Felix was asked whether the following equation is an identity:
(X^2+1)^2 = (x^2-1)^2+(2x)^2

Felix started that the equation is a true identity.

Is Felix correct? If not, in which step did he make a mistake?

Plz hurry (will mark Brainliest)

idk

step-by-step explanation:

use that smart brain

3. (5x + 1) - (-10x + 6)

simplify the answer choice. combine like terms. remember to distribute the negative to the terms inside the second parenthesis. also remember that two negatives = one positive, and one of each sign = negative.

- (-10x + 6) = + 10x - 6

simplify. combine like terms:

5x + 1 + 10x - 6

5x + 10x + 1 - 6

15x - 5

∴ it is your answer

~

the answer is d

for problems like this theres an app, its photomath that explains the problem and gives the answer hope i

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.