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What is the right choice for question 6) ? http://postimg.org/image/z4jswp8tv/

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jayson94 | Salutatorian

Posted August 21, 2013 at 4:19 PM via web

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What is the right choice for question 6) ?

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llltkl | College Teacher | Valedictorian

Posted August 21, 2013 at 4:41 PM (Answer #1)

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`f(x)=x^3+3x^2-kx+10`

It is divided by `x-5`

So, `f(5)=5^3+3*5^2-k*5+10=125+75-5k+10=210-5k`

By condition:

`210-5k=15`

`rArr 5k=210-15=195`

`rArr k=195/5=39`

Therefore, the right option is a).

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 21, 2013 at 5:49 PM (Answer #2)

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You may use as alternative method, the reminder theorem, such that:

`x^3 + 3x^2 - kx + 10 = (x - 5)(ax^2 + bx + c) + 15`

You need to notice that the quotient is second order polynomial whose coefficents are a,b,c.

`x^3 + 3x^2 - kx + 10 = ax^3 + bx^2 + cx - 5ax^2 - 5bx - 5c + 15`

Grouping the terms yields:

`x^3 + 3x^2 - kx + 10 = ax^3 + x^2(b - 5a) + x(c - 5b) - 5c + 15`

Equating the coefficients of like powers yields:

`a = 18`

`b - 5a = 3 => b - 5 = 3 => b = 8`

`15 - 5c = 10 => -5c = -5 => c = 1`

`c - 5b = -k => 1 - 40 = -k => -39 = -k => k = 39`

Hence, evaluating k, using the reminder theorem, yields `k = 39.`

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