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Math Help - Help solving polynomials

  1. #1
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    Help solving polynomials

    I am drawing a blank on these problems :

    (x+2)(x+1)=x^2 + 11

    My first thoughts were to multiply what's left of the equal sign...

    So I got:

    x^2 + 3x + 3 = x^2 + 11

    Next, I moved the x^2 to the right of the "=" to the left and it was eliminated.

    Next, I moved the 3 to the right side and got - 3x = 8 and then x = 8/3

    Is that correct?
    ==============================
    Next I have to solve by factoring:

    (x+1)^2 - 4 = 0

    My first instincts are to rewrite (x+1)^2 as (x+1) (x+1) and then multiply.

    I got: x^2 + 2x - 2 = 0. Do I use the quadratic formula to solve? Thanks in advance!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lilrhino View Post
    I am drawing a blank on these problems :

    (x+2)(x+1)=x^2 + 11

    My first thoughts were to multiply what's left of the equal sign...

    So I got:

    x^2 + 3x + 3 = x^2 + 11

    Next, I moved the x^2 to the right of the "=" to the left and it was eliminated.

    Next, I moved the 3 to the right side and got - 3x = 8 and then x = 8/3

    Is that correct?
    ==============================
    yes, you are correct. you made a typo though. you should have 3x = 8 not -3x = 8

    Next I have to solve by factoring:

    (x+1)^2 - 4 = 0

    My first instincts are to rewrite (x+1)^2 as (x+1) (x+1) and then multiply.

    I got: x^2 + 2x - 2 = 0. Do I use the quadratic formula to solve? Thanks in advance!
    it makes no sense to factor here, or expand.

    (x + 1)^2 - 4 = 0

    \Rightarrow (x + 1)^2 = 4

    \Rightarrow x + 1 = \pm \sqrt {4} = \pm 2

    \Rightarrow x = -1 \pm 2

    \Rightarrow x = 1 \mbox { or } x = -3


    Also, if they say "by factoring" using the quadratic formula would be wrong, that's not factoring.

    Let's do it the factoring way, which i think is harder than what i did in this case.

    (x + 1)^2 - 4 = 0 ..........expand as you did before

    \Rightarrow x^2 + 2x - 3 = 0 ......now we factor

    \Rightarrow (x + 3)(x - 1) = 0 ........this is factoring, do you understand how to get to this step?

    \Rightarrow x = -3 \mbox { or } x = 1
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    yes, you are correct. you made a typo though. you should have 3x = 8 not -3x = 8

    it makes no sense to factor here, or expand.

    (x + 1)^2 - 4 = 0

    \Rightarrow (x + 1)^2 = 4

    \Rightarrow x + 1 = \pm \sqrt {4} = \pm 2

    \Rightarrow x = -1 \pm 2

    \Rightarrow x = 1 \mbox { or } x = -3


    Also, if they say "by factoring" using the quadratic formula would be wrong, that's not factoring.

    Let's do it the factoring way, which i think is harder than what i did in this case.

    (x + 1)^2 - 4 = 0 ..........expand as you did before

    \Rightarrow x^2 + 2x - 3 = 0 ......now we factor

    \Rightarrow (x + 3)(x - 1) = 0 ........this is factoring, do you understand how to get to this step?

    \Rightarrow x = -3 \mbox { or } x = 1
    Jhevon, thanks for the response but...
    What the heck am I doing wrong? Because when I factor I don't get:
    \Rightarrow x^2 + 2x - 3 = 0

    I get...
    \Rightarrow x^2 + 2x - 2 = 0

    When I subtract the 2 from -4 that's what I get
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lilrhino View Post
    Jhevon, thanks for the response but...
    What the heck am I doing wrong? Because when I factor I don't get:
    \Rightarrow x^2 + 2x - 3 = 0

    I get...
    \Rightarrow x^2 + 2x - 2 = 0

    When I subtract the 2 from -4 that's what I get
    Let's do it the long way:

    (x + 1)^2 - 4 = (x + 1)(x + 1) - 4

    = x^2 + x + x + 1^2 - 4

    = x^2 + 2x + 1 - 4

    = x^2 + 2x - 3

    it seems you added 1 and 1 and got the 2. its 1 times 1, which is 1
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    Let's do it the long way:

    (x + 1)^2 - 4 = (x + 1)(x + 1) - 4

    = x^2 + x + x + 1^2 - 4

    = x^2 + 2x + 1 - 4

    = x^2 + 2x - 3

    it seems you added 1 and 1 and got the 2. its 1 times 1, which is 1
    I get it. I was adding vs. multiplying. Go figure. Thanks a bunch!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lilrhino View Post
    I get it. I was adding vs. multiplying. Go figure. Thanks a bunch!
    you may also want to remember a little shortcut.

    In general:

    (x \pm y)^2 = x^2 \pm 2xy + y^2

    So here, (x + 1)^2 = x^2 + 2(x)(1) + 1^2 = x^2 + 2x + 1
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    you may also want to remember a little shortcut.

    In general:

    (x \pm y)^2 = x^2 \pm 2xy + y^2

    So here, (x + 1)^2 = x^2 + 2(x)(1) + 1^2 = x^2 + 2x + 1
    I'll add this shortcut to my notes. If I follow it I won't get mixed up...thank you!
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  8. #8
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    Quote Originally Posted by lilrhino View Post
    Next I have to solve by factoring:

    (x+1)^2 - 4 = 0

    My first instincts are to rewrite (x+1)^2 as (x+1) (x+1) and then multiply.

    I got: x^2 + 2x - 2 = 0. Do I use the quadratic formula to solve? Thanks in advance!
    Call me crazy, but is it possible that they want this:

    a^2 - b^2 = (a + b)(a - b)

    In your problem let
    a = x + 1
    b = 2

    So
    (x+1)^2 - 4 = 0

    ([x + 1] + 2)([x + 1] - 2) = 0

    (x + 3)(x - 1) = 0

    etc.

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Call me crazy, but is it possible that they want this:

    a^2 - b^2 = (a + b)(a - b)

    In your problem let
    a = x + 1
    b = 2

    So
    (x+1)^2 - 4 = 0

    ([x + 1] + 2)([x + 1] - 2) = 0

    (x + 3)(x - 1) = 0

    etc.

    -Dan
    why, that's not crazy at all Dan. In fact, that's great! saves us the trouble of expanding
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  10. #10
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    Quote Originally Posted by topsquark View Post
    Call me crazy, but is it possible that they want this:

    a^2 - b^2 = (a + b)(a - b)

    In your problem let
    a = x + 1
    b = 2

    So
    (x+1)^2 - 4 = 0

    ([x + 1] + 2)([x + 1] - 2) = 0

    (x + 3)(x - 1) = 0

    etc.

    -Dan
    Thanks Dan, that was a really helpful tip .
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