What is the product?
(x^2-16)/(2x+8) x (x^3-2x^2+x)/(x^2+3x-4)
a. x(x-4)(x-1)/2(x+4)
b. x(x-1)/2
c. (x+4)(x-4)/2x(x-1)
d. (x-4)(x-1)/2x(x+4)

Option A is correct.

Step-by-step explanation:

We need to find the product of

We know (a^2-b^2) = (a+b)(a-b)

so, (x^2-16) = (x)^2-(4)^2 = (x-4)(x+4)

2x+8 Taking 2 common from this term:

2x+8 = 2(x+4)

(x^3-2x^2+x) Taking x common from this term

x(x^2-2x+1) = x(x-1)^2 = x(x-1)(x-1)

(x^2+3x-4) factorizing this term

x^2+4x-x-4 = x(x+4)-1(x+4)

= (x-1)(x+4)

Now, Putting these simplified terms in the given equation:

Now cancelling the same terms that are in numerator and denominator

So, Option A is correct.

=x(x-4)(x-1)/2(x+4)

Step-by-step explanation:

=x^2-4^2/2(x+4) * x^3-2x^2+x/x^2+3x-4

=(x+4)(x-4)/2(x+4) * x(x^2-2x+1)/x^2+3x-4

Factor x^2-2x+1 using the perfect square root

=(x+4)(x-4)/2(x+4) * x(x-1)^2/x^2+3x-4

Factor x^2+3x-4 using AC method.

=(x+4)(x-4)/2(x+4) * x(x-1)^2/(x-1)(x+4)

Cancel the common factor of x+4 and x-1

=(x-4)/2(x+4) * x(x-1)/1

=(x-4)x(x-1)/2(x+4)

Reorder the terms

=x(x-4)(x-1)/2(x+4)

y=sec x is the correct answer

~batmans wife dun dun

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.