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(17c^6+12c^5-3c^3+2c^2+9)+(5c^6-8c^4+9c^3+11c^2-5c-6)simplify and solve each...

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yyy | eNotes Newbie

Posted December 7, 2008 at 10:31 AM via web

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(17c^6+12c^5-3c^3+2c^2+9)+(5c^6-8c^4+9c^3+11c^2-5c-6)

simplify and solve each polynomial.write it in standard form.

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cburr | Middle School Teacher | (Level 2) Associate Educator

Posted December 7, 2008 at 11:42 AM (Answer #1)

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Since the two parenthetical expressions are added, we can simply ignore the parentheses.  (If they were subtracted, all the +s and -s in the second one would have to be switched.)

Now just find the ones that have the same exponent, since you can't combine ones that are raised to different powers.

17c^6 + 5c^6 = 22c^6

12c^5 stands alone - there is no other c^5

-8c^4 stands alone - there is no other c^4

-3c^3 + 9c^3 = 6c^3

2c^2 + 11c^2 = 13c^2

-5c stands alone - there is no other c

9 - 6 = 3

So, we are left with:

22c^6 + 12c^5 - 8c^4 + 6c^3 + 13c^2 - 5c + 3

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted December 7, 2008 at 8:06 PM (Answer #2)

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After opening the parentheses, we try to group terms which have the same exponents, like this:

c^6(17+5) + 12c^5 - 8c^4+ c^3(-3+9) + c^2 (2+11) - 5c + (9-6)=

22c^6 + 12c^5 - 8c^4 + 6c^3 + 3c^2 - 5c + 3 

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dracomath | Student, Grade 11 | (Level 1) Honors

Posted December 28, 2008 at 12:40 AM (Answer #3)

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In order to solve this problem, add numbers with same variables.

 (17c^6+12c^5-3c^3+2c^2+9)+(5c^6-8c^4+9c^3+11c^2-5c-6)

22c^6+1c^5-8c^4++6c^3+13c^2-5c+3

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revolution | College Teacher | (Level 1) Valedictorian

Posted July 17, 2010 at 10:11 PM (Answer #4)

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(17c^6+12c^5-3c^3+2c^2+9)+(5c^6-8c^4+9c^3+11c^2-5c-6)

= (17+5)c^6+12c^5-8c^4 (-3+9)c^3+(2+11)c^2-5c+9-6

= 22c^6+12c^5-8c^4+6c^3+13c^2-5c+3

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