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 October 28th, 2014, 02:37 AM #1 Member   Joined: Sep 2014 From: Kansas Posts: 34 Thanks: 1 Confused on factoring question This one threw me for a loop: Find all integers b so that the polynomial 2x^2 +bx+3 can be factored. If you wouldn't mind explaining your answer? Thank you! Last edited by skipjack; October 31st, 2014 at 12:27 AM.
 October 28th, 2014, 03:18 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra Consider the determinant, $b^2 - 4ac$. Thanks from topsquark
October 28th, 2014, 05:57 AM   #3
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Quote:
 Find all integers $b$ so that the polynomial $2x^2 +bx+3$ can be factored.

Hint: consider the factors.

$\qquad (2x \pm 3)(x \pm 1)$

$\qquad (2x \pm 1)(x \pm 3)$

October 28th, 2014, 08:10 AM   #4
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Quote:
 Originally Posted by soroban Hint: consider the factors. $\qquad (2x \pm 3)(x \pm 1)$ $\qquad (2x \pm 1)(x \pm 3)$
That assumes that the roots are rational. Which may have been intended by the question, but it's not stated.

Mind you, it's not stated that the roots are real either, but I think we can assume that they are.

 October 28th, 2014, 08:57 AM #5 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms It also doesn't say that the roots are Mersenne primes, why not try that? I think soroban's solution is the intended one. I think the most amusing answer would be if you assume the roots are allowed to be anywhere in C, in which case the fundamental theorem of algebra gives an easy answer. Thanks from topsquark
October 28th, 2014, 10:00 AM   #6
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Quote:
 Originally Posted by v8archie Consider the determinant, $b^2 - 4ac$.
That's "discriminant".

October 28th, 2014, 10:13 AM   #7
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Quote:
 Originally Posted by badmathdog This one threw me for a loop: Find all integers b so that the polynomial 2x^2 +bx+3 can be factored. If you wouldn't mind explaining your answer? Thank you!
You look for two numbers whose product is $\displaystyle a\times c=2\times3=6$. There are two possibilities: 2 and 3, or 1 and 6 (sorry, I just list the positive ones), thus $\displaystyle b=2+3=5$ or $\displaystyle b=1+6=7$.

If $\displaystyle b = 5$
$\displaystyle 2x^2+5x+3$
$\displaystyle 2x^2+2x+3x+3$
$\displaystyle 2x(x+1)+3(x+1)$
$\displaystyle (2x+3)(x+1)$

If $\displaystyle b = 7$
$\displaystyle 2x^2+7x+3$
$\displaystyle 2x^2+x+6x+3$
$\displaystyle x(2x+1)+3(2x+1)$
$\displaystyle (x+3)(2x+1)$

Last edited by skipjack; October 31st, 2014 at 12:33 AM.

October 28th, 2014, 12:59 PM   #8
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Quote:
 Originally Posted by badmathdog Find all integers b so that the polynomial 2x^2 +bx+3 can be factored.
$\displaystyle If\ \ x\in\mathbb{R} \ and\ \ f(x)= 2x^2+bx+3, \ \ \ then \\\;\\ f(x)=2(x-x_1)(x-x_2), \ with\ x_1,\ x_2\ \ the\ roots\ of\ equation\ \ f(x)=0 . \\\;\\ x_1,\ x_2\ \in \mathbb{R} \ \ only\ if\ the\ discriminant\ \ b^2-4\cdot2\cdot3 \geq0\ \Longrightarrow\ b^2\geq24\ \ \ (1) \\\;\\ But,\ \ b\in\mathbb{Z}\ \ \ (2) \\\;\\ (1), \ (2)\ \Longrightarrow\ b\leq-5\ \ and\ \ b\geq 5. \\\;\\Therefore\ \ b\in\mathbb{Z}-\{-4,\ -3, \ -2,\ -1,\ 0,\ 1,\ 2,\ 3,\ 4 \}$

 October 30th, 2014, 12:10 PM #9 Member   Joined: Sep 2014 From: Kansas Posts: 34 Thanks: 1 Thanks for all the great input!

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