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 Post subject: Stumped on Zeroes of a Function ProblemPosted: Thu, 26 Jan 2012 04:48:46 UTC
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Joined: Thu, 26 Jan 2012 04:04:32 UTC
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Granted this is a new topic for our class, but I am stumped with this question:

"A math textbook asks students to find all the zeroes of the function h(x)=x^4-2x^3+14x^2+6x-51 and gives a hint that 1+4i is a zero."

I am supposed to name one other zero.

I have calculated possible roots/zeroes of 1,3,17,51. I can't seem to grasp how to test these to see if they are a zero . I have tried just inserting the number(s) in for (x) and none of them are working.

Could someone please help with this problem? It would be most appreciated if you could give me the steps as well.

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 Post subject: Re: Stumped on Zeroes of a Function ProblemPosted: Thu, 26 Jan 2012 05:06:55 UTC
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HINT: [1 + 4*SQRT(-1)] * [1 - 4*SQRT(-1)] = 17

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 Post subject: Re: Stumped on Zeroes of a Function ProblemPosted: Thu, 26 Jan 2012 05:21:38 UTC
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Joined: Thu, 26 Jan 2012 04:04:32 UTC
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Denis,

You're a prince! Thanks.. I was going about it all the wrong way... and have seen the light...

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 Post subject: Re: Stumped on Zeroes of a Function ProblemPosted: Thu, 26 Jan 2012 23:17:42 UTC
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Joined: Fri, 1 Jul 2011 01:17:26 UTC
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jillmc107 wrote:
Granted this is a new topic for our class, but I am stumped with this question:

"A math textbook asks students to find all the zeroes of the function h(x)=x^4-2x^3+14x^2+6x-51 and gives a hint that 1+4i is a zero."

I am supposed to name one other zero.

I have calculated possible roots/zeroes of 1,3,17,51. I can't seem to grasp how to test these to see if they are a zero . I have tried just inserting the number(s) in for (x) and none of them are working.

Could someone please help with this problem? It would be most appreciated if you could give me the steps as well.

Since all the coefficients are real numbers, any complex roots must appear as conjugate pairs. Therefore 1-4i is also a root. To get the remaining roots, construct the quadratic with roots 1+4i and 1-4i. Divide (using synthetic division) the original equation by this quadratic. The quotient will be a quadratic, whose roots are the other two roots of the original equation.

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 Post subject: Re: Stumped on Zeroes of a Function ProblemPosted: Fri, 27 Jan 2012 01:15:09 UTC
 S.O.S. Oldtimer

Joined: Fri, 27 Jul 2007 10:17:26 UTC
Posts: 278
Location: Chandler, AZ, USA
How do you use synthetic division to divide by a quadratic? I think you meant to say long division.
You could use synthetic division with each of the two known factors (one at a time), although it gets a bit messy when complex numbers are involved.

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 Post subject: Re: Stumped on Zeroes of a Function ProblemPosted: Fri, 27 Jan 2012 06:36:22 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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Location: Austin, TX
jillmc107 wrote:
Granted this is a new topic for our class, but I am stumped with this question:

"A math textbook asks students to find all the zeroes of the function h(x)=x^4-2x^3+14x^2+6x-51 and gives a hint that 1+4i is a zero."

I am supposed to name one other zero.

I have calculated possible roots/zeroes of 1,3,17,51. I can't seem to grasp how to test these to see if they are a zero . I have tried just inserting the number(s) in for (x) and none of them are working.

Could someone please help with this problem? It would be most appreciated if you could give me the steps as well.

Every time you have a complex root to a polynomial with real coefficients, you are guaranteed that the complex conjugate is also a root. This takes some more advanced machinery than you have to prove, but it's given as a fact to many in their algebra classes.

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 Post subject: Re: Stumped on Zeroes of a Function ProblemPosted: Fri, 27 Jan 2012 06:39:59 UTC
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Well, since we have 17, and want to end with -51, then has to be 17 * -3;
and to get -3, we have 3 choices: -3 * 1 or 3 * -1 or SQRT(3) * -SQRT(3).
In other words: (x-1)(x+3) or (x+1)(x-3) or (x+SQRT(3))(x-SQRT(3)).

Or am I over-simplifying this?

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 Post subject: Re: Stumped on Zeroes of a Function ProblemPosted: Fri, 27 Jan 2012 06:41:23 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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Denis wrote:
Well, since we have 17, and want to end with -51, then has to be 17 * -3;
and to get -3, we have 3 choices: -3 * 1 or 3 * -1 or SQRT(3) * -SQRT(3).
In other words: (x-1)(x+3) or (x+1)(x-3) or (x+SQRT(3))(x-SQRT(3)).

Or am I over-simplifying this?

It depends, if the original question asked for zeros among the choices the op gave (I don't think that's the case) then your way makes sense, however I think the op implicitly assumed all the roots were integers, which just isn't true.

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 Post subject: Re: Stumped on Zeroes of a Function ProblemPosted: Sat, 28 Jan 2012 00:55:07 UTC
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Joined: Fri, 1 Jul 2011 01:17:26 UTC
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alstat wrote:
How do you use synthetic division to divide by a quadratic? I think you meant to say long division.
You could use synthetic division with each of the two known factors (one at a time), although it gets a bit messy when complex numbers are involved.

It is a matter of terminology. What I mean by synthetic division, you call long division.

http://en.wikipedia.org/wiki/Synthetic_division

Here the term used is "expanded synthetic division" for long division.

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