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# Assessment Lesson 03_09 Factoring Activity

Design an instruction guide explaining the various methods of factoring.
by

## Justin Hancock

on 26 September 2012

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#### Transcript of Assessment Lesson 03_09 Factoring Activity

Factoring: A Guide Greatest Common Factor Difference of Squares and Perfect Trinomials Special Products Factoring Trinomials Four-Term Polynomials and Trinomials Factoring by Grouping Sum and Difference of Cubes Appendices By: Justin Hancock GCF What is a GCF? The greatest common factor (also called greatest common divisor or highest common factor) of two values is the greatest factor that both values share. A factor/divisor is a value that can divide the value evenly, or multiply with another factor to produce the value. e.g. The GCF of 800 and 90 is 10. How do you use GCFs? You use GCFs when you simplify fractions, such as 90/800. To put the fraction in "simplest form", you must divide both numbers by the greatest common factor, 10, resulting in 9/80. GCFs in Factoring When factoring polynomials (including binomials and trinomials), you want to "factor out" the GCF. If you had the binomial 10x^3+25x, the GCF of the binomial would be the GCF of the two terms, 5x. You would then divide 5x from both terms, using the distributive property. The answer would look like this: 5x(2x^2+5). Always try to factor out the GCF of a polynomial before using other methods. For more information on GCFs, see Appendix A. Difference of Squares Whenever you see a binomial (two terms), where both terms are perfect squares, and they are being subtracted, you can factor them easily. For example, b^2-49. Both are perfect squares, and fit a^2-b^2. Whenever you see a binomial like this, remember that a^2-b^2=(a+b)(a-b). The factored form of b^2-49 is (b+7)(b-7). How is the product of two binomials a binomial? When multiplying (b+7)(b-7) using the FOIL method,
you begin with ^2-7b+7b-49. The middle two terms cancel out, leaving the first and last terms. For more information on conjugates and the difference of squares, see Appendix B. Perfect Square Trinomials Perfect square trinomials are trinomials of the form a^2+2ab+b^2 or a^2-2ab+b^2. Perfect square trinomials can always be factored into the square of a binomial. a^2+2ab+b^2=(a+b)^2. a^2-2a+b^2=(a-b)^2. e.g. 9x^2+28x+196=(3x+14)^2 and 9x^2-28x+196=(3x-14)^2. x^2+bx+c When factoring trinomials of the form x^2+bx+c, remember that x^2+bx+c=(x+p)(x+q) where pq=c and p+q=b. In other words, find the factors of c that add to equal b, then write as (x+p)(x+q). p and q In the two binomial factors, p and q can both be positive, both be negative, or differ. To check, see if c is positive. If c is positive and b is positive, both p and q are positive. If c is positive and b is negative, both p and q are negative. If c is negative, either p or q is negative.
For more information on factoring trinomials, see Appendix C. exempli gratia x^6+22x^3+72=(x^3+4)(x^3+18) x^2-9x+8=(x-1)(x-8) x^4+8x^2-20=(x^2-2)(x^2+10) ax^2+bx+c exempli gratia 2x^6+44x^3+144=2(x^3+4)(x^3+18) 5x^2-45x+40=5(x-1)(x-8) 3x^4+24x^2-60=3(x^2-2)(x^2+10) (If a can be factored out) To factor ax^2+bx+c where a can be factored out, first factor out the GCF of the trinomial (a), and then follow the steps detailed in x^2+bx+c. For more information on factoring trinomials, see Appendix C. Four-Term Polynomials To factor four-term polynomials using grouping, first attempt to factor out the GCF from the polynomial, then group similar terms and factor out GCFs from the two sets of terms. Then factor out the common binomial from the two terms.
e.g. 8x^2+40x+6xy+30y=2(x+5)(4x+3y) ax^2+bx+c To factor ax^2+bx+c where a cannot be factored out, first attempt to factor out a GCF, then assume mn=a, pq=c, and pn+mq=b (find the factors of ac that add to equal b). Then rewrite b using the factors, (ax^2+pnx+mpx+c). Then group similar terms and factor as if a four-term polynomial. (If a cannot be factored out) exempli gratia 24x^2-23x+5=(8x-5)(3x-1) 8x^4+18x^2+4=2(4x+1)(x^2+2) 3x^2+8x-3=(x+3)(3x-1) For more information on factoring trinomials, see Appendix C. What is a cube? Similar to perfect squares, a perfect cube is the third power of a number. 27 is a perfect cube, its cube root is 3. For more information on cubes, see Appendix D. How do you factor the sums and differences of cubes? Like the difference of squares is a^2-b^2, the difference of cubes is a^3-b^3.
Assume a^3-b^3=(a-b)(a^2+ab+b^2)
Assume a^3+b^3=(a+b)(a^2-ab+b^2)
Substitute values appropriately. exempli gratia x^3-343=(x-7)(x^2+7x+49) x^6+64=(x^2+4)(x^4-4x^2+16) Appendix A GCFs If two values share a GCF of 1, they are considered "coprime". To compute GCFs of values, one can use prime factorization (to write as a product of prime numbers), to find common primes and then multiply, or one can use Euclid's algorithm which is:
gcd(a, 0)=a
gcd(a, b)=gcd(b, a mod b)
"mod" is the Modulo operation, which finds the remainder after division of two values. The modulo operation can be simplified to a-b(floor a/b)
Floor meaning floor function (the largest integer less than the value, normally represented by brackets without the top horizontals) Appendix B Differences of Squares Differences of squares show the commutative property through ba-ab=0. One can factor the sum of squares when you take the imaginary number into account. a^2+b^2=(a+bi)(a-bi). The imaginary term makes the product of bi and -bi positive. Another usage of conjugates is the rationalization of irrational denominators. To remove a binomial with a radical in the denominator, simply multiply the denominator and numerator by the conjugate of the denominator, which squares both terms of the binomial, removing the radical. Appendix C Factoring Trinomials To determine if a trinomial is factorable over the integers, use the discriminant (b^2-4ac). If the discriminant of a trinomial is a perfect square, then it is factorable over the integers. Appendix D Cubes Waring's problem: every integer can be written as a sum of nine or fewer positive cubes. Fermat's last theorem: x^3+y^3=z^3 has no non-trivial (non-zero) integer solutions. 1^3+2^3...+n^3=(1+2...+n)^2=(n(n+1)/2)^2
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