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1) Let P(x) be a polynomial such that \(P(x) = P(0) + P(1) x + P(2) x^2\)and P(-1) = 1. Find P(x).

2) Let a,b,c be positive real numbers such that \(\log_a b + \log_b c + \log_c a = 0.\)Find \((\log_a b)^3 + (\log_b c)^3 + (\log_c a)^3.\)

3) Let a,b and c be the roots of \(x^3 - 5x + 7 = 0.\)Find the monic polynomial, in x, whose roots are a - 2, b - 2, and c - 2. 

 Jun 27, 2019
 #1
avatar+8406 
+2

1)   P(x)  =  P(0) + P(1)x + P(2)x2

 

Plug in  1  for  x  to get:

 

 

 
P(1)  =  P(0) + P(1) + P(2)

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Subtract  P(1)  from both sides of the equation
0  =  P(0) + P(2)

 

 

Subtract  P(2)  from both sides
-P(2)  =  P(0)

 

 

 

Plug in  -1  for  x  to get:

P(-1)  =  P(0) - P(1) + P(2)

 

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Since  P(-1)  =  1  we can substitute  1  in for  P(-1)
1  =  P(0) - P(1) + P(2)   Add  P(1)  to both sides and subtract  1  from both sides.
P(1)  =  P(0) + P(2) - 1

 

 

Substitute  0  in for  P(0) + P(2)
P(1)  =  -1    

 

Plug in  2  for  x  to get:

P(2)  =  P(0) + 2P(1) + 4P(2)

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Substitute  -1  in for  P(1)  and subsitute  -P(2)  in for  P(0)
P(2)  =  -P(2) + 2(-1) + 4P(2)   Simplify the right side.
P(2)  =  3P(2) - 2

 

 

Subtract  3P(2)  from both sides.
-2P(2)  =  -2    Divide both sides by  -2
P(2)  =  1

 

 

 
Now we can find  P(0):

 

 

 
P(0)  =  -P(2)       Substitute  1  in for  P(2)

P(0)  =  -1

 

 

 

So we have found that     P(x)  =  -1 - x + x2

 Jun 27, 2019
 #2
avatar+22527 
+2

2)
Let \(a,b,c\) be positive real numbers such that \(\log_a b + \log_b c + \log_c a = 0\).
Find \((\log_a b)^3 + (\log_b c)^3 + (\log_c a)^3\).

 

\(\text{Let $\log_a b = \mathbf{x} $} \\ \text{Let $\log_b c = \mathbf{y} $} \\ \text{Let $\log_c a = \mathbf{z} $} \)

 

\(\begin{array}{|rcll|} \hline \log_a b + \log_b c + \log_c a &=& 0 \\ \mathbf{x+y+z} &=& \mathbf{0} \qquad (1) \\ \hline \end{array}\)

 

\(\begin{array}{|rclrcl|} \hline \log_a b &=&\dfrac{\log_c b}{\log_c a} \quad&| \quad \log_c b &=&\dfrac{\log_b b}{\log_b c} \\\\ \log_a b &=&\dfrac{\log_b b}{\log_b c\log_c a} \quad&| \quad \log_b b = 1 \\\\ \log_a b &=&\dfrac{1}{\log_b c\log_c a} \\\\ \log_a b\log_b c\log_c a &=& 1 \\\\ \mathbf{xyz} &=& \mathbf{1} \qquad (2) \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline (x+y+z)^3 &=& (x+y+z)^2(x+y+z) \\ &=& \left(x^2+y^2+z^2+2(xy+yz+xz)\right)(x+y+z) \\ &=& (x^2+y^2+z^2)(x+y+z)+ 2(x+y+z)(xy+yz+xz) \\\\ &=& x^3+y^3+z^3 \\ && +x^2y+x^2z+y^2x+y^2z+z^2x+z^2y + 2(x+y+z)(xy+yz+xz) \\\\ &=& x^3+y^3+z^3 \\ && +(x+y+z)(xy+yz+xz) -3xyz + 2(x+y+z)(xy+yz+xz) \\\\ (x+y+z)^3&=& x^3+y^3+z^3 -3xyz + 3(x+y+z)(xy+yz+xz) \\ \hline \mathbf{x^3+y^3+z^3} &=& \mathbf{(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz} \qquad (3) \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{x^3+y^3+z^3} &=& \mathbf{(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz} \\ && \boxed{x+y+z = 0} \\ x^3+y^3+z^3 &=& 0^3-3\cdot 0\cdot (xy+yz+xz)+3xyz \\ x^3+y^3+z^3 &=& 3xyz \\ && \boxed{xyz = 1} \\ x^3+y^3+z^3 &=& 3\cdot 1 \\ x^3+y^3+z^3 &=& 3 \\ \mathbf{(\log_a b)^3 + (\log_b c)^3 + (\log_c a)^3} &=& \mathbf{3} \\ \hline \end{array}\)

 

laugh

 Jun 28, 2019

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