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### Three Polynomial Questions

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Date: 7 Aug 1995 09:00:07 -0400
From: Greg Sharpe
Subject: Polynomial Questions (3)

Hi!

I have 3 interesting polynomial questions from a past trial HSC paper.  I
am in year 12 in NSW, Australia.  Are there any 'intereactive' real time
discussion groups for maths problems?  Currently, I have www and email
access.

Greg Sharpe  <gsharpe@pioneer.as.edu.au>

1) The polynomial  2x^3 + 3x^2 + ax - 6  has (x + 3) and (2x + b) as factors.
Find a and b.

2) In the polynomial  x^3 - 4x^2 + Ax + 4 = 0, one root is equal to
the sum of the other two roots.  Solve the equation and find A.

3) Find the roots of  2x^3 - 3x^2 - 3x + 2 = 0  given that they are in
geometric progression.
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```
Date: 7 Aug 1995 09:44:23 -0400
From: Dr. Ken
Subject: Re: Polynomial Questions (3)

Hello there!

Nice questions.

>1) The polynomial  2x^3 + 3x^2 + ax - 6  has (x + 3) and (2x + b) as factors.
>   Find a and b.

To do this one, try doing polynomial long division, dividing (x + 3) into
the cubic polynomial.  Since it goes in evenly without remainder, that
should tell you what a is.  Then once you've done that, you'll have a
quadratic polynomial, and you can divide (2x + b) into it to find out what b
is, or you could use the quadratic formula to find out what the roots are
and then use that to figure out what b is.

>2) In the polynomial  x^3 - 4x^2 + Ax + 4 = 0, one root is equal to
>   the sum of the other two roots. Solve the equation and find A.

To figure this out, you'll most likely have to use the fact that the
coefficient on x^2 is the opposite of the sum of the roots.  So if the roots
are p, q, and r, then you have the equation

p + q + r = 4.

Also, the constant term in the polynomial is the opposite of the product of
the roots, so we have the equation

pqr = -4.

Since you said that one root is the sum of the other two, we have the
equation

p + q = r.

That's three equations and three unknowns.  That should do it for you!
(especially look at the first and the last equation together)

>3) Find the roots of  2x^3 - 3x^2 - 3x + 2 = 0  given that they are in
>   geometric progression.

Let's call a, ar, and ar^2 the three roots, with common ratio r.  Then expand
the polynomial

2(x - a)(x - ar)(x - ar^2)

and see how that expansion matches up with the given polynomial.

Good luck!

-K
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Associated Topics:
High School Polynomials

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