Using the Factor Theorem, which of the polynomial functions has the zeros 2, radical 3 , and negative radical 3 ? f (x) = x3 – 2x2 – 3x + 6 f (x)= x3 – 2x2 + 3x + 6 f (x) = x3 + 2x2 – 3x + 6 f (x) = x3 + 2x2 – 3x – 6

A

Step-by-step explanation:

According to the Factor Theorem, if (x - k) is a factor of a polynomial P(x), then P(k) must equal zero.

We are given that a polynomial function has the zeros 2, √3, and -√3. So, we can let k = 2, √3, -√3.

So, according to the Factor Theorem, P(2), P(√3) and P(-√3) must equal 0.

Testing each choice, we can see that only A is true:

Testing all three values yields that:

Hence, our answer is A.

(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

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2