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# a simple q, I think im overcopmplicating this. Help is appreciated quick.

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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
+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
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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
#6
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There ya go...... 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
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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
#10
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Yeppers !   c = 3

ElectricPavlov  Jul 4, 2020