+0  
 
0
140
2
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A:-7

B:-5

C:5

D:-4

 

 

 

I am lost with this problem.

Guest Mar 13, 2018
 #1
avatar+20009 
+1

What is the value of

\(k\) if \(x^3 - kx^2 + 2x + 1\) is divided by \(x - 4\)

and gives a remainder of \( -7\) ?


polynomial long division:
\(x^3 - kx^2 + 2x + 1 : x - 4 = x^2 - x·k + 4·x - 4·k + 18 - \dfrac{16·k - 73}{x - 4}\)

 

remainder  is -7:

\(\begin{array}{|rcll|} \hline -(16·k - 73) &=& -7 \\ 16·k - 73 &=& 7 \\ 16·k &=& 7+73 \\ 16·k &=& 80 \\ \mathbf{ k } & \mathbf{=} & \mathbf{5} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline x^3 - 5x^2 + 2x + 1 : x-4 &=& x^2 - x - 2 -7 \\ \hline \end{array}\)

 

or:

\(\begin{array}{|rcll|} \hline P(4) = -7 \\\\ P(4) = 4^3-k\cdot 4^2 + 2\cdot 4 + 1 &=& - 7 \\ 64-16k + 8 + 1 &=& - 7 \\ 73-16k &=& - 7 \\ 16k &=&73+7 \\ 16k &=& 80 \\ \mathbf{ k } & \mathbf{=} & \mathbf{5} \\ \hline \end{array}\)

 

laugh

heureka  Mar 13, 2018
edited by heureka  Mar 13, 2018
edited by heureka  Mar 13, 2018
 #2
avatar+88891 
+1

Using synthetic division, we have

 

4  [   1      -k       2           1         ]

                4     16 - 4k    72 -16k

       ______________________

        1    4-k     18 -4k    73  - 16k

 

So   73 -16k  is the remainder.....and this =  -7....so....

 

73  - 16k  =  -7       rearrange as

 

80  = 16k     divide both sides by 16

 

5   = k

 

 

 

cool cool cool

CPhill  Mar 13, 2018

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