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I simplified an expression to: 4(x^2-2x+2)-7(x^3-3x+1). 

 

The question asks: "what is the sum of the squares of the coefficients of the terms?" 

 

What would it be?

 

Thanks

 Jul 4, 2020
 #1
avatar+26620 
+2

4 x^2 -8x+8       - 7x^3 + 21x - 7      collect like terms

 

-7 x^3 + 4x^2 + 13x + 1              -7^2 +4^2 + 13^2   + 1^2      

    (some will argue '1' is not a coefficient....but it could be the coefficient of  x...depends on what you have been taught )

 Jul 4, 2020
 #2
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Sorry I wrote the problem wrong. The simplification was -7x^3 + 4x^2 + 13x +1. 

 

I need to find the sum of the squares of the coefficients of the terms. Im a tad confused as to what you meant. Sorry!

Guest Jul 4, 2020
 #3
avatar+26620 
+1

That is what I simplified the expression to also (see above)     the SUM of the SQUARES of the coefficients is given in my answer above......what part do you not understand?  

  -7^2 +4^2 + 13^2   + 1^2      

    (some will argue '1' is not a coefficient....but it could be the coefficient of  xo  ...depends on what you have been taught ...is the '1' a coefficient in your teachings or is it considered a 'constant' ?   )

ElectricPavlov  Jul 4, 2020
 #4
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wait I made a trivial mistake. Would the final answer be 11?

Guest Jul 4, 2020
 #5
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nvm I think it is 235. 

Guest Jul 4, 2020
 #6
avatar+26620 
0

There ya go......cheeky

ElectricPavlov  Jul 4, 2020
 #7
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since you were a great help, is it ok if you could help me on the next one? thx

 

Find the constant c such that (x^2-4x+3)(x+5) - (x^2+4x-5)(x-c)= 0, for all x. 

Guest Jul 4, 2020
 #8
avatar+26620 
+2

Factor 

(x^2-4x+3)(x+5) - (x^2+4x-5)(x-c)= 0

(x-3)(x-1) (x+5)           - (x +5)(x-1)(x-c) = 0     divide out  (x+5)  and  (x-1)

 

(x-3)                  -   (x-c )  = 0         think you can take it from here?

ElectricPavlov  Jul 4, 2020
edited by ElectricPavlov  Jul 4, 2020
 #9
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+1

is c = 3?

Guest Jul 4, 2020
 #10
avatar+26620 
+1

Yeppers !   c = 3

ElectricPavlov  Jul 4, 2020

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