hard PS factors

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hard PS factors

by jsl » Thu Sep 11, 2008 2:17 pm
What is the value of b if 2x-1 is a factor of 2x^2 + bx + 3

A. -7
B. -3
C. -1
D. 3
E. 7

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Re: hard PS factors

by parallel_chase » Thu Sep 11, 2008 9:14 pm
jsl wrote:What is the value of b if 2x-1 is a factor of 2x^2 + bx + 3

A. -7
B. -3
C. -1
D. 3
E. 7
I have solved it using Factor theorem.
2x-1 = 0

x=1/2

2* 1/4 + b/2 +3 = 0

1/2 + 3 + b/2 = 0

7/2 + b/2 = 0

therefore, b should be -7.

Whats the OA?

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Re: hard PS factors

by sudhir3127 » Thu Sep 11, 2008 9:53 pm
jsl wrote:What is the value of b if 2x-1 is a factor of 2x^2 + bx + 3

A. -7
B. -3
C. -1
D. 3
E. 7

try to factor out 2x^2 + bx + 3
we know one factor is 2X-1

( 2x-1 ) ( x-3)

2X^2- 7X +3

so we know b would be -7

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by boysangur » Mon Nov 15, 2010 4:13 pm
parallel_chase wrote:
jsl wrote:What is the value of b if 2x-1 is a factor of 2x^2 + bx + 3

A. -7
B. -3
C. -1
D. 3
E. 7
I have solved it using Factor theorem.
2x-1 = 0

x=1/2

2* 1/4 + b/2 +3 = 0

1/2 + 3 + b/2 = 0

7/2 + b/2 = 0

therefore, b should be -7.

Whats the OA?
I'm not even familiar with factor theorem, could you explain how it fits in here? Thank you

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by pkan51 » Mon Nov 15, 2010 6:19 pm
sudhir3127 wrote:
jsl wrote:What is the value of b if 2x-1 is a factor of 2x^2 + bx + 3

A. -7
B. -3
C. -1
D. 3
E. 7
2X^2- 7X +3

so we know b would be -7
Should b be (7)? 2X^2-(7)X+3

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by Rezinka » Mon Nov 15, 2010 10:57 pm
@ boysangur :
Whenever any expression 'a' is a factor of expression 'b', you can equate expression 'a' to 0 and put the value in expression 'b' and the result would be 0.
E.g.
Say, x-1 is a factor of a*(x^2) + b*x + c
Now,
equate x - 1 to 0
i.e.
x-1=0
or, x=1
Put this value in the given expression and the result should be 0.
i.e.
Put x = 1 in a*(x^2) + b*x + c and equate to 0.
or,
a*(1^2) + b*1 + c = 0
or, a + b + c = 0
If the valus of any 2 variables are given, we can find the value of the third.

Coming to the question that is stated :
we are given
2x - 1 is a factor of 2x^2 + bx + 3

Equate 2x-1 to 0
or, 2x - 1 = 0
or, x = 1/2

Now put this valus of x in the expression;
2*{(1/2)^2)} + b*(1/2) + 3 = 0
or, 2*(1/4) + b/2 + 3 = 0
or; 1/2 + b/2 = -3
or, b = -7