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

Guest 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^{o }...depends on what you have been taught )

ElectricPavlov Jul 4, 2020

#2**0 **

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**+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 x^{o} ...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

#7**0 **

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**+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