Factoring Review Dividing GCF Factor by grouping Factor
Factoring Review -Dividing -GCF -Factor by grouping -Factor by sum and product rule -a*c method -Difference of squares and cubes -Quadratic Formula -Quadratic Equations Chapter 8 Involves Factoring in most of the questions presented. We simply reviewing the quickest factoring techniques/concepts today. This is not all of the different ways you can factor but it is the set of most frequently used techniques. If you are looking for questions to practice factoring with they will come from Chapter 7. Most of these powerpoints (as they change backgrounds) can be paired with one or two sections of Chapter 7. YOU MUST HAVE DEVELOPED A PROFICIENT SKILL OF FACTORING BEFORE YOU ARE TO BE SUCCESSFUL INSIDE THIS COURSE.
Objectives ■ Common Factors ■ Factoring Trinomials ■ Special Factoring Formulas ■ Factoring an Expression Completely ■ Factoring by Grouping Terms 2
Factoring We use the Distributive Property to expand algebraic expressions. We sometimes need to reverse this process (again using the Distributive Property) by factoring an expression as a product of simpler ones. For example, we can write x 2 – 4 = (x – 2)(x + 2) We say that x – 2 and x + 2 are factors of x 2 – 4. 3
Common Factors 4
Common Factors The easiest type of factoring occurs when the terms have a common factor. 5
Example 1 – Factoring Out Common Factors Factor each expression. (a) 3 x 2 – 6 x (b) 8 x 4 y 2 + 6 x 3 y 3 – 2 xy 4 Solution: (a) The greatest common factor of the terms 3 x 2 and – 6 x is 3 x, so we have 3 x 2 – 6 x = 3 x(x – 2) (b) We note that 8, 6, and – 2 have the greatest common factor 2 x 4, x 3, and x have the greatest common factor x y 2, y 3, and y 4 have the greatest common factor y 2 6
Example 1 – Solution cont’d So the greatest common factor of the three terms in the polynomial is 2 xy 2, and we have 8 x 4 y 2 + 6 x 3 y 3 – 2 xy 4 = (2 xy 2)(4 x 3) + (2 xy 2)(3 x 2 y) + (2 xy 2)(–y 2) = 2 xy 2(4 x 3 + 3 x 2 y – y 2) 7
Factoring Trinomials 8
Factoring Trinomials To factor a trinomial of the form x 2 + bx + c, we note that (x + r)(x + s) = x 2 + (r + s)x + rs so we need to choose numbers r and s so that r + s = b and rs = c. 9
Example 3 – Factoring x 2 + bx + c by Trial and Error Factor: x 2 + 7 x + 12 Solution: We need to find two integers whose product is 12 and whose sum is 7. By trial and error we find that the two integers are 3 and 4. Thus the factorization is x 2 + 7 x + 12 = (x + 3)(x + 4) 10
Factoring Trinomials To factor a trinomial of the form ax 2 + bx + c with a 1, we look for factors of the form px + r and qx + s: ax 2 + bx + c = (px + r)(qx + s) = pqx 2 + (ps + qr)x + rs Therefore we try to find numbers p, q, r, and s such that pq = a, rs = c, ps + qr = b. If these numbers are all integers, then we will have a limited number of possibilities to try for p, q, r, and s. 11
Example 5 – Recognizing the Form of an Expression Factor each expression. (a) x 2 – 2 x – 3 (b) (5 a + 1)2 – 2(5 a + 1) – 3 Solution: (a) x 2 – 2 x – 3 = (x – 3)(x + 1) Trial and error (b) This expression is of the form where represents 5 a + 1. This is the same form as the expression in part (a), so it will factor as 12
Example 5 – Solution cont’d = (5 a – 2)(5 a + 2) 13
Special Factoring Formulas 14
Special Factoring Formulas Some special algebraic expressions can be factored by using the following formulas. The first three are simply Special Product Formulas written backward. 15
Special Factoring Formulas A trinomial is a perfect square if it is of the form A 2 + 2 AB + B 2 or A 2 – 2 AB + B 2 So we recognize a perfect square if the middle term (2 AB or – 2 AB) is plus or minus twice the product of the square roots of the outer two terms. 16
Example 7 – Recognizing Perfect Squares Factor each trinomial. (a) x 2 + 6 x + 9 (b) 4 x 2 – 4 xy + y 2 Solution: (a) Here A = x and B = 3, so 2 AB = 2 x 3 = 6 x. Since the middle term is 6 x, the trinomial is a perfect square. By the Perfect Square Formula we have x 2 + 6 x + 9 = (x + 3)2 17
Example 7 – Solution cont’d (b) Here A = 2 x and B = y, so 2 AB = 2 2 x y = 4 xy. Since the middle term is – 4 xy, the trinomial is a perfect square. By the Perfect Square Formula we have 4 x 2 – 4 xy + y 2 = (2 x – y)2 18
Example 8 – Factoring Differences and Sums of Cubes Factor each polynomial. (a) 27 x 3 – 1 (b) x 6 + 8 Solution: (a) Using the Difference of Cubes Formula with A = 3 x and B = 1, we get 27 x 3 – 1 = (3 x)3 – 13 = (3 x – 1)[(3 x)2 + (3 x)(1) + 12] = (3 x – 1)(9 x 2 + 3 x + 1) 19
Example 8 – Solution cont’d (b) Using the Sum of Cubes Formula with A = x 2 and B = 2, we have x 6 + 8 = (x 2)3 + 23 = (x 2 + 2)(x 4 – 2 x 2 + 4) 20
Factoring an Expression Completely 21
Factoring an Expression Completely When we factor an expression, the result can sometimes be factored further. In general, we first factor out common factors, then inspect the result to see whether it can be factored by any of the other methods of this section. We repeat this process until we have factored the expression completely. 22
Example 9 – Factoring an Expression Completely Factor each expression completely. (a) 2 x 4 – 8 x 2 (b) x 5 y 2 – xy 6 Solution: (a) We first factor out the power of x with the smallest exponent. 2 x 4 – 8 x 2 = 2 x 2(x 2 – 4) = 2 x 2(x – 2)(x + 2) Common factor is 2 x 2 Factor x 2 – 4 as a difference of squares 23
Example 9 – Solution cont’d (b) We first factor out the powers of x and y with the smallest exponents. Common factor is xy 2 x 5 y 2 – xy 6 = xy 2(x 4 – y 4) = xy 2(x 2 + y 2)(x 2 – y 2) = xy 2(x 2 + y 2)(x + y)(x – y) Factor x 4 – y 4 as a difference of squares Factor x 2 – y 2 as a difference of squares 24
Factoring by Grouping Terms 25
Factoring by Grouping Terms Polynomials with at least four terms can sometimes be factored by grouping terms. The following example illustrates the idea. 26
Example 11 – Factoring by Grouping Factor each polynomial. (a) x 3 + x 2 + 4 x + 4 (b) x 3 – 2 x 2 – 9 x + 18 Solution: (a) x 3 + x 2 + 4 x + 4 = (x 3 + x 2) + (4 x + 4) = x 2(x + 1) + 4(x + 1) = (x 2 + 4)(x + 1) Group terms Factor out common factors Factor x + 1 from each term 27
Example 11 – Solution (b) x 3 – 2 x 2 – 9 x + 18 = (x 3 – 2 x 2) – (9 x – 18) cont’d Group terms = x 2(x – 2) – 9(x – 2) Factor common factors = (x 2 – 9)(x – 2) Factor (x – 2) from each term = (x – 3)(x + 3)(x – 2) Factor completely 28
Dividing a Polynomial by a Monomial Dividing a Polynomial by a Polynomial The Greatest Common Factor, Factoring by Grouping
• When dividing a polynomial by a monomial you will divide every single term in the polynomial by the monomial. • Remember to use your rules for exponents when dividing like bases.
The Greatest Common Factor, Factoring by Grouping
• When we find products we are multiplying together several terms. Those terms are referred to as the FACTORS for that PRODUCT. • 3 x 15 = 45 3 and 15 are the factors • (x+2)(x+3) = x 2+5 x+6. . (x+2) and (x+3) are the factors
Greatest Common Factor • The greatest common factor for a polynomial is the largest monomial that divides (is a factor of) each term of the polynomial. • The largest monomial is the monomial with the largest coefficient and largest power that is a common factor for each term in the polynomial.
• Find the greatest common factor for… – Look at the coefficients first, then the variable
• Find the greatest common factor for… – Look at the coefficients first, then the variable
• When asked to factor the greatest common factor from a quantity, we are reversing the distributive property. So when you are done you should be able to distribute and get back to what you started with. • Factor the greatest common factor from 5 x+15
Factor the greatest common factor from
Factor the greatest common factor from
Factor the greatest common factor from
Factor by grouping • At times we may have a group of terms in which there is not a greatest common factor other than 1 amongst the group. But we still want to factor. Our next strategy will be to group the terms in pairs and focus on factoring something out of each group.
Example
Example
Example
Example
Example same problem but alternative approach
• As we saw in the last example the grouping really didn’t matter we still achieved the same result. This is a consistent occurrence. Sometimes you may have to re-write the polynomial and re-arrange terms to recognize how terms should be grouped.
Factoring Trinomials of the Form x 2+bx+c Factoring Trinomials of the Form ax 2+bx+c
• Because of the FOIL METHOD we know that when we multiply together two binomials such as (x+a) and (x+b) we get the following… (x+a)(x+b) = x 2+ax+bx+ab = x 2+(a+b)x+ab sum of a and b Product of a and b
• When we are factoring we reverse the idea… • Every trinomial will have a middle term that has the variable x. Its coefficient is a particular sum of two numbers. • The last term (with no variable) will be a product of the exact same two numbers.
So what does this mean to us • There must be two numbers that multiply to give 12 and those same two numbers must add to give 8. Can we figure those out? If so then we know what a and b are and they get placed in the product of factors (x+a)(x+b) to give then the factored form of the trinomial
• You can always FOIL to check to see if your factored polynomial is equivalent to the original (expanded) polynomial.
FACTOR
FACTOR IF YOU GET STUCK A CHART OF PRODUCTS AND SUMS MAY BE OF SOME USE a b Product Sum
FACTOR
FACTOR
• On occasions the first term (x 2) may not have a coefficient of 1. If this is the case check to see if you can divide everything by that number and factor it out. Place that number on the outside of your binomials.
FACTOR
• On other occasions there may be a Greatest Common Factor (GCF) present that needs to be factored out first (place this GCF in front of your binomials) Always try to make the leading coefficient become a 1. This will make things easier.
Factor
Factor
Trinomials with 2 variables
Trinomials with 2 variables
Trinomials with 2 variables
Factoring Trinomials in the Form ax 2+bx+c • In these trinomials the coefficient of the x 2 will not be a 1 and it will not divide/factor out of each term evenly. Thus we need a new approach. • Example: 2 x 2+7 x+3 • The text book uses a trial and error method…And that is a waste of our time • We will use “FACTOR BY GROUPING”
• We have factored by grouping already, but if you remember the polynomial had 4 terms so that we could have two groups of two terms. Thus we must transition the current trinomial that is given into a polynomial of 4 terms.
Steps for factor by grouping • • • Start with a trinomial in the form of ax 2+bx+c Multiply a and c. Determine two factors of “ac” that sum to give b. Lets call those two numbers “m” and “n”. Re-write the trinomial but instead of using bx as the middle term, write it as the sum of mx and nx. It will look like this ax 2 + mx + nx + c You can now group the first two terms It will look like this (ax 2+mx) + (nx+c) Factor the GCF out of each group. The remaining set of (___) will provide you with a factor The terms outside will be your other factor. Write the factored polynomial in (_____) form.
Example • Factor by grouping
Example • Factor by grouping
Example • Factor by grouping
Example • Factor by grouping
Example • Factor by grouping
Example • Factor by grouping
Factor out the GCF first
Factor out the GCF first
• Make sure you keep doing your homework, the number 1 way to get good at these and to be able to complete them quickly is to replicate the process as many times as you can. Practice makes perfect with these.
The Difference of Two Squares and Perfect Square Trinomials The Sum and Difference of Two Cubes
Difference of Two Squares • This is a difference of squares. It can be seen that the first term is a perfect square and so is the second. It may better be seen written the following way. • The completed factoring would look like this • As always you can check the factorization by multiplying.
Example 2 • Factor the following
• Sometimes you may not initially recognize a quantity being squared but re-writing it using your exponent rules may make it more obvious • This is a difference of squares but you cannot see it initially. (this is in difference of squares form) • So now it is a quantity squared minus a quantity squared. Can this be factored any further?
• Rewrite so that this is in difference of squares form. And then factor. Now this result does not provide a difference of perfect squares, but a difference of cubes. This can eventually be factored further we will learn in the next section
Perfect Square Trinomials • These trinomials can obviously be factored using previous learned techniques. However, if we recognize that the first and last terms are perfect squares we can quickly develop the possible factored form and then just multiply to check to see if the factoring worked. Pay attention to signs. If – then + the binomial will be (a-b). If + then + the binomial will be (a+b).
Example
Before recognizing a perfect square trinomial, you may have to first factor out a GCF.
• In this example there is a a 2 -b 2 form, however the a is a binomial expression.
Example 2
• Take a moment and complete the factorization of the following on your own.
A few more examples
The Sum and Difference of Two Cubes • We may want to factor something like the following.
These are the formulas for factoring difference and sums of cubes The most troublesome part of these formulas is maintaining the correct signs. You will notice that the first binomial has a sign that is congruent to that of the cube. The last sign in the trinomial is always a + and the first sign in the trinomial is always that opposite of what the cube has.
• We can multiply these formulas out to make sure that they are true.
Factor
• Going back to an earlier slide, lets factor this completely now
Example : rewrite so it looks like a sum of cubes first
Example
Factoring: A General Review Solving Quadratic Equations by Factoring
Checklist for factoring polynomials of any type. • 1. If the polynomial has a greatest common factor other than 1, factor out this GCF • 2. If the polynomial has two terms (binomial), check to see if it is a difference of two squares or sum/difference of two cubes and factor accordingly, if it is a sum of squares it will not factor • 3. If the polynomial has three terms (trinomial), then either it is a perfect square trinomial, which will factor into a binomial squared, or it is not a perfect square trinomial, in which you can try to factor using methods developed in 7. 2 and 7. 3 • 4. If the polynomial has more than three terms, try to factor by grouping • 5. As a final check, see if any of the factors you have written can be factored again or further. If you have overlooked a common factor you can catch it here. Remember, some polynomials cannot be factored, in which case they are considered prime.
Practice
Solving Quadratic Equations by Factoring This will be the exact same thing that we have been doing, except Now there will be an equals sign and another expression on the other side, the overall goal is to make one side of the equal sign a zero and then factor the remaining side. Then use what we call the ZEROFACTOR PROPERTY
Definition: Quadratic Equations • Any equation that can be put in the form ax 2+bx+c=0, where a, b, and c are real numbers (a cannot equal 0), is called a quadratic equation. The equation ax 2+bx+c=0 is called standard form for a quadratic equation.
Zero-Factor Property • Let a and b represent real numbers. If a x b =0, then a=0 or b=0. • Thus if we factor x 2+5 x+6=0, we are left with (x+2)(x+3)=0. Because there is multiplication and our product is zero, one or both factors, must indeed be zero themselves, thus we make the equations x+2=0 and x+3=0 and solve for x. This will then provide the only values for x that will force the quadratic equation to equate to zero.
Solve the quadratic equation
Solve
What to do when a GCF is factored out and the GCF is just a constant.
What to do when the GCF has a variable term.
• On occasions you may have to multiply something out and combine like terms before you can factor. (initially the problem may actually look pre-factored but it is not because the other side is not a zero).
Example
Harder example
Solve: this example is not a quadratic, (we call it a cubic) but it can be solved in the same manner
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