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24x^2 + bx + c=(dx+e)(fx+g)

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24x^2 + bx + c=(dx+e)(fx+g) [#permalink] New post   25 Dec 2019, 00:54
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\(24x^2 + bx + c=(dx+e)(fx+g)\)

\(b,c,d,e,f,\) and \(g\) are integers


Quantity A
Quantity B
\(d+f\)
\(26\)



A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

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B
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Re: 24x^2 + bx + c=(dx+e)(fx+g) [#permalink] New post   04 Apr 2020, 06:31
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Carcass wrote:
\(24x^2 + bx + c=(dx+e)(fx+g)\)

\(b,c,d,e,f,\) and \(g\) are integers


Quantity A
Quantity B
\(d+f\)
\(26\)



Given: 24x² + bx + c =(dx + e)(fx + g)
Use FOIL to expand the right side to get: 24x² + bx + c = dfx² + gdx + efx + eg
On the left side, the coefficient of x² is 24.
On the right side, the coefficient of x² is df
So, we can conclude that: 24 = df

At this point, the key word in the given information is that d and f are INTEGERS
Since we know that 24 = df, we can see that there aren't many pairs of INTEGER values that have a product of 24.
We get: (1)(24) = 24, (2)(12) = 24, (3)(8) = 24, and (4)(6) = 24

Quantity A = d + f
Notice that the MAXIMUM possible value of d + f occurs when the two numbers are 1 and 24, in which case we get us some of 25
This means d + f will always be less than 26

Answer: B

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Brent
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Re: 24x^2 + bx + c=(dx+e)(fx+g) [#permalink] New post   25 Dec 2019, 01:30
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You can solve this very easily. Since 24x2+bx+c can be expressed as (x-x1)(x-x2), you can consider d and f equal to 1. Therefore d+f=2 < 26. Regards!
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Re: 24x^2 + bx + c=(dx+e)(fx+g) [#permalink] New post   26 Dec 2019, 07:29
I still don't understand the explanation.. would really appreciate if someone could clarify. thank you in advance.
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Re: 24x^2 + bx + c=(dx+e)(fx+g) [#permalink] New post   26 Dec 2019, 07:39
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find the roots of the equation.
Let's denote the roots as x1 and x2. The equation can now be written (using the formula of factorization) as (x-x1)*(x-x2); The coefficients d,f are on the left side of x. in this case d=1 and f=1. So d+f=2;
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Re: 24x^2 + bx + c=(dx+e)(fx+g) [#permalink] New post   03 Apr 2020, 21:49
Can someone else please explain how to solve?
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Re: 24x^2 + bx + c=(dx+e)(fx+g) [#permalink] New post   Updated on: 04 Apr 2020, 08:20
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24x^2 + bx + c = dfx^2 + dgx +efx + ge
comparing the two equations,
df = 24
since d and f are integers ,possible values of d * f are as follows
d * f = 1*24 ,24*1 , 3*8, 8*3, 6*4, 4*6, 2*12 ,12*2
from the above set of possible values,
d+f<26
hence B is answer

Originally posted by dare90 on 04 Apr 2020, 03:53.
Last edited by dare90 on 04 Apr 2020, 08:20, edited 1 time in total.
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Re: 24x^2 + bx + c=(dx+e)(fx+g) [#permalink] New post   07 Nov 2024, 14:10
Hello from the GRE Prep Club BumpBot!

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Re: 24x^2 + bx + c=(dx+e)(fx+g) [#permalink]
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