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 September 13th, 2013, 11:05 AM #1 Newbie   Joined: Sep 2013 Posts: 24 Thanks: 0 Help with polynomials I have never solved something like this, can you please show me step by step? The polynomial P4 (x) = x4 + x3-x2 + x-2 is divisible by x2 +1. Enter zeros polynomial. I used google translate to translate it so please bear with me If you could comment every step as well as present all the laws/rules you were using, it would be great because I really want to learn how to solve these.
September 13th, 2013, 11:49 AM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
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Re: Help with polynomials

Hello, n777l!

Quote:
 $\text{The polynomial }\,f(x) \:=\:x^4\,+\,x^3\,-\,x^2\,+\,x\,-\,2\,\text{ is divisible by }x^2\,+\,1$ $\text{Find the zeros of }f(x).$

$\text{W\!e want to solve: }\:x^4\,+\,x^3\,-\,x^2\,+\,x\,-\,2\:=\:0$

$f(x)\text{ is divisible by }(x^2\,+\,1).$

$\text{Do the division:}$
[color=beige]. . [/color]$\begin{array}{cccccccccccc}
&&&&&& x^2 &+& x &-& 2 \\
&& --&--&--&--&--&--&--&--&-- \\
x^2\,+\,1 & ) & x^4 &+& x^3 &-& x^2 &+& x &-& 2 \\
&& x^4 &&& + & x^2 \\
&& --&--&--&--&-- \\
&&&& x^3 &-& 2x^2 &+& x \\
&&&& x^3 &&& + & x \\
&&&& --&--&--&--&-- \\
&&&&& - & 2x^2 &&& -&2 \\
&&&&& - & 2x^2 &&& -&2 \\
&&&&& --&--&--&--&--&-- \end{array}$

$\text{Hence: }\:(x^2\,+\,x\,-\,2)(x^2\,+\,1) \:=\:0$

[color=beige]. . . . . . [/color]$(x\,-\,1)(x\,+\,2)(x^2\,+\,1) \:=\:0$

$\text{Therefore: }\:x \;=\;1,\,-2,\,\pm\,\!i$

 September 13th, 2013, 05:12 PM #3 Newbie   Joined: Sep 2013 Posts: 24 Thanks: 0 Re: Help with polynomials Thank you, but what does the +-i stand for? How did you come up with the zeros 1, -2 and +-i just by looking at: (x-1)(x+2)(x2+1)=0?
 September 13th, 2013, 06:48 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,805 Thanks: 2150 (x - 1)(x + 2)(x² + 1) = 0 The product on the left-hand side can't be zero unless one of the factors is zero. If x - 1 = 0, x = 1. If x + 2 = 0, x = -2. If x² + 1 = 0, x² = -1 and so x = ?(-1) or -?(-1). As ?(-1) is not a real number, it's convenient to use $^i$ to represent it.
 September 14th, 2013, 03:13 PM #5 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Help with polynomials If you are not familiar with "i" then you probably are only expected to give the real number roots. But still the fact that $x^4+ x^3- x^2+ x- 2$ is divisible by $x^2+ 1$ (and leaves $x^2+ x- 2$) tells you that $x^4+ x^3- x^2+ x- 2= (x^2+ 1)(x^2+ x- 2)$. And since $x^2+ 1$ is not 0 for any real number value of x, any root must satisfy $x^2+ x- 2= (x+ 2)(x- 1)= 0$. Now, use the fact that if the product of two numbers is 0, one or both must be 0. So either x+ 2= 0 or x- 1= 0. If the first is true, then x= -2, if the second is true, then x= 1.
 September 14th, 2013, 03:25 PM #6 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Help with polynomials If you are not familiar with "i" then you probably are only expected to give the real number roots. But still the fact that $x^4+ x^3- x^2+ x- 2$ is divisible by $x^2+ 1$ (and leaves $x^2+ x- 2$) tells you that $x^4+ x^3- x^2+ x- 2= (x^2+ 1)(x^2+ x- 2)$. And since $x^2+ 1$ is not 0 for any real number value of x, any root must satisfy $x^2+ x- 2= (x+ 2)(x- 1)= 0$. Now, use the fact that if the product of two numbers is 0, one or both must be 0. So either x+ 2= 0 or x- 1= 0. If the first is true, then x= -2, if the second is true, then x= 1.

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