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quad eqn factors

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quad eqn factors

by kushal.adhia » Mon Nov 01, 2010 11:33 am
If r - 5 is a factor of r^2-cr+30 , what is the value of c ?

A. - 6
B. - 3
C. 6
D. 9
E. 11

Please help

Kushal

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by Rahul@gurome » Mon Nov 01, 2010 11:56 am
kushal.adhia wrote:If r - 5 is a factor of r^2-cr+30 , what is the value of c ?

A. - 6
B. - 3
C. 6
D. 9
E. 11

Please help

Kushal
This question can be solved in two methods.

Method 1: (Methodical one)
(r - 5) is factor (r^2 - cr + 30).
Thus, (r^2 - cr + 30) = (r - 5)*p(r) , where p(r) is linear polynomial in r, i.e. p(r) doesn't contain r^2, r^3... etc. Without loss of generality we can assume p(r) = (r - a), where a is constant.

So, (r^2 - cr + 30) = (r - 5)*(r - a) = r^2 - (a + 5)*r + 5a
Comparing the coefficients of the expressions in LHS and RHS,
5a = 30 and (a + 5) = c

Thus, a = 6 => c = 5 + 6 = 11

The correct answer is E.


Method 1: (Mechanical one)
Try to factorize (r^2 - cr + 30) in such a way that (r - 5) will be factor.

.. (r^2 - cr + 30)
= (r^2 - 5r + 5r - cr + 30)
= [r(r - 5) + (5 - c)r + 30]
= [r(r - 5) + (5 - c)r - 5(5 - c) + 5(5 - c) + 30]
= [r(r - 5) + (5 - c)(r - 5) + 5(5 - c) + 30]
= [(r - 5)(r + 5 - c) + 25 - 5c + 30]
= [(r - 5)(r + 5 - c) + 55 - 5c]

Thus, we see that for (r - 5) to be a factor of (r^2 - cr + 30), (55 - 5c) must be zero.
So, 55 - 5c = 0 => c = 11

The correct answer is E.
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by Rezinka » Mon Nov 01, 2010 9:10 pm
The way I did it :
(r - 5) is a factor.
So, r - 5 = 0
or r = 5

Put this value in the equation given.
r^2 - cr + 30
5^2 - c*5 + 30 = 0
25 - 5c +30 = 0
5c = 55
c = 11

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by waltz2salsa » Tue Nov 02, 2010 11:26 pm
Another way:

r^2 - cr + 30; let r1, r2 be factors

sum of roots, r1+r2 = c
product of roots, r1*r2 = 30

if r1=5, solve to get r2 = 6
hence c= 11 :)

Regards,
Shashwat