Precalculus Factoring Review.

Precalculus Factoring Review

What is Factorization? Factoring is un-multiplying.
You multiply factors together to get a product. Example: 3 × 4 = 12 You factor a value to find its factors. Example: factors of 12 are; ±1, ±2, ±3, ±4, ±6, ±12 Use the distributive property to multiple and also to factor (in reverse)

Factoring Numbers To factor a number is to write it as a product of two or more other numbers, each of which is called a factor 12 = (3)(4) 3 & 4 are factors 12 = (6)(2) 6 & 2 are factors 12 = (12)(1) 12 and 1 are factors 12 = (2)(2)(3) 2, 2, and 3 are factors In the last case we say the 12 is “completely factored” because all the factors are prime numbers

Prime Numbers A Natural Number, greater than 1, whose only factors are itself and 1 (the number 1 is not prime). 2, 3, 5, 7, 11, 13, 17, 19, 23, etc.

Factoring Polynomials
Apply the distributive property to 3c(4c – 2) In this example, the result of the multiplication, 12c2 – 6c, is the product. The factors in this problem were the monomial, 3c, and the polynomial, (4c-2). 3c(4c-2) is the factored form of 12c2 – 6c. Polynomials can be written in factored form by reversing the process of the distributive property.

Factoring Groups of Numbers to Find the Greatest Common Factor
Factor each number completely Construct the Greatest Common Factor by including all factors that are common to all groups Example: Find the GCF of: 30, 12, and 18 30 = (2)(3)(5), 12= (2)(2)(3), 18 = (2)(3)(3) GCF = (2)(3) = 6

Find the GCF of the terms:

Procedure for factoring the GCF
3 Steps Factor all terms of the polynomial and find the GCF of the terms. Divide each term of the polynomial by the GCF. Write the polynomial as the product of the GCF and the remaining factors of each term using the distributive property.

Factor the following polynomial.
Step 1: Factor and find the GCF for both terms. 9m3n2 = 3 × 3 × m × m × m × n × n 24mn4 = 2 × 2 × 2 × 3 × m × n × n × n × n The GCF is 3mn2.

Step 2: Divide each term of the polynomial by the GCF.

Step 3: Write the polynomial as the product of the GCF and the remaining factor of each term using the distributive property.

Example: Factor 4x 2 – 12x + 20. GCF = 4. = 4(x 2 – 3x + 5) Check the answer. 4(x 2 – 3x + 5) = 4x 2 – 12x + 20 A common binomial factor can be factored out of certain expressions. Example: Factor the expression 5(x + 1) – y(x + 1). 5(x + 1) – y(x + 1) = (x + 1) (5 – y) Check. (x + 1) (5 – y) = 5(x + 1) – y(x + 1)

Your Turn: a. 4x2 + 6xy b. 60a2 + 30ab – 90ac a. 2x(2x +3y)
1. Find the GCF of the terms in each expression and factor. a. 4x2 + 6xy b. 60a2 + 30ab – 90ac a. 2x(2x +3y) b. 30a(2a + b – 3c)

VERY IMPORTANT When factoring, always factor out the GCF first.

Polynomial A finite sum of terms Examples of polynomials:
Term: a number, a variable, or any product of numbers and variables Examples of polynomials:

Prime Polynomials A polynomial is called prime, if it is not 1, and if its only factors are itself and 1 Just like we learn to identify certain numbers as being prime we will learn to identify certain polynomials as being prime We will also completely factor polynomials by writing them as a product of prime polynomials

Importance of Factoring
If you don’t learn to factor polynomials you can’t pass precalculus or more advanced math classes It is essential that you memorize the following procedures and become proficient in using them

5 Steps in Completely Factoring a Polynomial
(#1) Write the polynomial in descending powers of one variable (if there is more than one variable, pick any one you wish) 2x + 3x2 – 1 would be written: 3x2 + 2x – 1 3xy2 + y3 + 4x2y – 3 could be written: y3 + 3xy2 + 4x2y – 3 (powers of y) 4x2y + 3xy2 + y3 – 3 (powers of x)

(#2) Look at each term of the polynomial to see if every term contains a common factor other than 1, if so, use the distributive property in reverse to place the greatest common factor outside a parentheses and other terms inside parentheses that give a product equal to the original polynomial In the previous two examples what was the greatest common factor found in all terms: 3x2 + 2x – 1 y3 + 3xy2 + 4x2y – 3

Factoring the Greatest Common Factor from Polynomials
9y5 + y2 y2( ) y2(9y3 + 1) 6x2t + 8xt + 12t 2t( ) 2t(3x2 + 4x + 6)

Factoring the Greatest Common Factor from Polynomials
14m4(m + 1) – 28m3(m + 1) – 7m2(m + 1) 7m2(m + 1)( ) 7m2(m + 1)(2m2 – 4m – 1)

(#3) After factoring out the greatest common factor, look at the new polynomial factors to determine how many terms each one contains (The fourth step will depend on the number of terms in each of the factors)

(#4) Use the method appropriate to the number of terms in the polynomial: 4 or more terms: “Factor by Grouping” 3 terms: PRIME UNLESS they are of the form “ax2 + bx + c”. If of this form, factor as “Perfact Square Trinomial” or factor trinomial using “Trial and Error FOIL” / “Splitting the Middle Term”. 2 terms: Always PRIME UNLESS they are: “difference of squares”: a2 – b2 “difference of cubes”: a3 – b3 “sum of cubes”: a3 + b3 In each of these cases factor by a formula

(#5) Cycle through step #4 as many times as necessary until all factors are “prime”

Procedure: Factor by Grouping (Used for 4 or more terms)
(#1) Group the terms by underlining: If there are exactly 4 terms try: 2 & 2 grouping, 3 & 1 grouping, or 1 & 3 grouping If there are exactly 5 terms try: 3 & 2 grouping, or 2 & 3 grouping (#2) Factor each underlined group as if it were a factoring problem by itself (#3) Now determine if the underlined and factored groups contain a common factor, if they contain a common factor, factor it out if they don’t contain a common factor, try other groupings, if none work, the polynomial is prime (#4) Once again count the terms in each of the new polynomial factors and return to step 4.

Example of Factoring by Grouping
Factor: ax + ay + 6x + 6y (1) Group the terms by underlining (start with 2 and 2 grouping): ax + ay + 6x + 6y (2) Factor each underlined group as if it were a factoring problem by itself: a(x + y) + 6(x + y) [notice sign between groups gets carried down]

Factoring by Grouping Example Continued
(3) Now determine if the underlined and factored groups contain a common factor, if they do, factor it out: a(x + y) + 6(x + y) (x + y)(a + 6) ax + ay + 6x + 6y = (x + y)(a + 6) (4) Once again count the terms in each of the new polynomial factors and return to step 4. Each of these polynomial factors contains two terms, return to step 4 to see if these will factor (SINCE WE HAVE NOT YET DISCUSSED FACTORING POLYNOMIALS WITH TWO TERMS WE WILL NOT CONTINUE AT THIS TIME)

Example of Factoring by Grouping
(1) Group the terms by underlining (Try 2 and 2 grouping): Factor each underlined group as if it were a factoring problem by itself: [notice sign between groups gets carried down and you have to be careful with this sign]

(3) Now determine if the underlined and factored groups contain a common factor, if they do, factor it out: (4) Once again count the terms in each of the new polynomial factors and return to step 4. Each of these polynomial factors contains two terms, return to step 4 to see if these will factor (AGAIN WE HAVE LEARNED TO FACTOR BINOMIALS YET, SO WE WON’T CONTINUE ON THIS EXAMPLE)

Note on Factoring by Grouping
It was noted in step 3 of the factor by grouping steps that sometimes the first grouping, or the first arrangement of terms might not result in giving a common factor in each term – in that case other groupings, or other arrangements of terms, must be tried Only after we have tried all groupings and all arrangement of terms can we determine whether the polynomial is factorable or prime

Try Factoring by Grouping Without First Rearranging
(1) Group the terms by underlining (Try 2 and 2): Factor each underlined group as if it were a factoring problem by itself: .

Now Try Same Problem by Rearranging
Factor: Rearrange: (1) Group the terms by underlining: Factor each underlined group as if it were a factoring problem by itself: .

(3) Now factor out the common factor: (4) Once again count the terms in each of the new polynomial factors and return to step 4. Each of these polynomial factors contains two terms, return to step 4 to see if these will factor (AGAIN WE TO WAIT UNTIL WE LEARN TO FACTOR BINOMIALS BEFORE WE CAN CONTINUE)

Example - Grouping Factor 4ab + 2ac + 8xb + 4xc
Group terms which have something in common. Often, this can be done in more than one way. For example: or Next, find the greatest common factor for the polynomial in each set of parentheses. The GCF for (4ab+2ac) is 2a. The GCF for (8xb + 4xc) is 4x. The GCF for (4ab +8xb) is 4b. The GCF for (2ac + 4xc) is 2c.

Example - Grouping Continued: Write each of the polynomials in parentheses as the product of the GCF and the remaining polynomial. Apply the distributive property to any common factors. Factor further if necessary. Notice that it did not matter how the terms were originally grouped, the factored forms of the polynomials are identical.

Your Turn: 1. Factor 2xy + 3y – 4x – 6. Group terms.
Notice the sign! 2xy + 3y – 4x – 6 = (2xy + 3y) – (4x + 6) Group terms. = y (2x + 3) – 2(2x + 3) Factor each pair of terms. = (2x + 3) ( y – 2) Factor out the common binomial. 2. Factor 2a 2 + 3bc – 2ab – 3ac. 2a 2 + 3bc – 2ab – 3ac = 2a 2 – 2ab + 3bc – 3ac Rearrange terms. = (2a 2 – 2ab) + (3bc – 3ac) Group terms. = 2a (a – b) + 3c (b – a) Factor. = 2a (a – b) – 3c (a – b) b – a = – (a – b). = (a – b) (2a – 3c) Factor.

Additional Note on Factoring by Grouping
In all of our examples we tried this method on polynomials with four terms and used only “2 and 2 grouping” Sometimes with four terms we must use “3 and 1 grouping” or “1 and 3 grouping” On polynomials with more than four terms other combinations of groupings must be tried In any case, all combinations of grouping must be tried before we can determine of the polynomial is factorable or is prime We will not deal with any of these situations at this time

More on Factoring In the overall scheme of “factoring polynomials completely” we need to know how to factor polynomials containing various numbers of terms Thus far we have learned that regardless of the number of terms, we should always attempt, as a first step, to factor out the GCF We have also learned that for a polynomial with four or more terms we can try the “factor by grouping” method We next learn methods for factoring polynomials with three terms (trinomials)

Factoring Polynomials with 3 Terms
#1 Trinomial form x2+bx+c=0 (i.e., a=1)

Factoring Simple Trinomial
To factor a simple trinomial of the form x 2 + bx + c (leading coefficient equals one), express the trinomial as the product of two binomials. For example, x x + 24 = (x + 4)(x + 6). Factoring these trinomials is based on reversing the FOIL process.

Multiplying Binomials (FOIL)
Here is a way to help you remember how to multiply two binomials. This only works for 2 binomials! It is called F.O.I.L. which stands for “First, Outer, Inner, Last” F = first terms in each binomial O = the two outer terms I = the two inner terms L = last terms in each There is a way to remember the process you found in the previous examples. THIS ONLY WORKS FOR MULTIPLYING TWO BINOMIALS!!! It is called F.O.I.L. If you want practice, go back and simplify the ones you did with the tiles using FOIL. This is really just a double distribution. F and O come from distributing the first x over the second binomial. The I and L come from distributing the second term of the first binomial over the second binomial.

Factoring Trinomials of the Form x 2 + bx + c
Recall by using the FOIL method that F O I L (x + 2)(x + 4) = x 2 + 4x + 2x + 8 = x 2 + 6x + 8 To factor x 2 + bx + c into (x + one #)(x + another #), note that b is the sum of the two numbers and c is the product of the two numbers. So we’ll be looking for 2 numbers whose product is c and whose sum is b. Note: there are fewer choices for the product, so that’s why we start there first.

List the factors of 20: Select the pairs from which 12 may be obtained Write the two binomial factors: Check using FOIL: 𝑥 2 +12𝑥+20 20 1×20 2×10 4×5

Factoring Trinomials TIP
If the last term of the trinomial is positive and the middle sign is positive, both binomials will have the same “middle” sign as the second term.

If the last term of the trinomial is positive and the middle sign is negative, both binomials will have the same “middle” sign as the second term.

If the last term of the trinomial is negative, both binomials will have one plus and one minus “middle” sign.

Example: Factor the polynomial x 2 – 2x – 35. Since our two numbers must have a product of – 35 and a sum of – 2, the two numbers will have different signs. Factors of – 35 Sum of Factors – 1, 1, – – 34 – 5, 5, – – 2 So x 2 – 2x – 35 = (x + 5)(x – 7).

Example: Factor the polynomial x x + 30. Since our two numbers must have a product of 30 and a sum of 13, the two numbers must both be positive. Positive factors of Sum of Factors 1, 2, 3, Note, there are other factors, but once we find a pair that works, we do not have to continue searching. So x x + 30 = (x + 3)(x + 10).

Your Turn: Factor the polynomial x 2 – 11x + 24. Since our two numbers must have a product of 24 and a sum of –11, the two numbers must both be negative. Negative factors of 24 Sum of Factors – 1, – – 25 – 2, – – 14 – 3, – 8 – 11 So x 2 – 11x + 24 = (x – 3)(x – 8).

Prime Polynomials Example: Factor the polynomial x 2 – 6x + 10.
Since our two numbers must have a product of 10 and a sum of – 6, the two numbers will have to both be negative. Negative factors of 10 Sum of Factors – 1, – – 11 – 2, – – 7 Since there is not a factor pair whose sum is – 6, x 2 – 6x +10 is not factorable and we call it a prime polynomial.

Prime Polynomials Factor : The factors of 5: 1 and 5.
Since –2 cannot be obtained from 1 and 5, the polynomial is prime. A PRIME POLYNOMIAL cannot be factored using only integer factors.

Factoring Trinomials - 2 Variables
The factors of 8 are: 1,8 & 2,4, & -1,-8 & -2, -4 Choose the pairs from which –6 can be obtained: -2 & -4 Use y in the first position and z in the second position Write the two binomial factors and check your answer

Factoring Trinomials—with a GCF
If there is a greatest common factor? If yes, factor it out first. Factor: 3z4 – 15z3 +18z2 First factor GCF: 3z2 (z2 – 5z + 6) 3z2 (z2 – 5z + 6) factors of 6: 1×6 & 2×3 Choose 2×3: 3z2 (z – 2)(z – 3)

Factoring Polynomials with 3 Terms
#2 Trinomial form ax2 + bx + c (i.e., a≠1)

Factoring Trinomials Trial and Error FOIL (Used for 3 terms of form ax2 + bx + c)
Given a trinomial if this form, experiment to try to find two binomials that could multiply to give that trinomial. Remember that when two binomials are multiplied: First times First = First Term of Trinomial Outside times Outside + Inside times Inside = Middle Term of Trinomial Last times Last = Last Term of Trinomial

Steps in Using Trial and Error FOIL
Given a trinomial of the form: Write two blank parentheses that will each eventually contain a binomial Use the idea that “first times first = first” to get possible answers for first term of each binomial

Continuing Steps in Trial and Error FOIL
Given a trinomial of the form: Next use the idea that “last times last = last” to get possible answers for last term of each binomial Finally use the idea that “Outside times Outside + Inside times Inside = Middle Term of Trinomial” to get the final answer for two binomials that multiply to give the trinomial

Factoring Trinomials of the Form ax 2 + bx + c
Example: Factor the polynomial 25x x + 4. Possible factors of 25x 2 are {x, 25x} or {5x, 5x}. Possible factors of 4 are {1, 4} or {2, 2}. We need to methodically try each pair of factors until we find a combination that works, or exhaust all of our possible pairs of factors. Keep in mind that, because some of our pairs are not identical factors, we may have to exchange some pairs of factors and make 2 attempts before we can definitely decide a particular pair of factors will not work. Continued.

Example continued: We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to 20x. Factors of 25x 2 Resulting Binomials Product of Outside Terms Product of Inside Terms Sum of Products Factors of 4 {x, 25x} {1, 4} (x + 1)(25x + 4) x x x (x + 4)(25x + 1) x x x {x, 25x} {2, 2} (x + 2)(25x + 2) x x x {5x, 5x} {2, 2} (5x + 2)(5x + 2) x x x Continued.

Example continued: Check the resulting factorization using the FOIL method [OUTER (product outer numbers) + INNER (product inner numbers) = MIDDLE (middle number)]. F 5(2) O + 2(5) I L (5x + 2)(5x + 2) = = 10 So our final answer when asked to factor 25x x + 4 will be (5x + 2)(5x + 2) or (5x + 2)2.

Example: Factor the polynomial 21x 2 – 41x + 10. Possible factors of 21x 2 are {x, 21x} or {3x, 7x}. Since the middle term is negative, possible factors of 10 must both be negative: {-1, -10} or {-2, -5}. We need to methodically try each pair of factors until we find a combination that works, or exhaust all of our possible pairs of factors. Continued.

Example continued: We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to 41x. Factors of 21x 2 Resulting Binomials Product of Outside Terms Product of Inside Terms Sum of Products Factors of 10 {x, 21x} {1, 10} (x – 1)(21x – 10) –10x 21x – 31x (x – 10)(21x – 1) –x 210x – 211x {x, 21x} {2, 5} (x – 2)(21x – 5) –5x 42x – 47x (x – 5)(21x – 2) –2x 105x – 107x Continued.

Example continued: Factors of 21x 2 Resulting Binomials Product of Outside Terms Product of Inside Terms Sum of Products Factors of 10 {3x, 7x} {1, 10} (3x – 1)(7x – 10) 30x 7x 37x (3x – 10)(7x – 1) 3x 70x 73x {3x, 7x} {2, 5} (3x – 2)(7x – 5) 15x 14x 29x (3x – 5)(7x – 2) 6x 35x 41x Continued.

Factoring Trinomials of the Form ax2 + bx + c
Example continued: Check the resulting factorization using the FOIL method [OUTER (product outer numbers) + INNER (product inner numbers) = MIDDLE (middle number)]. F 3(-2) O + (-5)(7) I L (3x – 5)(7x – 2) = = (– 6) + (– 35) = - 41 So our final answer when asked to factor 21x 2 – 41x + 10 will be (3x – 5)(7x – 2).

Your Turn: Factor: 12x2 + 11x – 5 (4x + 5)(3x – 1) = 12x2 +11x – 5

Your Turn: Factor the polynomial 6x 2y 2 – 2xy 2 – 60y 2.
Solution: Remember that the larger the coefficient, the greater the probability of having multiple pairs of factors to check. So it is important that you attempt to factor out any common factors first. 6x 2y 2 – 2xy 2 – 60y 2 = 2y 2(3x 2 – x – 30) The only possible factors for 3 are 1 and 3, so we know that, if we can factor the polynomial further, it will have to look like 2y 2(3x )(x ) in factored form. Continued.

Solution continued: Since the product of the last two terms of the binomials will have to be –30, we know that they must be different signs. Possible factors of –30 are {–1, 30}, {1, –30}, {–2, 15}, {2, –15}, {–3, 10}, {3, –10}, {–5, 6} or {5, –6}. We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to –x. Continued.

Solution continued: Factors of -30 Resulting Binomials
Product of Outside Terms Product of Inside Terms Sum of Products {-1, 30} (3x – 1)(x + 30) x x x (3x + 30)(x – 1) Common factor so no need to test. {1, -30} (3x + 1)(x – 30) x x x (3x – 30)(x + 1) Common factor so no need to test. {-2, 15} (3x – 2)(x + 15) x x x (3x + 15)(x – 2) Common factor so no need to test. {2, -15} (3x + 2)(x – 15) x x x (3x – 15)(x + 2) Common factor so no need to test. Continued.

Solution continued: Factors of –30 Resulting Binomials
Product of Outside Terms Product of Inside Terms Sum of Products {–3, 10} (3x – 3)(x + 10) Common factor so no need to test. (3x + 10)(x – 3) –9x x x {3, –10} (3x + 3)(x – 10) Common factor so no need to test. (3x – 10)(x + 3) x –10x –x Continued.

Solution continued: Check the resulting factorization using the FOIL method [OUTER (product outer numbers) + INNER (product inner numbers) = MIDDLE (middle number)]. F 3(3) O + (–10)(1) I L (3x – 10)(x + 3) = = 9 + (–10) = – 1 So our final answer when asked to factor the polynomial 2y 2(3x 2 – x – 30) will be 2y 2(3x – 10)(x + 3).

Prime Trinomials Example: Factor the polynomial 3x 2 – 7x + 6.
The only possible factors for 3 are 1 and 3, so we know that, if factorable, the polynomial will have to look like (3x )(x ) in factored form, so that the product of the first two terms in the binomials will be 3x 2. Since the middle term is negative, possible factors of 6 must both be negative: {1,  6} or { 2,  3}. We need to methodically try each pair of factors until we find a combination that works, or exhaust all of our possible pairs of factors. Continued.

Prime Trinomials Example continued:
We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to 7x. Factors of 6 Resulting Binomials Product of Outside Terms Product of Inside Terms Sum of Products {1, 6} (3x – 1)(x – 6) 18x x 19x (3x – 6)(x – 1) Common factor so no need to test. {2, 3} (3x – 2)(x – 3) 9x 2x 11x (3x – 3)(x – 2) Common factor so no need to test. Continued.

Prime Trinomials Solution continued:
Now we have a problem, because we have exhausted all possible choices for the factors, but have not found a pair where the sum of the products of the outside terms and the inside terms is –7. So 3x 2 – 7x + 6 is a prime polynomial and will not factor.

Alternate Method of Factoring trinomials of the form ax 2 + bx + c, (a  1) can be done by decomposition or splitting the middle term method. Example: Factor 3x 2 + 8x + 4. 3  4 = 12 Decomposition Method 1. Find the product of first and last terms 2. We need to find factors of 12 whose sum is 8 1, 12 2, 6 3, 4 3. Rewrite the middle term decomposed into the two numbers 3x 2 + 2x + 6x + 4 = (3x 2 + 2x) + (6x + 4) = x(3x + 2) + 2(3x + 2) 4. Factor by grouping in pairs = (3x + 2) (x + 2) 3x 2 + 8x + 4 = (3x + 2) (x + 2)

Splitting the Middle Term
Example: Factor 8x 2 - 2x - 3. 8  3 = 24 We need to find factors of 24 whose difference is -2 1, 24 2, 12 3, 8 4, 6 Rewrite the middle term decomposed into the two numbers 8x 2 + 4x - 6x – 3 = (8x 2 + 4x) + (-6x – 3) Factor by grouping in pairs = 4x(2x + 1) + (-3)(2x + 1) = (2x + 1) (4x - 3) 8x 2 - 2x - 3 = (2x +1)(4x – 3)

Splitting the Middle Term
Example: Factor 3x 2 - 5x + 2. 3  2 = 6 We need to find factors of 6 whose sum is -5 1, 6 2, 3 Rewrite the middle term decomposed into the two numbers 3x 2 - 2x - 3x + 2 = (3x 2 - 2x) + (-3x + 2) Factor by grouping in pairs = x(3x - 2) + (-1)(3x - 2) = (3x - 2) (x - 1) 3x 2 - 5x + 2 = (3x -2)(x – 1)

Your Turn: Factor 4x 2 + 8x – 5. We need to find factors of 20
4  5 = 20 We need to find factors of 20 whose difference is 8 1, 20 2, 10 4, 5 Rewrite the middle term decomposed into the two numbers 4x x - 2x – 5 = (4x x) + (-2x – 5) Factor by grouping in pairs = 2x(2x + 5) + (-1)(2x + 5) = (2x + 5) (2x – 1) 4x 2 + 8x – 5 = (2x + 5)(2x –1)

Factor the X-Box Method
ax2 + bx + c Combine the x-factor and a 2⨯2 box to create the x-box method. Factor Col. 1 Factor Col. 2 Product ac=mn First and Last Coefficients 1st Term GCF Row 1 Factor n n m Middle Last term Factor m Factor Row 2 b=m+n Sum

X-Box Procedure ax2 + bx + c 1st Term Factor n Factor m Last term
Create an x-factor with the product ac on the top, the middle term b on the bottom and the factors m & n on the sides. Create a 2x2 box. In the top left, put the 1st term. In the bottom right corner, put the last term. Put the two factors m and n times the variable x in the open boxes. Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 1st Term Factor n Factor m Last term

Factoring ax2+bx+c Factor 5x2+11x+2 5•2= 10 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 1 10 11 x 2 5x x 10x 5x2 1 2 (x+2)(5x+1)

Factor 3x2 – x – 4 3 -4 -1 3x -4 x 3x -4x 3x2 1 -4 3•-4= -12
Factoring ax2+bx+c Factor 3x2 – x – 4 3•-4= -12 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 3 -4 -1 3x -4 x 3x -4x 3x2 1 -4 (x+1)(3x – 4)

Factor 12x2 +5x – 2 8 -3 5 4x -1 3x 12x2 8x -3x 2 -2 12•-2= -24
Factoring ax2+bx+c Factor 12x2 +5x – 2 12•-2= -24 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 8 -3 5 4x -1 3x 12x2 8x -3x 2 -2 (3x+2)(4x – 1)

Factor 15x2 +7x – 2 10 -3 7 5x -1 3x 10x -3x 15x2 2 -2 15•-2= -30
Factoring ax2+bx+c Factor 15x2 +7x – 2 15•-2= -30 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 10 -3 7 5x -1 3x 10x -3x 15x2 2 -2 (3x+2)(5x – 1)

Factoring ax2+bx+c Factor 10x2 +9x +2 10•2= 20 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 5 4 9 5x 2 2x 5x 4x 10x2 1 2 (2x+1)(5x +2)

Your Turn: Factor: 2x² - 13x + 6

Factor 2x2 – 13x +6 -12 -1 -13 2x -1 x -12x -x 2x2 -6 6 2•6= 12
Factoring ax2+bx+c Factor 2x2 – 13x +6 2•6= 12 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. -12 -1 -13 2x -1 x -12x -x 2x2 -6 6 (2x – 1)(x – 6)

Your Turn: Factor: 4x² + 11x – 3

Factor 4x²+11x – 3 12 -1 11 4x -1 x 12x -x 4x2 3 -3 4•-3= -12
Factoring ax2+bx+c Factor 4x²+11x – 3 4•-3= -12 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 12 -1 11 4x -1 x 12x -x 4x2 3 -3 (4x – 1)(x + 3)

Your Turn: Factor: 3x²+8x+4

Factor 3x²+8x+4 6 2 8 3x 2 x 6x 2x 3x2 2 4 4•3= 12 Factoring ax2+bx+c
Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 6 2 8 3x 2 x 6x 2x 3x2 2 4 (3x + 2)(x + 2)

Your Turn: Factor: -5x²+6x-1

Factor -5x²+6x-1 5 1 6 -5x 1 x 5x x -5x2 -1 -1 -5•-1= 5
Factoring ax2+bx+c Factor -5x²+6x-1 -5•-1= 5 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 5 1 6 -5x 1 x 5x x -5x2 -1 -1 (-5x + 1)(x – 1)

Factoring Trinomials of the Form -ax 2 + bx + c , (a<0)
When the leading coefficient is negative, it is easier to first factor the negative out of the trinomial, then complete factoring the remaining trinomial. Example: Factor: -x2 + 7x + 18 = -(x2 - 7x - 18) = -(x – 9)(x + 2) or (9 - x)(x + 2)

Factoring “Perfect Square Trinomials”
A trinomial is a “perfect square trinomial” if it has resulted from squaring a binomial: Perfect square trinomials have the characteristic that the first and last terms are perfect squares, and the middle term is twice the product of the square roots of the first and last terms When this characteristic is seen in a trinomial, we automatically factor it as a binomial squared being careful to place the correct sign in the middle

Perfect Square Trinomials
(a + b)2 = a 2 + 2ab + b 2 (a – b)2 = a 2 – 2ab + b 2 So if the first and last terms of our polynomial to be factored can be written as expressions squared, and the middle term of our polynomial is twice the product of the square root of these two expressions, then we can use these two previous equations to easily factor the polynomial. a 2 + 2ab + b 2 = (a + b)2 a 2 – 2ab + b 2 = (a – b)2

So how do we factor a Prefect Square Trinomial
When you have to factor a perfect square trinomial, the patterns make it easier Product Doubled Example: Factor 36x2 + 60x + 25 Perfect Square 6x 30x 5 Perfect Square Square Root Product Square Root First you have to recognize that it’s a perfect square trinomial And so, the trinomial factors as: (6x + 5)2

Perfect Square Trinomials
Example: Factor the polynomial 16x 2 – 8xy + y 2. Since the first term, 16x 2, can be written as (4x)2, and the last term, y 2 is obviously a square, we check the middle term. 8xy = 2(4x)(y) (twice the product of the expressions that are squared to get the first and last terms of the polynomial) Therefore 16x 2 – 8xy + y 2 = (4x – y)2. Note: You can use FOIL method to verify that the factorization for the polynomial is accurate.

Try factoring these… m2 – 6m + 9 4w2 + 28w + 49 81x2 – 18x + 1
25h2 – 100h + 64 9a2 + 12a + 4 = (m – 3)2 = (2w + 7)2 = (9x – 1)2 = (5h – 16)(5h – 4) = (3a + 2)2 m 3m 3 2w 14w 7 9x 9x 1 Not a perfect square trinomial! 5h 40h 8 3a 6a 2

More on Factoring In the overall scheme of “factoring polynomials completely” we need to know how to factor polynomials containing various numbers of terms Thus far we have learned that regardless of the number of terms, we should always attempt, as a first step, to factor out the GCF We have also learned that for a polynomial with four or more terms we can try the “factor by grouping” method We have also learned that we should try to factor trinomials of the form ax2+bx+c by either “trial and error ” or “splitting the middle term” We next learn methods for factoring binomials

Note on Factoring Binomials
Binomials are factorable only if they are a: Difference of Squares: Difference of Cubes: Sum of Cubes: In each of these cases, factoring is done by means of a formula that needs to be memorized All other binomials are prime (In saying this, we assume that any GCF has already been factored out)

Factoring Binomials by Formula
Factor by using formula appropriate for the binomial: “difference of squares”: a2 – b2 = (a – b)(a + b) “difference of cubes”: a3 – b3 = (a – b)(a2 + ab + b2) Trinomial is prime “sum of cubes”: a3 + b3 = (a + b)(a2 – ab + b2) Trinomial is prime If none of the formulas apply, the binomial is prime BINOMIALS ARE PRIME UNLESS THEY ARE ONE OF THESE

Difference of Two Squares
D.O.T.S.

Conjugate Pairs (3x + 6) (3x - 6) (r - 5) (r + 5) (2b - 1) (2b + 1)
The following pairs of binomials are called conjugates. Notice that they all have the same terms, only the sign between them is different. (3x + 6) (3x - 6) and (r - 5) (r + 5) and (2b - 1) (2b + 1) and (x2 + 5) (x2 - 5) and

Multiplying Conjugates
When we multiply any conjugate pairs, the middle terms always cancel and we end up with a binomial. (3x + 6)(3x - 6) = 9x2 - 36 (r - 5)(r + 5) = r2 - 25 (2b - 1)(2b + 1) = 4b2 - 1

Only TWO terms (a binomial)
Difference of Two Squares Binomials that look like this are called a Difference of Squares: Only TWO terms (a binomial) 9x2 - 36 A MINUS between! The first term is a Perfect Square! The second term is a Perfect Square!

A binomial is the difference of two square if both terms are squares and the signs of the terms are different. 9x 2 – 25y 2 – c 4 + d 4

Factoring the Difference of Two Squares
A Difference of Squares! A Conjugate Pair!

Example: Factor the polynomial x 2 – 9. The first term is a square and the last term, 9, can be written as 32. The signs of each term are different, so we have the difference of two squares Therefore x 2 – 9 = (x – 3)(x + 3). Note: You can use FOIL method to verify that the factorization for the polynomial is accurate.

Example: Factor x2 - 64 = (x + 8)(x - 8) x2 = x • x 64 = 8 • 8 Example: Factor 9t2 - 25 = (3t + 5)(3t - 5) 9t2 = 3t • 3t 25 = 5 • 5

Your Turn: Factor x 2 – 16. Since this polynomial can be written as x 2 – 42, x 2 – 16 = (x – 4)(x + 4). Factor 9x2 – 4. Since this polynomial can be written as (3x)2 – 22, 9x 2 – 4 = (3x – 2)(3x + 2). Factor 16x 2 – 9y 2. Since this polynomial can be written as (4x)2 – (3y)2, 16x 2 – 9y 2 = (4x – 3y)(4x + 3y).

Your Turn: Factor: 25x2 – 9y2 Note that this binomial is a difference of squares: (5x)2 – (3y)2 Using formula gives: (5x – 3y)(5x + 3y)

A Sum of Squares? A Sum of Squares, like x2 + 64, can NOT be factored! It is a PRIME polynomial.

Your Turn: Factor: 4x2 + 9 Note that this is not a difference of squares, difference of cubes, or sum of cubes, therefore it is prime (4x2 + 9) To show factoring of a polynomial that is prime, put it inside parentheses

Example: Factor 36x 2 – 64. Remember that you should always factor out any common factors, if they exist, before you start any other technique. Factor out the GCF. 36x 2 – 64 = 4(9x 2 – 16) Since the polynomial can be written as (3x)2 – (4)2, (9x 2 – 16) = (3x – 4)(3x + 4). So our final result is 36x 2 – 64 = 4(3x – 4)(3x + 4).

Your Turn: Example: Factor (x + 1)2 – 25y 4. (x + 1)2 – 25y 4

Factoring Difference of Cubes
Factor: 8x3 – 27 Note that this is a difference of cubes: (2x)3 – (3)3 Formula: a3 – b3 = (a – b)(a2 + ab + b2) Using formula gives: (2x – 3)(4x2 + 6x + 9)

Your Turn: Factor: 27x8-64 Formula: a3 – b3 = (a – b)(a2 + ab + b2) (3x2-4)(9x4+12x2+16)

Sum of Two Cubes Factor: y3 + 8 Note that this is a sum of cubes:
Formula: a3 + b3 = (a + b)(a2 - ab + b2) Using formula gives: (y + 2)(y2 – 2y + 4)

Your Turn: Factor: 125x3+27y6 Formula: a3 + b3 = (a + b)(a2 - ab + b2) (5x+3y2)(25x2-15xy2+9y4)

Completely Factoring Polynomials
We now have all the skills necessary to either factor polynomials completely, or to determine if they are prime Summary of steps: Arrange polynomial in descending powers of one variable Factor out the GCF (also factor out a negative if highest degree term has a negative coefficient) For each polynomial factor in the expression, try to factor it by using the method appropriate for the number of terms it has Continue factoring each new polynomial factor until all polynomial factors are prime We now apply this procedure in completely factoring polynomials

Example of Completely Factoring Polynomials Using Five Steps
Factor: 2x3 – 8x + 2x6 – 8x4 (1) Write the polynomial in descending powers of one variable (if there is more than one variable, pick any one you wish) 2x6 – 8x4 + 2x3 – 8x (2) Look at each term of the polynomial to see if every term contains a common factor, if so, use the distributive property in reverse to place the greatest common factor outside a parentheses 2x(x5 – 4x3 + x2 – 4)

2x(x5 – 4x3 + x2 – 4) (3) After factoring out the greatest common factor, look at the new polynomial factors to determine how many terms each one contains The polynomial in parentheses has 4 terms (4) Use the method appropriate to the number of terms in the polynomial Since there are 4 terms we will try “factor by grouping”: x5 – 4x3 + x2 – 4 x3(x2 – 4) + 1(x2 – 4) (x2 – 4)(x3 + 1) So far we have factored the original polynomial as: 2x(x2 – 4)(x3 + 1)

2x(x2 – 4)(x3 + 1) (5) Cycle through step 4 as many times as necessary until all factors are “prime” (count terms and use appropriate method) The first binomial is a difference of squares, and the second is a sum of cubes so they must be factored by formulas to get the final complete factoring of: 2x(x – 2)(x + 2)(x + 1)(x2 – x + 1) COMPLETELY FACTORED!

Factoring Completely Your Turn: Factor completely: 6x 2  6x  36
6x 2  6x  36 = 6(x 2  x  6) Factor out the GCF, 6. 3 terms, the coefficient of x 2 is 1. = 6(x  3)( x + 2) Factor. – 3(2) = – 6 and – = – 1

Factoring Completely Your Turn: Factor completely: 1  16x 4
1 and 16x4 are both perfect squares so we have the difference of two squares. 1  16x4 = [12  (4x 2)2] = (1  4x 2)(1 + 4x 2) Factor. a = 1, b = 4x 2 This is also a difference of two squares. = (1  2x)(1 + 2x)(1 + 4x 2) Factor.

Precalculus Factoring Review.

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