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anonymous

  • 5 years ago

if f(x)=(x+1)(1-x^2), evaluate f(a+3)

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  1. anonymous
    • 5 years ago
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    some help me

  2. anonymous
    • 5 years ago
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    Plug in (a+3) for every value of x in f(x).

  3. anonymous
    • 5 years ago
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    uh huh i got =((a+3)+1)(1-(a+3)^2) dont know what to do after that

  4. anonymous
    • 5 years ago
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    \[f(a+3) = (a+3+1)(1-(a+3)^2)\] just like pjschlotter said.

  5. anonymous
    • 5 years ago
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    Got it!

  6. anonymous
    • 5 years ago
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    Solve it like: (a+3+1) = (a+4) (a+4) (1-(a^2 +6a + 9)) (a+4) (-a^2 -6a -8)

  7. anonymous
    • 5 years ago
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    You can also multiply (a+4) for the rest, but I don't think it's necessary... this form is clearer.

  8. anonymous
    • 5 years ago
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    thank you where is the 6a from?

  9. anonymous
    • 5 years ago
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    It's from the development of (a+3)^2, right? (b+c)^2 = b^2 + 2.b.c + c^2 So, (a+3)^2 = a^2 + 2.a.3 + 3^2 = a^2 + 6a + 3^2

  10. anonymous
    • 5 years ago
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    ok kinda got it so if the number is 6 instead of 3 then it would be 12a instead of 6a correct?

  11. anonymous
    • 5 years ago
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    Correct!

  12. anonymous
    • 5 years ago
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    so is the solution -12a^2-32a-32

  13. anonymous
    • 5 years ago
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    It's (a+4) (-a^2 -6a -8), as I said!

  14. anonymous
    • 5 years ago
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    oh i though you have to combine them. but thanks alot how do i become a fan?

  15. anonymous
    • 5 years ago
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    can you help me with another problem?

  16. anonymous
    • 5 years ago
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    Of course, just send it. :)

  17. anonymous
    • 5 years ago
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    Erm, actually, I gotta go sleep. lol Class tomorrow. See ya.

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