Find the solutions to the system of equations. select all that apply.

y=x^2-4
y=-2x-5

a. (-2,0)
b. (-5,0)
c. (2,0)
d. (-1,-3)

Step-by-step explanation:

Hello there!

The system of equation given is:

y = x^2-4

y = -2x-5

And we want to solve this system:

the first step we can do is replace y in the first equation by the given in the second equation:

y = x^2-4 = -2x - 5

now we can solve it for x:

x^2-4 +2x + 5 = 0

x^2 +2x + 1 = 0

and now we can use the cuadratic equation forumla:

this is, if we have a function of the form:

ax^2 + bx + c = 0

the solutions are:

in our equation we have a=1, b = 2 and c= 1

we replace it in the formula given, and we obtain:

Now we got the value of x, and we could replace it in one of the equations and get the value of y:

y = -2x-5

y = -2*(-1) - 5 = 2 -5 = -3

then the pair is (-1, -3)

and the correct option is D.

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0