BdMO National Secondary 2007#9

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samiul_samin
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BdMO National Secondary 2007#9

Unread post by samiul_samin » Sat Feb 23, 2019 9:17 am

If $x^2+3x-4$ is a factor of $x^3+bx^2+cx+11$,then find the values of $b$ and $c$.

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samiul_samin
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Re: BdMO National Secondary 2007#9

Unread post by samiul_samin » Sat Feb 23, 2019 11:31 am

Answer: $b=\dfrac {-23}{4},c=\dfrac {17}{4}$

Solution:
$x^2+3x-4=0\Rightarrow x^2+4x-x-4=0\Rightarrow(x+4)(x-1)=0\Rightarrow x=-4,1$
The given equation $x^3+bx^2+cx+11=0$
Let $x=x_1,x_2,x_3$
So,$x_1=-4,x_2=1$

From the equation $x_1\times x_2\times x_3= 11\Rightarrow x_3=\dfrac {11}{-4}$

Hence,$b=-4+1+\dfrac {11}{-4}=\dfrac{-23}{4}$

And $c=-4\ times \dfrac {11}{-4}+\dfrac {11}{-4}\times 1+-4\times 1=\dfrac {17}{4}$

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samiul_samin
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Re: BdMO National Secondary 2007#9

Unread post by samiul_samin » Sun Mar 10, 2019 11:03 pm

samiul_samin wrote:
Sat Feb 23, 2019 11:31 am
Answer: $b=\dfrac {23}{4},c=\dfrac {17}{4}$

Solution:
$x^2+3x-4=0\Rightarrow x^2+4x-x-4=0\Rightarrow(x+4)(x-1)=0\Rightarrow x=-4,1$
The given equation $x^3+bx^2+cx+11=0$
Let $x=x_1,x_2,x_3$
So,$x_1=-4,x_2=1$

From the equation $x_1\times x_2\times x_3= 11\Rightarrow x_3=\dfrac {11}{-4}$

Hence,$b=-(-4+1+\dfrac {11}{-4})=\dfrac{23}{4}$

And $c=-4\times \dfrac {11}{-4}+\dfrac {11}{-4}\times 1+-4\times 1=\dfrac {17}{4}$
Corrected

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