Find function from remainders

**arze** · September 21st 2010, 07:56 PM

Determine $g(x)\equiv (px+q)^2(2x^2-5x-3)$ where p,q are constants, given that, on division by $(x-2)^2$ , the remainder is $-39-3x$ .

I found
$g(2)=(2p+q)^2(-5)=-45$
$(2p+q)^2=9$
$2p+q=\pm 3$

I don't think I'm on the right track, but I don't know how else to work it out.
Thanks!

**emakarov** · September 22nd 2010, 03:30 PM

Good start. I am not sure if the following is the simplest way, but it seems to lead to finding p and q, at least in theory.

So, we have $(px+q)^2(2x^2-5x-3)=(x-2)^2f(x)-39-3x$ for some function $f(x)$ . Now, looking at degrees of polynomials, $f(x)$ is a quadratic polynomial, let's say $ax^2+bx+c$ .

Next, we look at both sides' coefficients of $x^4$ , $x^3$ , as well as the free terms. Equating the leading coefficients, we see that $a=2p^2$ . Similarly, equating the constant terms (or putting x = 0), we get that $c=(39-3q^2)/4$ . Finally, equating the coefficients at $x^3$ , we get that $b=3p^2+4pq$ . Therefore, $f(x)=2p^2+(3p^2+4pq)x+(39-3q^2)/4$ .

Next, one of the roots of $2x^2-5x-3=0$ is -1/2. Substituting $x=-1/2$ into the original equation, we get $f(-1/2)=-15$ , or $4f(-1/2)=-60$ . Substitute the expression for $f(x)$ above; this gives (after lots of simplifications) $4p^2+3q^2+8pq=99$ .

Finally, combining this with $2p+q=\pm 3$ , one can find $p$ and $q$ . In my calculations, this boils down to a linear (not even a quadratic) equation.

Disclaimer: my calculations can be wrong.

**arze** · September 22nd 2010, 04:32 PM

i solved it, but there were 2 possible values for p and 4 possible values for q. which is too many.

**emakarov** · September 22nd 2010, 04:49 PM

In my calculations, it's p = -6 and q = 15 or p = 6 and q = -15. It does not matter which is chosen since px+q is squared.

**arze** · September 22nd 2010, 05:13 PM

the answer is p=1, q=-1. what i got was p=1, q=1.

**emakarov** · September 22nd 2010, 05:29 PM

p=1, q=-1 does not fit 2p + q = 3.

**sa-ri-ga-ma** · September 22nd 2010, 06:13 PM

Given problem is

$(px + q)^2(2x^2 - 5x - 3)$

$(px + q)^2[(2x^2 - 8x +8) +3x -11]$

Since (2x^2 - 8x +8) is divisible by (x-2)^2

$(px + q)^2(3x - 11) = (x - 2)^2(ax + b) - 39 - 3x$

Compare the coefficients of x^3....> 3p^2 = a...(1)

Compare the coefficients of x^2...> 6pq - 11p^2 = -4a + b....(2)

Compare the coefficients of x....> 3q^2 - 22q = 4a - 4b - 3....(3)

Compare the constants....> -11q^2 = 4b - 39...(4)

Adding them you get

-8(p+q)^2 = a + b - 42

To satisfy equation, if p+q = 2, a+b should be 10.

from eq(1) if p = 1 , a = 3.

from eq.(2) if q = 1 , b = 7.

Hence p = 1 and q = 1.

**arze** · September 22nd 2010, 07:30 PM

This is the answer given: $(x-1)^2(2x^2-5x-3)$
So it means that p=1, and q=-1.

**sa-ri-ga-ma** · September 22nd 2010, 09:08 PM

You can cross check by dividing the given polynomial by ( x-2)^2 and finding the reminder.

**arze** · September 23rd 2010, 02:44 AM

Originally Posted by sa-ri-ga-ma

To satisfy equation, if p+q = 2, a+b should be 10.

How do I know p+q=2?

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