Question

1. How many complex roots does the polynomial equation have?

Answer 1

Answer:

1) There are two complex roots. A is correct.

2) The function has three real zeros. The graph of the function intersects the x-axis at exactly three locations. D is correct

3) There are four complex roots. B is correct.

4) C is correct. $f(x)=(x+2)(x-(-1+i\sqrt{2})(x-(-1-i\sqrt{2})$

5) $(x+3)(x+4i)(x-4i)$

Step-by-step explanation:

1) Number of complex root.

$-3x^4-7x+17=0$

We can not factor it completely. Because it's prime equation. We will make the graph of $-3x^4-7x+17=0$ and then see the x-intercept of the graph.

Because number x-intercepts are the zeros of polynomial. Please see the attachment for graph. In graph we can see it cuts only two points but their highest degree is 4. Maximum possible roots are 4 but only 2 are real and 2 are complex.  

Thus, There are two complex roots.

2) This is cubic polynomial function.

Maximum number of zeros should be 3.

Please see the graph with attachment.

Zeros is a value of x where graph cuts x-axis. In graph we can see graph cuts x-axis at three points.

Thus, The function has three real zeros. The graph of the function intersects the x-axis at exactly three locations.

3) This is 4th degree polynomial.

Maximum number of zeros should be 4.

Please see the graph of polynomial.

Zeros is a value of x where graph cuts x-axis. In graph we can see doesn't cut x-axis at any points.

Thus, The function has four complex zeros. The graph of the function does not intersects the x-axis at any point.

4) We are given 3 degree polynomial. We need to factor it. First we split middle term and then factor it.

$f(x)=x^3+4x^2+7x+6$

$f(x)=x^3+2x^2+2x^2+4x+3x+6$

$f(x)=x^2(x+2)+2x(x+2)+3(x+2)$

$f(x)=(x+2)(x^2+2x+3)$

Now we factor quadratic part using quadratic formula

$x^2+2x+3$

$x=-1+i\sqrt{2},-1-i\sqrt{2}$

Complete factor of f(x)

Thus, $f(x)=(x+2)(x-(-1+i\sqrt{2})(x-(-1-i\sqrt{2})$

5) This is cubic polynomial. We need to factor it.

$f(x)=x^3+3x^2+16x+48$

First we make two group and then factor it

$f(x)=x^2(x+3)+16(x+3)$

$f(x)=(x+3)(x^2+16)$

$f(x)=(x+3)(x+4i)(x-4i)$

Thus, complete factor of f(x) is (x+3)(x+4i)(x-4i)

Answer 2

1. A.) 2

2. D.) The function has three real zeros. The graph of the function intersects the x-axis at exactly three locations.

3. A.) 3

4. C.) f(x) + (x+2)(x-(-1$\sqrt{} 2$)(x-(-1- $\sqrt{} 2$))

(5.)
1.) x = -3 
2.)x = 4
 3.) x = -4