Chapter 6.1: Introduction to parabolas and solving equations by factoring

Transcription

1 Chapter 6 Solving Quadratic Equations and Factoring Chapter 6.1: Introduction to parabolas and solving equations by factoring If you push a pen off a table, how does it fall? Does it fall like this? Or like this? The second picture is the correct one! The pen will fall in an arc, and that arc is part of a parabola. We get the whole parabola if we start from the ground and throw an object both upward and horizontally: In this picture, the ball starts at 0 feet away, at a height of 0. This is the point (0,0). Then it goes up to a height of 6 feet. It ends up on the ground again 2 feet away, at (2,0). The equation is yy = 6xx or, in factored form, yy = 6xx(xx 2) Can you see how the factored form could give us x=0 and x= 2, for where the ball hits the ground?

2 In our basketball example, the equation yy = 6xx(xx 2) gives us the two places where the ball is on the ground when the equation is equal to zero, since a height of zero, or y=0, is when the ball is on the ground. So, we have 6xx(xx 2) = 0 To solve this equation, we will use the Zero-Product Property, which says that if the product of two (or more) numbers is equal to zero, then either the first number equals zero, the second number equals zero, or they both are equal to zero. More formally, we write, if ab=0, then either a=0, or b=0, or both a and b are zero. In our problem, above, if 6xx(xx 2) = 0, then we know that 6xx = 0, or xx 2 = 0, or both. We solve each to find the two answers 6xx = 0 xx 2 = 0 /-6 / xx = 0 xx = 2 Thus, the ball will be on the ground (a height of y = zero) at two places: 0 feet away and 2 feet away. To use the zero factor property in more problems, we ll first do some factoring review, because to use the property, we must insure that all the factors are set equal to zero. We will first practice factoring out the greatest common factor and factoring simple trinomials. Then we will come back to solving equations using factoring and the zero-product principle. Wait! I still don t really get the zero product property! Explain it again? Okay. I m thinking of two numbers that multiply to get zero. What are my two numbers? Silly, how would I know that? Well, you know what one of them is. OH! One of them has to be zero! You got it!

3 To solve the problem, 6xx = 0, we first had to factor out the greatest common factor, or GCF. On the previous page, the factoring was already given to you as 6xx(xx 2) = 0. How do we know that factoring is correct? We know it s correct, because if we multiply back through using the distributive property, we get back the original problem: 6xx(xx 2) = 6xx xx FACTORING is the OPPOSITE of MULTIPLYING! When we factor, we are trying to get the greatest common factor outside the parenthesis. Then we have to figure out what to put inside the parenthesis, so that if we multiplied, we would get our original problem back. Thus, in factoring, we start with 6xx xx. The greatest common factor of the two terms is 6xx. Then we write 6xx( ) and try to figure out what goes inside the ( ) to get back 6xx xx. The next section gives some review of how to find the GCF and then what goes inside the parenthesis. Factoring Practice Greatest Common Factor We begin by discussing factoring using what we call the Greatest Common Factor, or the GCF for short. Think about all three of the words in GCF: Greatest: The biggest one Common: One that is shared Factor: A part of a multiplication problem So, when we are looking for the GCF, we are looking for the largest factor shared by all of the terms. Take, for example, finding the GCF of 25xx 2 yy 2 and 10xxyy 3. Let s start with finding the GCF of the coefficients, then we will move on to the variables. The factors of 25 are: 1, 5, 25 These are all the numbers that go into 25. The factors of 10 are: 1, 2, 5, 10 These are all the numbers that go into and 10 share two factors, 1 and 5. The greatest factor they share is 5. Now let s find the GCF of each one of the variable terms. For 25xx 2 yy 2, the factors of xx 2 yy 2 are xx, xx, yy, yy For 10xxyy 3, the factors of xxyy 3 are xx, yy, yy, yy They each share factors of x and y: the greatest amount of factors of x that it shares is one, and the greatest amount of factors of y that it shares is two. Therefore, xx 2 yy 2 and xxyy 3 have a GCF of xxyy 22 (one factor of x and two factors of y). Putting it all together, the GCF of 25xx 2 yy 2 and 10xxyy 3 is 5555yy 22.

4 EXAMPLES Find the GCF. a. 36aa 2 bb 2, 48aa 3 bb 4, and 24aa 3 bb 3 STEP 1: Find the GCF of the coefficients Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 Factors of 48: 1,2, 3, 4, 6, 8, 12, 16, Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24 Greatest Common Factor is 12 STEP 2: Find the GCF of the variables 36aa 2 bb 2 : a,a,b,b 48aa 3 bb 4 : a,a,a,b,b,b,b 24aa 3 bb 3 : a,a,a,b,b,b Greatest Common Factor is a,a,b,b or aa 2 bb 2 Therefore, the complete GCF is 12aa 2 bb 2. b. 18aa 2, 25 and 10aa 3 STEP 1: Find the GCF of the coefficients Factors of -18: 1,2, 3, 6, 9, 18 Factors of 25: 1, 5, 10 Factors of 10: 1, 2, 5, 10 Greatest Common Factor is 1 STEP 2: Find the GCF of the variables 25: no variables 18aa 2 : a, a 10aa 3 : a, a, a The variable a is not common to all of the terms, so it cannot be part of the GCF. So the GCF of the variables is also 1. Therefore, the complete GCF is 1. The factors of -18 are actually ± 1, ±2, ±3, ±6, ±9, and ±18. We will discuss when to consider the negative in the next part. Terms that have a complete GCF of 1 are called relatively prime. This does not mean the individual numbers are prime, but that compared to each other, they have no factors in common except one. Now that you know how to find the GCF, let s discuss how to factor out the Greatest Common Factor. Factoring out the GCF consists of finding the GCF, placing it in front of parenthesis, and using the reverse of the distributive property to divide each term of the polynomial by the GCF. c. Factor: 24xx 3 yy 2 18xxyy 3 STEP 1: Find the GCF of both terms. 6xxyy 2 ( ) The GCF is 6xxyy 2. Place it in front of ( ). STEP 2: Imagine what would go inside the parenthesis, so that if you multiplied, you would get back the original problem. 6xxyy 2 (?? ) = 24xx 3 yy 2 18xxyy 3 See if you can get this before you turn the page!

5 6xxyy 2 (4xx 2 3yy) = 24xx 3 yy 2 18xxyy 3 If you had a hard time figuring out what to put inside, there is another way! Since factoring and multiplication are opposites, factoring is like dividing. So you can divide each term by the GCF to get what should go in the ( ). Divide: 24xx 3 yy 2 6xxyy 2, then divide: 18xxyy 3 6xxyy 2 You can write the division as fractions: 24xx3 yy 2 Division means we subtract exponents. We get: 24xx3 1 yy xx1 1 yy = 24xx2 yy 0 6 6xxyy 2 18xx0 yy xxyy3 6xxyy 2. Divide the numbers by 6, and anything to the zero power is 1, so we get: 4xx 2 3yy 6xxyy 2 (4xx 2 3yy) = 24xx 3 yy 2 18xxyy 3 Yes! STEP 3: Check your answer by multiplying. d. Factor: 5xx xx 2 15xx 5xx(? ) STEP 1: Find the GCF of all three terms. The GCF is either 5xx or 5x. Let s use 5xx here because it may be easier to have the result inside the ( ) be positive. Place 5xx in front of ( ). STEP 2: Imagine what would go inside the parenthesis, so that if you multiplied, you would get back the original problem OR divide each term by 5xx: 5xx xx2 15xx 5xx 5xx 5xx Notice what happens to the signs in each term. In the first, the negatives cancel. In the second, we are adding a negative, so we have a negative. In the third, we are subtracting a negative, so we have a positive. xx 2 2xx + 3 STEP 3: Check your answer by multiplying. 5xx(xx 2 2xx + 3) = 5xx xx 2 15xx Yes! In the next part of this section, we will look at trinomials and how to factor them, so that we can see if the inside part of our answer, (xx 2 2xx + 3), can be factored further.

6 Factoring Practice Trinomials PATTERN EXAMPLES When you multiply binomials, a pattern emerges in the answers that we will use in factoring. Look at the examples, below, and see if you see the pattern. a. (xx + 2)(xx + 5) = xx 2 + 2xx + 5xx + 10 = xx 2 + 7xx + 10 How do these two numbers relate to these? What do you do to 2 and 5. to get 7?.to get 10? b. (xx + 4)(xx + 2) = xx 2 + 4xx + 2xx + 8 = xx 2 + 6xx + 8 How do these two numbers relate to these? What do you do to 4 and 2. to get 6?.to get 8? c. Does the same pattern hold here? (xx + 3)(xx + 7) = xx 2 + 3xx + 7xx + 21 = xx xx + 21 d. What about when we include negatives, does the pattern still hold? (xx 3)(xx + 7) = xx 2 3xx + 7xx + 21 = xx 2 + 4xx 21 What do you do to 3 and 7. to get 4?.to get 21? Once you see the pattern, see if you can use the pattern to find the answers to the following multiplication problems, without actually multiplying out the whole problem or using FOIL just use the pattern! e. (xx 4)(xx + 2) = f. (xx 5)(xx 2) = g. (xx 5) 2 = h. (xx 5)(xx + 5) = Try each of these before you turn the page.

7 In each of the examples on the previous page, we add the two numbers in the binomial to get the middle coefficient in the trinomial, and we multiply them to get the constant on the end. a. (xx + 2)(xx + 5) = xx 2 + 2xx + 5xx + 10 = xx 2 + 7xx + 10 Add 2 and 5 to get the middle number, 7; multiply 2(5) to get the end number, 10. We can also see this in our work, since we got2x and 5x, which we added together, since they are like terms. b. (xx + 4)(xx + 2) = xx 2 + 4xx + 2xx + 8 = xx 2 + 6xx + 8 Add 4 and 2 to get the middle number, 6; multiply 4(2) to get the end number, 8. c. (xx + 3)(xx + 7) = xx 2 + 3xx + 7xx + 21 = xx xx + 21 Add 3 and 7 to get the middle number, 10; multiply 3(7) to get the end number, 21. d. (xx 3)(xx + 7) = xx 2 3xx + 7xx + 21 = xx 2 + 4xx 21 Add 3 and 7 to get the middle number, 4; multiply 3(7) to get the end number, -21. e. (xx 4)(xx + 2) = xx 2 2xx (2) f. (xx 5)(xx 2) = xx 2 7xx (-2) g. (xx 5) 2 = (xx 5)(xx 5) = xx 2 10xx (-5) Remember, in this problem we write (xx 5) twice, since it is squared. We also know by the pattern that we should get a middle term, since we are adding -5 to -5 for the middle. h. (xx 5)(xx + 5) = x (5) In this problem, the middle terms cancels out because -5+5 = 0 If we write out the work, we can see (xx 5)(xx + 5) = x 2 5x + 5x 25 the terms in the middle are -5x and 5x, which are opposites when added. Now let s use the pattern to factor! STEPS for factoring trinomials: 1. Factor out any GCF. 2. Write the start of the binomials. For these examples, we will write (x ) (x ), because we know xx xx = xx 2, the first term. If the first term is not xx 2, this will not work more on this in future sections! 3. Look for two numbers that would multiply to equal the last number in the trinomial and add to equal the middle number. Tip: start by listing all the possible things that multiply to get the last number. 4. Check your answer!

8 EXAMPLES factor a. xx 2 + 9xx + 20 STEP 1: The GCF is 1 (xx )(xx ) STEP 2: Write the start of the binomials STEP 3: What multiplies to get 20 and adds to get 9? Start by listing the multiplication: 2(10), 4(5) Which pair adds up to 9? 2+10 =12, no good; 4+5 =9, so 4 and 5 work! (xx + 4)(xx + 5) STEP 4: check! (xx + 4)(xx + 5) = xx 2 + 4xx + 5xx + 20 = xx 2 + 9xx + 20 Yes! Note: Order does not matter here; we could also write (xx + 5)(xx + 4). b. xx 2 + 7xx 30 STEP 1: The GCF is 1 (xx )(xx ) STEP 2: Write the start of the binomials STEP 3: What multiplies to get -30 and adds to get 7? Start by listing the multiplication, and careful: to get a negative result, one of the two numbers must be negative! So, we could have 2(-15), or -2(15), or 3(-10) or -3(10). Which pair adds up to 7? The last one! (xx 3)(xx + 10) STEP 4: check! (xx 3)(xx + 10) = xx 2 3xx + 10xx 30 = xx 2 + 7xx 30 Yes! c. xx 2 7xx 30 STEP 1: The GCF is 1 (xx )(xx ) STEP 2: Write the start of the binomials STEP 3: What multiplies to get -30 and adds to get -7? We can use our previous result to see it must be 3 and -10, so that this time, when we add them, we get -7 instead of +7. (xx + 3)(xx 10) STEP 4: check! (xx + 3)(xx 10) = xx 2 + 3xx 10xx 30 = xx 2 7xx 30 Yes! d. xx 2 8xx + 16 STEP 1: The GCF is 1 (xx )(xx ) STEP 2: Write the start of the binomials STEP 3: What multiplies to get +16 and adds to get -8? 16 =2(8) or 4(4). But none of these will add to get -8! We must have two negative numbers multiplying to get 16, so that when we add them, we get a negative. 16 =-2(-8) or -4(-4). The second pair works: -4+-4=-8 (xx 4)(xx 4) STEP 4: check! (xx 4)(xx 4) = xx 2 4xx 4xx + 16 = xx 2 8xx + 16 Yes! We can also write (xx 4) 2.

9 e. 5xx xx 2 15xx STEP 1: The GCF is not 1! We have to first factor out the GCF. Luckily, we had this problem back a few pages, so we can save some work! 5xx(xx 2 2xx + 3) 5xx (xx )(xx ) STEP 2: Write the start of the binomials. Notice that we Put our GCF out front. 5xx(xx 2 2xx + 3) STEP 3: What multiplies to get +3 and adds to get -2? (3)(1) gets us 3, but doesn t add to get -2. (-3)(-1) gets us 3, but adds to get -4, not -2. This can t be factored more! So we stop. STEP 4: We checked this one in the previous example. f. xx 2 36 STEP 1: The GCF is 1. STEP 2: Wait a second, that s not a trinomial! What is it doing here? Well, it s true, xx 2 36 is not a trinomial unless we think of it as xx 2 + 0xx 36, which seems a little silly but might come in handy. xx 2 36 is a special kind of binomial, called a difference of squares. square square difference (minus sign) square root of xx 2 = xx square root of 36=6 xx 2 36 = (xx + 6)(xx 6) plus minus Any difference of squares will factor into two nearly identical factors! Each factor will have the square root of each original term, but one will have addition in the middle and the other will have subtraction. STEP 3: check if we multiply back through, the middle term cancels out: (xx + 6)(xx 6) = xx 2 + 6xx 6xx 36 = xx 2 + 0xx 36 = xx 2 36 So, it actually does make sense to think of xx 2 36 as being the same as xx 2 + 0xx 36, since +6xx 6xx gives us that middle term of 0x.

10 Factoring to Solve We will now use factoring and the zero-product principle to solve quadratic equations. A quadratic equation is an equation that has a squared term as its highest term. A quadratic equation is considered to be in standard form if it is written aaxx 2 + bbbb + cc = 0, where a, b and c are constants, and aa 0. To solve quadratic equations: 1. Express the equation in standard form. 2. Factor the equation appropriately, first factoring out a GCF if needed. 3. Use the Zero-Product Principle to set each factor equal to Solve each resulting equation. 5. Check the solutions. EXAMPLES Solve each equation. a. xx 2 xx 6 = 0 STEP 1: The equation is already in standard form, since it is set equal to zero and written with the xx 2 term first, then x. STEP 2: Factor. There is no GCF to factor out, so let s see what multiplies to get 6 on the end and adds to get -1 in the middle ( xx is the same as 1xx). 2( 3) = 6, 2(3) = 6, 1( 6) = 6, 1(6) = 6 (xx + 2)(xx 3) = 0 The first pair will also add to get -1 xx + 2 = 0 or xx 3 = 0 STEP 3: Set each factor equal to zero STEP 4: Solve each resulting equation. xx = 2 or xx = 3 STEP 5: Check each solution one at a time. xx 2 xx 6 = 0 for xx = 2 gives ( 2) 2 ( 2) 6 = 0? = 0 0 = 0 Yes! xx 2 xx 6 = 0 for xx = 3 gives (3) 2 (3) 6 = 0? = 0 0 = 0 Yes! b. 2xx 3 16xx xx = 0 STEP 1: The equation is already in standard form. STEP 2: Factor. There is a GCF to factor out first! Look for what each term has in common. Each term has an x, and each term is divisible by 2, so factor 2x out before continuing.

11 2xx(xx 2 8xx + 7) = 0 Now see what multiplies to get +7 on the end and adds to -8 in the middle. ( 7)( 1) = 7 and 7 + ( 1) = 8 2xx(xx 7)(xx 1) = 0 STEP 3: Set each factor equal to zero. Notice that there are now three factors to set equal to zero. STEP 4: Solve each resulting equation. 2x = 0 or xx 7 = 0 or xx 1 = 0 /2 / xx = 0 or xx = 7 or xx = 1 STEP 5: Check each solution one at a time in the original (unfactored) equation. 2xx 3 16xx xx = 0 for x=0 gives 2(0) 3 16(0) (0) = = 0 Yes! 2xx 3 16xx xx = 0 for x=7 gives 2(7) 3 16(7) (7) = = 0 Yes! 2xx 3 16xx xx = 0 for x=1 gives 2(1) 3 16(1) (1) = = 0 Yes! Any time you check your work in any algebra you do, it s a good idea to use your original equation, in case you made a mistake when you rewrote the equation. c. 5xx = 0 STEP 1: The equation is already in standard form. 5(xx 2 36) = 0 5(xx 6)(xx + 6) = 0 STEP 2: Factor. There is a GCF to factor out first, since both terms are divisible by 5 Now we see that we have a difference (subtraction) of squares, since 36 is a square and xx 2 is a square. STEP 3: Set each factor equal to zero and STEP 4, solve. 5 = 0 oooo (xx 6) = 0 oooo (xx + 6) = Not true x = 6 x=-6 STEP 5: Since 5=0 is not a solution, we have two remaining solutions, 6 or -6, which we can check in the original equation: 5xx = 0 for x = 6 gives 5(6) = 5(36) 180 = = 0 5xx = 0 for x = -6 gives 5( 6) = 5(36) 180 = = 0 The two answers are x = 6 or -6.

12 d. 2xx = 0 STEP 1: The equation is already in standard form. STEP 2: Factor. There is a GCF to factor out first, since both terms are divisible by 2. 2(xx ) = 0 Now we see that we have a difference (subtraction) of squares, since 49 is a square and xx 2 is a square. Wait a second, that s not a difference of squares! Difference means subtraction! That s the sum of two squares! That s true! We actually can t factor anymore! You can t factor xx ! You could try, but (xx 7)(xx 7) = xx 2 14xx + 49 and (xx + 7)(xx + 7) = xx xx + 49 and (xx + 7)(xx 7) = xx 2 49 so none of them work to get xx ! We stop here and say there is no solution for the real numbers. If we allowed for complex numbers, we could have xx = 7ii as a solution.. But let s just say here that there is no real number solution. More on the complex number possibilities in later sections! WHEW! e. xx 2 8xx + 5 = 10 STEP 1: The equation is not in standard form, since it is not equal to zero. Add 10 to both sides to get it in standard form. xx 2 8xx + 5 = xx 2 8xx + 15 = 0 STEP 2: Factor. There is no GCF to factor out, so see what multiplies to get 15 on the end and adds to get -8 in the middle. (xx 5)(xx 3) = 0 (-5)(-3)=15 and =-8 xx 5 = 0 oooo xx 3 = 0 STEP 3: Set each factor equal to zero and STEP 4: solve xx = 5 oooo xx = 3 STEP 5: The check is left to you! How do you know when to only factor and when to keep going to get an answer? f. xx 2 2xx + 2 Do we just factor this problem, or do we find what x is? See if you can decide on your own before you turn the page!

13 (xx 1)(xx 1) STOP! This equation is not equal to zero! In fact, there is no equal sign at all! So, we can only simplify the problem by factoring, we can t solve it. If we had been given xx 2 2xx + 2 = 0, then we could solve by factoring. We would get: xx 1 = 0 or xx 1 = 0, which would give us the one solution, x=1. We can only solve an algebra problem if there is an equal sign = in it somewhere!