Question

solve polynomial equations
date hour

1. use the fundamental theorem of algebra to determine how many zeros the function has.

$f(x) = 16x - 22x^3 + 6x^6 + 19x^5 - 3$
a) 3
b) 6
c) 1
d) 5

1. a function $f(x)$ passes through points $(4,0)$, $(0,-2)$, $(3,-8)$. which of the following is a factor of $f(x)$?

a) $x - 2$
b) $x + 2$
c) $x - 4$
d) $x + 4$

1. it is known that the function $f(x) = 5x^3 + 2x^2 - 5x + 5$ has only one real zero. how many imaginary zeros does $f(x)$ have?

a) 0
b) 1
c) 3
d) 2

1. a quartic polynomial function $g(x)$ is graphed and all real zeros are shown. how many imaginary zeros does this polynomial have?

a) 0
b) 1
c) 2
d) 4
solve each equation, then find all zeros. use your graphing calculator to find the first zero.

1. $5x^3 + 31x^2 = -31x - 5$
2. $x^3 - 4x^2 - 15x = -50$

Explanation:

Step1: Identify polynomial degree

The function $f(x)=16x - 22x^3 + 6x^6 + 19x^5 - 3$ has a highest power of $x^6$, so degree $n=6$. By the Fundamental Theorem of Algebra, the number of zeros (counting multiplicities) equals the degree.

Step2: Apply Factor Theorem

For a point $(a,0)$ on $f(x)$, $x-a$ is a factor. The function passes through $(4,0)$, so $x-4$ is a factor.

Step3: Calculate imaginary zeros

The cubic function $f(x)=5x^3+2x^2-5x+5$ has degree 3, so total zeros = 3. With 1 real zero, imaginary zeros = $3-1=2$.

Step4: Find imaginary zeros of quartic

A quartic polynomial has degree 4, so total zeros = 4. The graph crosses/touches the x-axis 2 times, so 2 real zeros. Imaginary zeros = $4-2=2$.

Step5: Rearrange and factor cubic

Rearrange $5x^3 + 31x^2 + 31x + 5 = 0$. Use rational root theorem: $x=-1$ is a root. Factor out $(x+1)$:
$$(x+1)(5x^2+26x+5)=0$$
Factor quadratic: $5x^2+26x+5=(5x+1)(x+5)$
Solve for zeros: $x=-1, x=-\frac{1}{5}, x=-5$

Step6: Rearrange and factor quartic

Rearrange $x^3 - 4x^2 -15x + 50 = 0$. Use rational root theorem: $x=2$ is a root. Factor out $(x-2)$:
$$(x-2)(x^2-2x-25)=0$$
Solve quadratic with quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=1,b=-2,c=-25$:
$$x=\frac{2\pm\sqrt{4+100}}{2}=\frac{2\pm\sqrt{104}}{2}=1\pm\sqrt{26}$$
Zeros: $x=2, x=1+\sqrt{26}, x=1-\sqrt{26}$

Answer:

1. B) 6
2. C) $x - 4$
3. D) 2
4. C) 2
5. $x=-5,\ x=-1,\ x=-\frac{1}{5}$
6. $x=2,\ x=1+\sqrt{26},\ x=1-\sqrt{26}$