1. ## 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?
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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!

2. Originally Posted by lilrhino
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.

$\displaystyle (x + 1)^2 - 4 = 0$

$\displaystyle \Rightarrow (x + 1)^2 = 4$

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

$\displaystyle \Rightarrow x = -1 \pm 2$

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

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.

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

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

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

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

3. Originally Posted by Jhevon
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.

$\displaystyle (x + 1)^2 - 4 = 0$

$\displaystyle \Rightarrow (x + 1)^2 = 4$

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

$\displaystyle \Rightarrow x = -1 \pm 2$

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

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.

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

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

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

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

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

When I subtract the 2 from -4 that's what I get

4. Originally Posted by lilrhino
Jhevon, thanks for the response but...
What the heck am I doing wrong? Because when I factor I don't get:
$\displaystyle \Rightarrow x^2 + 2x - 3 = 0$

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

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

$\displaystyle (x + 1)^2 - 4 = (x + 1)(x + 1) - 4$

$\displaystyle = x^2 + x + x + 1^2 - 4$

$\displaystyle = x^2 + 2x + 1 - 4$

$\displaystyle = x^2 + 2x - 3$

it seems you added 1 and 1 and got the 2. its 1 times 1, which is 1

5. Originally Posted by Jhevon
Let's do it the long way:

$\displaystyle (x + 1)^2 - 4 = (x + 1)(x + 1) - 4$

$\displaystyle = x^2 + x + x + 1^2 - 4$

$\displaystyle = x^2 + 2x + 1 - 4$

$\displaystyle = x^2 + 2x - 3$

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!

6. Originally Posted by lilrhino
I get it. I was adding vs. multiplying. Go figure. Thanks a bunch!
you may also want to remember a little shortcut.

In general:

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

So here, $\displaystyle (x + 1)^2 = x^2 + 2(x)(1) + 1^2 = x^2 + 2x + 1$

7. Originally Posted by Jhevon
you may also want to remember a little shortcut.

In general:

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

So here, $\displaystyle (x + 1)^2 = x^2 + 2(x)(1) + 1^2 = x^2 + 2x + 1$
I'll add this shortcut to my notes. If I follow it I won't get mixed up...thank you!

8. Originally Posted by lilrhino
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:

$\displaystyle a^2 - b^2 = (a + b)(a - b)$

$\displaystyle a = x + 1$
$\displaystyle b = 2$

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

$\displaystyle ([x + 1] + 2)([x + 1] - 2) = 0$

$\displaystyle (x + 3)(x - 1) = 0$

etc.

-Dan

9. Originally Posted by topsquark
Call me crazy, but is it possible that they want this:

$\displaystyle a^2 - b^2 = (a + b)(a - b)$

$\displaystyle a = x + 1$
$\displaystyle b = 2$

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

$\displaystyle ([x + 1] + 2)([x + 1] - 2) = 0$

$\displaystyle (x + 3)(x - 1) = 0$

etc.

-Dan
why, that's not crazy at all Dan. In fact, that's great! saves us the trouble of expanding

10. Originally Posted by topsquark
Call me crazy, but is it possible that they want this:

$\displaystyle a^2 - b^2 = (a + b)(a - b)$

$\displaystyle a = x + 1$
$\displaystyle b = 2$

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

$\displaystyle ([x + 1] + 2)([x + 1] - 2) = 0$

$\displaystyle (x + 3)(x - 1) = 0$

etc.

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