FACTORING
Did somebody say reverse FOIL ? Oh -­‐ don’t forget about that common monomial What is this I hear about key numbers READ ON and Have some Fun Really ! Some pracGce problems for you to try Remember to try to solve each step before you click your mouse Given the following polynomial factor out the common monomial term Click one Gme for the next step Click one more Gme for the Answer FACTOR 4x5 + 6x3 + 6x2 + 9 Step 1 Combine like term ( there are none) Step 2Check for a common monomial ( there is none) Step3 Try to group the polynomial into two binomials such that each binomial has a common monomial Step 4 Factor the common monomial out of each binomial Step 5 If the resulGng binomials are the same -­‐factor it out. If the given polynomial factors the result will be the product of two binomials Click one Gme for the results of step 3&4 next step Click one more Gme for the Answer 2x³(2x²+3) +3(2x²+3) (2x³ + 3)(2x² + 3) Factor x² + 11x +10 Your answer will be the product of two binomials. In a way this is FOIL in reverse. You start by wriGng ( __ __ )(__ __ ) Now you fill in the four missing terms. The answers to the following quesGons will help. What Gmes what is x² and what Gmes what is 10 such that the mulGplicaGon of the outside terms added to the mulGplicaGon of the inside terms will result in 11x Procedure -­‐-­‐ Hum -­‐Trial and Error -­‐-­‐ or Success Hint: if the leading coefficient is 1 than all you need to do is find two numbers such that when you mulGply them you get 10, the last term , and when you add them you get 11, the coefficient of the middle term . Click one Gme for the possible numbers Pairs of factors Sum of factors 1 , 10 -­‐ 1, -­‐ 10 2 , 5 -­‐2 , -­‐5 11 -­‐11 7 -­‐7 Click one more Gme for the answer The numbers we want are 1 and 10 x2 + 11x + 10 = ( x + 1 )(x + 10)
More problems Factor the following trinomial Hint rewrite in descending order 4y – 45 +y2 Click one Gme for the answer 4y – 45 +y2 = y2 +4y -45
The numbers we need are -5 and 9
the answer is (y – 5)(y + 9)
Ok -­‐ next problem is a challenge. Factor the trinomial below. NoGce the leading coefficient is not a nice +1 -­‐ 2 x2 + 15 + x Hint: First rewrite the trinomial in descending order Next factor out a – 1 (it is easier if the leading coefficient is posiGve) Next try trial an error or factor by grouping. When a trinomial is factored by grouping the method is someGmes called the key number method. Please go to the next slide Factoring -­‐ 2 x2 + 15 + x Using Trial an Error (1)  We factor out a -­‐1 in order to have a trinomial with a posiGve leading coefficient. -­‐2x² + x + 15 = -­‐1(2x² -­‐ x – 15) Now factor by finding the missing terms. Hum, by now I think you can see how important it is that you understand how to FOIL. Trial an Error is like the reverse of FOIL -­‐ 1(2x ? )( x ? ) Your possible combinaGons are -­‐ 15 , 1 or 1, -­‐15 15, -­‐1 or -­‐1, 15 -­‐5, 3 or 3, -­‐5 5, -­‐3 or -­‐3, 5 The product of your Outside terms added to the product of your Inside terms Must equal your middle term of –x Click one Gme for the answer -­‐ 1(2 x2 – x – 15) = -­‐1(2x + 5)( x – 3 ) or -­‐ (2x + 5)(x – 3) You can ALWAYS check your answer by mulGplying. IMPORTANT COMMENT you must not forget to put in the – 1 or the negaGve sign ! If you wrote (2x + 5)(x – 3) it would be WRONG Quadratic Factoring
Using The
Key Number Method
Given ax² + bx +c [a,b,c are integers with no common factor other than one]
I. Multiply a time c. This is your KEY NUMBER
III. Now find two integers such that their product is the KEY NUMBER and their sum is b
V. Rewrite the expression using these two numbers in place of b
VII. Now factor by grouping
Example
6x² - 13x - 15
The Key Number is a*c
Hence, (6)*(-15) = -90 this is our Key Number
The factors that we need are the ones that when you multiply them
you get -90 but when you add them you get -13
The two numbers that satisfy this condition are -18 and 5
using this information you rewrite -13x as -18x +5x
Continue now using grouping procedures
6x² - 13x - 15
6x² - 18x + 5x - 15
6x(x - 3) + 5(x - 3)
(6x+5)(x-3)
Factoring -­‐ 2 x2 + 15 + x By grouping (Key Number method) (1)  We factor out a -­‐1 in order to have a trinomial with a posiGve leading coefficient. -­‐2x² + x + 15 = -­‐1(2x²-­‐ x – 15) Now we mulGply 2 Gmes -­‐15 the key number is -­‐30 Next we look for two numbers whose product is -­‐30 but whose sum is -­‐1 ( -­‐1 is the coefficient of the middle term) The two numbers are 5 and – 6 Now rewrite the middle term using these two numbers. -­‐1( 2x2 -­‐6x +5x -­‐15) Group the terms so that you can factor out a common factor -­‐1( 2x2 – 6x + 5x – 15) = -­‐1[ 2x(x – 3 ) + 5(x – 3)] The common term is (x – 3) Click one Gme for the answer Answer -­‐1(2x + 5)( x – 3 ) or -­‐ (2x + 5)(x – 3) You can ALWAYS check your answer by mulGplying. CREDITS Pg 311 #31 Pg311 #75 Pg 317 #9 4y – 45 +y2
Pg 317#29 -­‐ 2 x2 + 15 + x Pg 327 #31 The above examples (that were used in this presentaGon) came from the following text: Elementary & Intermediate Algebra 5th EdiGon Concepts & ApplicaGons By: M. Biknger, D. Ellenbogen, B. Johnson