Chapter 8.5. C OMBINING L IKE T ERMS  Lesson Objective: NCSCOS 1.01 – Write the equivalent forms of algebraic expressions to solve problems  Students.

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Presentation transcript:

Chapter 8.5

C OMBINING L IKE T ERMS  Lesson Objective: NCSCOS 1.01 – Write the equivalent forms of algebraic expressions to solve problems  Students will know how to add and subtract polynomials using the concept of “like terms.”

C OMBINING L IKE T ERMS  *Like Terms – Like terms must all have the same letters and each letter must have the same exponent.  Like terms can be added and subtracted from each other.  This is called “combining like terms”.  If they are not like terms they cannot be added or subtracted from each other.

C OMBINING L IKE T ERMS EExample 1: Solve 3x 2 + 5x 2 33x 2 and 5x 2 are “like terms” because they both contain an x and both of the x’s are squared WWhen we combine like terms either add or subtract the numbers, but the variable will not change TTherefore we can combine 3x 2 and 5x 2 together and get 8x 2 = 8x 2

C OMBINING L IKE T ERMS  Example 2: Solve 4x 3 + 4x 2  4x 3 and 4x 2 are not “like terms” because the x’s have different exponents  Therefore they cannot be combined = 4x 3 + 4x 2

C OMBINING L IKE T ERMS  Example 3: Solve 3x 2 y 3 + 5x 2 y 3  3x 2 y 3 and 5x 2 y 3 are “like terms” because they both have x’s and y’s and the exponents for both x and y are the same.  We can combine them = 8x 2 y 3

C OMBINING L IKE T ERMS  Example 4: Solve 4x 2 y 3 + 4x 2 y 5  4x 2 y 3 and 4x 2 y 5 are not like terms because the exponents on the y’s are different  These cannot be combined = 4x 2 y 3 + 4x 2 y 5

C OMBINING L IKE T ERMS  Example 5: Find (2x 2 + x – 8) + (3x – 4x 2 + 2)  Combine like terms: (2x 2 + x – 8) + (3x – 4x 2 + 2)  ( ) + ( ) + ( )  Add them and write your answer: -2x 2 + 4x – 6  Rule: When adding polynomials you just combine the like terms. Remember minus signs are a negative for the number right after it! 2x 2 – 4x 2 x + 3x-8 + 2

C OMBINING L IKE T ERMS  Example 5: Find (2x 2 + x – 8) + (3x – 4x 2 + 2)  Let’s try this a different way  We can set up this addition problems like:

C OMBINING L IKE T ERMS  Let’s do the same with our problem  Make sure the numbers line up properly when you do it!  Now add them up (2x 2 + x – 8)

C OMBINING L IKE T ERMS 1. 2x 2 + 3x x 3 + 2x x 2 y 3 – 3x 2 y 3 4. (7x 2 + 2x – 3) + (4x 2 – x + 5)

C OMBINING L IKE T ERMS 1. 2x 2 + 3x x 3 + 2x x 2 y 3 – 3x 2 y 3 4. (7x 2 + 2x – 3) + (4x 2 – x + 5) 5x 2 5x 3 + 2x 2 4x 2 y 3 11x 2 + x + 2

C OMBINING L IKE T ERMS  Example 6: Find (3x 2 + 2x – 6) – (3x + x 2 + 3)  Distribute the negative sign into the parenthesis: (3x 2 + 2x – 6) – (3x + x 2 + 3)  (3x 2 + 2x – 6) + (–3x – x 2 – 3)  Combine like terms: (3x 2 – x 2 ) + (2x – 3x) + (-6 – 3)  Add them and write your answer: 2x 2 – x – 9  Rule: When subtracting polynomials you must first distribute the negative into the parenthesis before combining like terms.

 Example 6: Find  First, distribute the negative into the parenthesis after it  Now all the parenthesis can go away

 Combine like terms  I always put the sign in with my squares or circles

 Or you can add them  Make sure to line them up properly!

C OMBINING L IKE T ERMS  Example: (2x 2 + 3x – 1) + (5x 2 – 3x + 5)  Combine like terms  *If two numbers add up to zero, do not write them in the final answer.

C OMBINING L IKE T ERMS 1. (7x 2 + 2x – 3) – (4x 2 – x + 5) 2. (5x 3 – 2x 2 + 4) – (3x 3 + 2x – 2) 3. (3x 2 + 4x – 2) – (6x 2 + 4x – 5)

C OMBINING L IKE T ERMS 1. (7x 2 + 2x – 3) – (4x 2 – x + 5) 2. (5x 3 – 2x 2 + 4) – (3x 3 + 2x – 2) 3. (3x 2 + 4x – 2) – (6x 2 + 4x – 5) 3x 2 + 3x – 8 2x 3 – 2x 2 – 2x x 2 + 3

C OMBINING L IKE T ERMS 1. 3x 2 + 5x – 2x 2 2. (3x 2 + 2x + 4) + (4x 2 – x + 6) 3. (2x 3 – 3x 2 + 5) – (x 3 – 2x – 2) 4. (6x 2 + 2x – 2) – (3x 2 + 2x – 5) 5. (5x 2 y 3 – 2xy + 4y 2 ) – (2x 2 y 3 + 5x 3 – 5xy)

C OMBINING L IKE T ERMS 1. 3x 2 + 5x – 2x 2 2. (3x 2 + 2x + 4) + (4x 2 – x + 6) 3. (2x 3 – 3x 2 + 5) – (x 3 – 2x – 2) 4. (6x 2 + 2x – 2) – (3x 2 + 2x – 5) 5. (5x 2 y 3 – 2xy + 4y 2 ) – (2x 2 y 3 + 5x 3 – 5xy) x 2 + 5x 7x 2 + x + 10 x 3 – 3x 2 + 2x + 7 3x x 3 + 3x 2 y 3 + 3xy + 4y 2