Stumped on Zeroes of a Function Problem

jillmc107 · **Joined:** Thu, 26 Jan 2012 04:04:32 UTC **Posts:** 2

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.

Thanks in advance!

Denis · **Posted:** Thu, 26 Jan 2012 05:06:55 UTC

HINT: [1 + 4*SQRT(-1)] * [1 - 4*SQRT(-1)] = 17

jillmc107 · **Joined:** Thu, 26 Jan 2012 04:04:32 UTC **Posts:** 2

Denis,

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

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 321

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.

Thanks in advance!

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.

alstat · **Posted:** Fri, 27 Jan 2012 01:15:09 UTC

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.

Shadow · **Posted:** Fri, 27 Jan 2012 06:36:22 UTC

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.

Thanks in advance!

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.

Denis · **Posted:** Fri, 27 Jan 2012 06:39:59 UTC

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?

Shadow · **Posted:** Fri, 27 Jan 2012 06:41:23 UTC

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.

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 321

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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Stumped on Zeroes of a Function Problem

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