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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: 7,403 Thanks: 2477 Math Focus: Mainly analysis and algebra 
Consider the determinant, $b^2  4ac$.

October 28th, 2014, 05:57 AM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Hello, badmathdog! Quote:
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  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,403 Thanks: 2477 Math Focus: Mainly analysis and algebra  Quote:
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. 
October 28th, 2014, 10:00 AM  #6 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,859 Thanks: 1080 Math Focus: Elementary mathematics and beyond  
October 28th, 2014, 10:13 AM  #7  
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry  Quote:
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  
Senior Member Joined: Apr 2014 From: Europa Posts: 571 Thanks: 176  Quote:
\\\;\\ f(x)=2(xx_1)(xx_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^24\cdot2\cdot3 \geq0\ \Longrightarrow\ b^2\geq24\ \ \ (1) \\\;\\ But,\ \ b\in\mathbb{Z}\ \ \ (2) \\\;\\ (1), \ (2)\ \Longrightarrow\ b\leq5\ \ 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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