 3 was the GCF.
2. If there are FOUR TERMS, try to factor by grouping (GR). Group two terms at a time, and factor out the greatest common factor from each group.
Example:
X3+6 x-2X-12 group the first two terms then the last two terms
X2(X+6)-2(X+6) factor the x+6 out of both terms
(x+6)(X2-2) this is the factored answer
3. If there are TWO TERMS, look for one of these patterns:
a. The difference of squares (DOS) factors into conjugate binomials (conjugate means
terms are separated by a plus sign in one binomial and a minus sign in the other binomial):
a2-b2= (a-b)(a+b)
Example: 9x4− 64y2= (3x2 − 8y)(3x2 + 8y)
Note: a variable is a perfect square if the exponent is even
b. The sum of squares does not factor: a2 + b2 is prime (doesn’t factor)
Example:
9x4 + 64y 2 does not factor because it is the SUM of squares
c. The sum of cubes (SOC) or difference of cubes (DOC) factors by these patterns:
(each type contains a binomial times a trinomial).
We came across these before:
[Sum of two cubes]: x3 + y3 = (x + y)(x2 − xy + y2)
[Difference of 2 cubes]: x3 − y3 = (x − y)(x2 + xy + y2)
Factor 64x3 + 125
Answer:
We use the Sum of 2 Cubes formula above.
64x3 + 125
= (4x)3 + (5)3
= (4x + 5)[(4x)2 − (4x)(5) + (5)2]
= (4x + 5)(16x2 − 20x + 25)
4. If there are THREE TERMS, look for these patterns:
a. Quadratic trinomials of the form ax2 + bx + c where a = 1 (QT a = 1) factor intothe product of two binomials (double bubble) where the factors of c must add to b.
Example: x2 − 4x −12 = (x − 6)(x + 2)
b. Quadratic trinomials of the form ax2 + bx + c where a ≠ 1 (QT a ≠ 1) eventually
factor into the product of two binomials (double bubble), but you must first find
the factors of ac that add to b, rewrite the original replacing b with these factors of ac, then factor
by grouping to finally get to the double bubble.
The Distributive Law is used in reverse to factorise a quadratic trinomial, as illustrated below.
Consider the expansion of (X+2) (X+3)
(X+2) (X+3)= X(X+3)+2(X+3)
= X2+5X+6
c. Quadratic square trinomials (QST) of the form ax2 + bx + c may factor into the
square of a binomial. Look for the pattern where two of the terms are perfect
squares, and the remaining term is twice the product of the square root of the
squares: a2 ± 2ab ± b2 = (a ± b)2
Example: 16x2 − 40x + 25 = (4x − 5)2
Note the pattern: Square root of 16x2 is 4x. Square root of 25 is 5.
Twice the product of the square roots: 2(4x)(5)=40x, which is the middle term
5. Factor all expressions completely. Sometimes, you will need to use two or three types of
factoring in a single problem.
Given:
|
5x2 + 11x + 2
|
Find the product ac:
|
(5)(2) = 10
|
Think of two factors of 10 that add up to 11:
|
1 and 10
|
Write the 11x as the sum of 1x and 10x:
|
5x2 + 1x + 10x + 2
|
Group the two pairs of terms:
|
(5x2 + 1x) + (10x + 2)
|
Remove common factors from each group:
|
x(5x + 1) + 2(5x + 1)
|
Notice that the two quantities in parentheses are now identical. That means we can factor out a common factor of (5x + 1):
|
(5x + 1)(x + 2)
|
RELATED ARTICLES
In number theory, integer factorization or prime factorization is the decomposition of a composite number into smaller non-trivial divisors, which when multiplied together equal the original integer.
When the numbers are very large, no efficient, non-quantum
integer factorization algorithm is
known; an effort concluded in 2009 by several researchers factored a 232-digit
number (RSA-768),
utilizing hundreds of machines over a span of 2 years.[1] The presumed difficulty of this problem
is at the heart of certain algorithms in cryptography such
as RSA. Many areas of mathematics and computer
science have been brought to bear on the problem, including elliptic
curves, algebraic number theory, and quantum
computing( Lance Fortnow 2002-09-13).
Not all numbers of a given length are equally hard to factor. The hardest instances of these problems (for currently known techniques) are semiprimes, the product of two prime numbers. When they are both large, for instance more than 2000 bits long, randomly chosen, and about the same size (but not too close, e.g. to avoid efficient factorization by Fermat's factorization method), even the fastest prime factorization algorithms on the fastest computers can take enough time to make the search impractical; that is, as the number of digits of the primes being factored increases linearly, the number of operations required to perform the factorization on any computer increases exponentially. (David Bressoud and Stan Wagon (2000).
Many cryptographic protocols are based on the difficulty of factoring large composite integers or a related problem, the RSA problem. An algorithm that efficiently factors an arbitrary integer would render RSA-based public-key cryptography insecure.( Kleinjung, et al 2010-02-18).
In order to factor a polynomial like f(x) = x3+ 2x – 3, you first need to be able to divide the polynomial by a factor. Once you divide by a factor, you can rewrite f(x) as the product of your divisor times the quotient obtained. So first, we must review polynomial division.
"Factors" are the numbers you multiply to get another number. For instance, the factors of 15 are 3 and5, because 3×5 = 15. Some numbers have more than one factorization (more than one way of being factored). For instance, 12 can be factored as 1×12, 2×6, or 3×4. A number that can only be factored as 1 times itself is called "prime". The first few primes are 2, 3, 5, 7, 11, and 13. The number 1 is not regarded as a prime, and is usually not included in factorizations, because 1 goes into everything. (The number 1 is a bit boring in this context, so it gets ignored.)
You most often want to find the "prime factorization" of a number: the list of all the prime-number factors of a given number. The prime factorization does not include 1, but does include every copy of every prime factor. For instance, the prime factorization of 8 is 2×2×2, not just "2". Yes, 2 is the only factor, but you need three copies of it to multiply back to 8, so the prime factorization includes all three copies.( http://www.purplemath.com/modules/factnumb.htm)
Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent of computers and the resultant development of error-correcting codes, factoring over finite fields (e.g., Zp, for p a prime number) has become important as well. To understand this discussion, you need to know what polynomials are, and how to add, subtract, multiply and divide them. Many of the theorems below will be familiar, but you may not have seen the proofs. Some may be new. All of them have been useful to various mathematicians. This is not an exhaustive list, but hopefully enough to give you an idea of the variety of information that can be gleaned and used. A field is a set F with two operations, usually denoted + and , such that F is an abelian group under the operation + with identity 0, F \ {0} is an Mabelian group under the operation with identity 1, and the distributive law, a(b+c) = ab+ac for all a, b, c ∈ F , holds. For example, the rational numbers, the real numbers, the complex numbers, and Zp, for p a prime number, are all fields. (For Zp, recall that x ∈ Zp has a multiplicative inverse (or generates the units) if and only if x = 0. So 6 Zp \ {0} is an abelian group, whenceMZp is a field.) So for this dicussion F will always denote a field, and F [x] the ring of polynomials with coefficients in the field F . Similarly, Z[x] is the ring of polynomials with integer coefficients. We call F or Z the ground ring. You need not know what a ring is to understand what follows, but for completeness, a commutative (which all of ours are) ring is a set R with two operations, usually denoted + and , such that F is an abelian group (1997–2003 Susan C. Geller).
Under the operation + with identity 0, is a commutative, associative binary operation, and the distributive law holds. From high school algebra we know that the set of polynomials is a commutative ring which has a multiplicative identity 1.
Recall that, for integers a, b with b = 0, there exist unique integers 6 q, rso that a = qb + r where 0 ≤ r < |b|, i.e., the division algorithm. Also recall Nthat a polynomial f(x) = anx n + an−1x n−1+ a0 where an = 0 is said 6 to have degree n. Note that the zero polynomial does not have a degree, so degree 0 polynomials are non-zero “constants”, i.e., are non-zero elements of he ground ring. So we can use the idea of the division algorithm on the degree of the polynomials to get a similar theorem for polynomials.
Theorem 1 Division Algorithm: Let f, g be polynomials with rational (or real or complex or any other field) coefficients where g = 0 6 . Then there exist unique polynomials q, r with coefficients in the same field as f and g so that f = qg + r where r = 0 or deg(r) < deg(g).
Proof: We start with existence. Let f(x) = anx n + an−1x n−1 + a0 and g(x) = bmx m + bm−1x m−1 + + b0 be polynomials where an = 0 and 6 bm = 0. (Note: If 6 f = 0, 0 = 0g + 0, so the 1`assumption that an = 0 6 is okay.) We proceed by induction on the degree of f. If n = 0, then either f = c = (cd −1 )g, where g = d has degree 0 or f = 0g + f where degree(g) > 0 satisfies the conditions. Assume that, if n < k, then there exist polynomials q, r with coefficients in the same field as f so that f = qg + r where r = 0 or deg(r) < deg(g). Now assume that the degree of Mf is k. If k < m, then f = 0g + f satisfies the conditions. If m ≤ n, then f(x) − ak(b −1 m )x n−m g(x) has degree at most k − 1 < k. So there exist q1, r so that f(x) − ak(b −1 m )x n−m g(x) = q1(x)g(x) + r(x) where r = 0 or deg(r) < deg(g). Thus f(x) = (akb −1 m x n−m + q1(x))g(x) + r(x). Letting q(x) = akb −1 m x n−m + q1(x), we see that f = qg + r with the coeffiecients of q, r in the same field as f, as required. Now suppose that f = pg + s where s = 0 or deg(s) < deg(g). Then 0 = (p − q)g + s − r. So (p − q)g = r − s. If r − s = 0 (so 6 p =6 q), then deg(r − s) < deg g ≤ deg((p − q)g). Contradiction. Therefore, r = s, and pg = qg. Since g is not a zero-divisor, p = q. 2NOTE: When we wrote b −1 m we used the fact that the coefficients came from a field, i.e., that all non-zero coefficients had inverses. We could just have easily allowed any coefficient ring (such as Z) and insisted that the leading coefficient, bm, of g be invertible (or ±1 in the case of Z).
We now turn to a special case of the division algorithm, that of the divisor having degree one. This case helps us prove many of the theorems we used in high school algebra, especially the correspondence between roots of a polynomial and factors of that polynomial, and the fact that a polynomial of degree n has at most n roots in the ground field. Corollary 1 Remainder Theorem: Let f be a polynomial with coeffi- cients in a field or in the integers or in any ring. Let a be a number in the ground ring. Then there exists a polynomial q with coefficients in the same field or ring as f such that f = (x − a)q + f(a).
Proof:Since the leading coefficient of x − a is 1, we may apply theDivision Algorithm to f(x) and (x − a) and get that f = q(x − a) + r where the coefficients of q, r are in the same field or ring as those of f and either r = 0 or deg r < deg(x − a) = 1. So r(x) is a constant. Evaluating at a, we get f(a) = q(a)(a − a) + r = r. By the uniqueness part of the Division Algorithm, f(x) = (x − a)q(x) + f(a).
Not all numbers of a given length are equally hard to factor. The hardest instances of these problems (for currently known techniques) are semiprimes, the product of two prime numbers. When they are both large, for instance more than 2000 bits long, randomly chosen, and about the same size (but not too close, e.g. to avoid efficient factorization by Fermat's factorization method), even the fastest prime factorization algorithms on the fastest computers can take enough time to make the search impractical; that is, as the number of digits of the primes being factored increases linearly, the number of operations required to perform the factorization on any computer increases exponentially. (David Bressoud and Stan Wagon (2000).
Many cryptographic protocols are based on the difficulty of factoring large composite integers or a related problem, the RSA problem. An algorithm that efficiently factors an arbitrary integer would render RSA-based public-key cryptography insecure.( Kleinjung, et al 2010-02-18).
In order to factor a polynomial like f(x) = x3+ 2x – 3, you first need to be able to divide the polynomial by a factor. Once you divide by a factor, you can rewrite f(x) as the product of your divisor times the quotient obtained. So first, we must review polynomial division.
"Factors" are the numbers you multiply to get another number. For instance, the factors of 15 are 3 and5, because 3×5 = 15. Some numbers have more than one factorization (more than one way of being factored). For instance, 12 can be factored as 1×12, 2×6, or 3×4. A number that can only be factored as 1 times itself is called "prime". The first few primes are 2, 3, 5, 7, 11, and 13. The number 1 is not regarded as a prime, and is usually not included in factorizations, because 1 goes into everything. (The number 1 is a bit boring in this context, so it gets ignored.)
You most often want to find the "prime factorization" of a number: the list of all the prime-number factors of a given number. The prime factorization does not include 1, but does include every copy of every prime factor. For instance, the prime factorization of 8 is 2×2×2, not just "2". Yes, 2 is the only factor, but you need three copies of it to multiply back to 8, so the prime factorization includes all three copies.( http://www.purplemath.com/modules/factnumb.htm)
Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent of computers and the resultant development of error-correcting codes, factoring over finite fields (e.g., Zp, for p a prime number) has become important as well. To understand this discussion, you need to know what polynomials are, and how to add, subtract, multiply and divide them. Many of the theorems below will be familiar, but you may not have seen the proofs. Some may be new. All of them have been useful to various mathematicians. This is not an exhaustive list, but hopefully enough to give you an idea of the variety of information that can be gleaned and used. A field is a set F with two operations, usually denoted + and , such that F is an abelian group under the operation + with identity 0, F \ {0} is an Mabelian group under the operation with identity 1, and the distributive law, a(b+c) = ab+ac for all a, b, c ∈ F , holds. For example, the rational numbers, the real numbers, the complex numbers, and Zp, for p a prime number, are all fields. (For Zp, recall that x ∈ Zp has a multiplicative inverse (or generates the units) if and only if x = 0. So 6 Zp \ {0} is an abelian group, whenceMZp is a field.) So for this dicussion F will always denote a field, and F [x] the ring of polynomials with coefficients in the field F . Similarly, Z[x] is the ring of polynomials with integer coefficients. We call F or Z the ground ring. You need not know what a ring is to understand what follows, but for completeness, a commutative (which all of ours are) ring is a set R with two operations, usually denoted + and , such that F is an abelian group (1997–2003 Susan C. Geller).
Under the operation + with identity 0, is a commutative, associative binary operation, and the distributive law holds. From high school algebra we know that the set of polynomials is a commutative ring which has a multiplicative identity 1.
Recall that, for integers a, b with b = 0, there exist unique integers 6 q, rso that a = qb + r where 0 ≤ r < |b|, i.e., the division algorithm. Also recall Nthat a polynomial f(x) = anx n + an−1x n−1+ a0 where an = 0 is said 6 to have degree n. Note that the zero polynomial does not have a degree, so degree 0 polynomials are non-zero “constants”, i.e., are non-zero elements of he ground ring. So we can use the idea of the division algorithm on the degree of the polynomials to get a similar theorem for polynomials.
Theorem 1 Division Algorithm: Let f, g be polynomials with rational (or real or complex or any other field) coefficients where g = 0 6 . Then there exist unique polynomials q, r with coefficients in the same field as f and g so that f = qg + r where r = 0 or deg(r) < deg(g).
Proof: We start with existence. Let f(x) = anx n + an−1x n−1 + a0 and g(x) = bmx m + bm−1x m−1 + + b0 be polynomials where an = 0 and 6 bm = 0. (Note: If 6 f = 0, 0 = 0g + 0, so the 1`assumption that an = 0 6 is okay.) We proceed by induction on the degree of f. If n = 0, then either f = c = (cd −1 )g, where g = d has degree 0 or f = 0g + f where degree(g) > 0 satisfies the conditions. Assume that, if n < k, then there exist polynomials q, r with coefficients in the same field as f so that f = qg + r where r = 0 or deg(r) < deg(g). Now assume that the degree of Mf is k. If k < m, then f = 0g + f satisfies the conditions. If m ≤ n, then f(x) − ak(b −1 m )x n−m g(x) has degree at most k − 1 < k. So there exist q1, r so that f(x) − ak(b −1 m )x n−m g(x) = q1(x)g(x) + r(x) where r = 0 or deg(r) < deg(g). Thus f(x) = (akb −1 m x n−m + q1(x))g(x) + r(x). Letting q(x) = akb −1 m x n−m + q1(x), we see that f = qg + r with the coeffiecients of q, r in the same field as f, as required. Now suppose that f = pg + s where s = 0 or deg(s) < deg(g). Then 0 = (p − q)g + s − r. So (p − q)g = r − s. If r − s = 0 (so 6 p =6 q), then deg(r − s) < deg g ≤ deg((p − q)g). Contradiction. Therefore, r = s, and pg = qg. Since g is not a zero-divisor, p = q. 2NOTE: When we wrote b −1 m we used the fact that the coefficients came from a field, i.e., that all non-zero coefficients had inverses. We could just have easily allowed any coefficient ring (such as Z) and insisted that the leading coefficient, bm, of g be invertible (or ±1 in the case of Z).
We now turn to a special case of the division algorithm, that of the divisor having degree one. This case helps us prove many of the theorems we used in high school algebra, especially the correspondence between roots of a polynomial and factors of that polynomial, and the fact that a polynomial of degree n has at most n roots in the ground field. Corollary 1 Remainder Theorem: Let f be a polynomial with coeffi- cients in a field or in the integers or in any ring. Let a be a number in the ground ring. Then there exists a polynomial q with coefficients in the same field or ring as f such that f = (x − a)q + f(a).
Proof:Since the leading coefficient of x − a is 1, we may apply theDivision Algorithm to f(x) and (x − a) and get that f = q(x − a) + r where the coefficients of q, r are in the same field or ring as those of f and either r = 0 or deg r < deg(x − a) = 1. So r(x) is a constant. Evaluating at a, we get f(a) = q(a)(a − a) + r = r. By the uniqueness part of the Division Algorithm, f(x) = (x − a)q(x) + f(a).
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