1. kalena was asked to prove ifx(x - 1)(x + 1) = x3 - x represents a polynomial identity. she
states that this relationship is not true and the work she used to justify her thinking is shown.
step 1: x(x - 1)(x + 1)
step 2: (x2 + x)(x - 1)
step 3: x3 - x2 - x2 - x
step 4: x3 - 2x2 - x
which statement correctly describes kalena's justification?
a. kalena is correct. there are no mistakes in her justification.
b. kalena made a mistake in step 2. the justification should state: (2x - x)(x + 1).
c. kalena made a mistake in step 3. the justification should state: x
- x.
d. kalena made a mistake in step 4. the justification should state: x3 + 2x2 - x

C. Kalena made a mistake in Step 3. The justification should state: -x²

+ x²

Step-by-step explanation:

Given the function x(x - 1)(x + 1) = x3 - X

To justify kelena proof

We will need to show if the two equations are equal.

Starting from the RHS with function x³-x

First we will factor out the common factor which is 'x' to have;

x(x²-1)

Factorising x²-1 using the difference of two square will give;

x(x+1)(x-1)

Note that for two real number a and b, the expansion of a²-b² using difference vof two square will give;

a²-b² = (a+b)(a-b) hence;

Factorising x²-1 using the difference of two square will give;

x(x+1)(x-1)

Factorising x(x+1) gives x²+x, therefore

x(x+1)(x-1) = (x²+x)(x-1)

(x²+x)(x-1) = x³-x²+x²-x

The function x³-x²+x²-x gotten shows that kelena made a mistake in step 3, the justification should be -x²+x² not -x-x²

i believe the correct answer is b.

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

we now create space between the x and y terms with parenthesis.

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

we add the squares to both sides.

simplify.

and write the quadratics in factored form.

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.