Find Zeros of Polynomials

Since $p(x)$ has a zero at $x = 2$, it means that $(x - 2)$ is a factor of $p(x)$. We divide $p(x)$ by $(x - 2)$ using polynomial long division or synthetic division:

$$ \frac{p(x)}{x - 2} = \frac {x^3 + 5x^2 - 2x - 24}{x - 2} = x^2 + 7x + 12 $$

Using the result of the division, $p(x)$ may now be written in a partially factored form:

$$ p(x) = (x - 2)(x^2 + 7x + 12) $$

Next, factor the quadratic expression $x^2 + 7x + 12$. We need two numbers that multiply to 12 and add to 7. Those numbers are 3 and 4:

$$ x^2 + 7x + 12 = (x + 3)(x + 4) $$

Substitute this back into $p(x)$ to get the completely factored form:

$$ p(x) = (x - 2)(x + 3)(x + 4) $$

The zeros are found by setting the equation to zero: $p(x) = 0$. This means:

$$ x - 2 = 0 \quad \text{or} \quad x + 3 = 0 \quad \text{or} \quad x + 4 = 0 $$

Final Zeros:
$$ x = 2, \quad x = -3, \quad x = -4 $$

The zeros at $x = \pm \sqrt{5}$ correspond to the linear factors:

$$ (x - \sqrt{5}) \quad \text{and} \quad (x + \sqrt{5}) $$

Multiply these factors together to create a quadratic divisor. Notice this is a difference of squares:

$$ (x - \sqrt{5})(x + \sqrt{5}) = x^2 - 5 $$

Hence, the polynomial $p(x)$ may be written as:

$$ p(x) = (x^2 - 5)Q(x) $$

Find $Q(x)$ using long division of polynomials:

$$ Q(x) = \frac{3x^4 + 5x^3 - 17x^2 - 25x + 10}{x^2 - 5} = 3x^2 + 5x - 2 $$

Now, factor the quadratic $Q(x) = 3x^2 + 5x - 2$:

$$ 3x^2 + 5x - 2 = (3x - 1)(x + 2) $$

We can now write $p(x)$ completely factored:

$$ p(x) = (x - \sqrt{5})(x + \sqrt{5})(3x - 1)(x + 2) $$

Final Zeros: Set each factor to zero:
$$ x = \pm \sqrt{5}, \quad x = \frac{1}{3}, \quad x = -2 $$

a) Prove Multiplicity 2:
If $x = 1$ is a zero of multiplicity 2, then $(x - 1)^2$ must be a factor of $p(x)$. Expanding $(x-1)^2$ gives $x^2 - 2x + 1$. A long division of $p(x)$ by $x^2 - 2x + 1$ must yield a remainder of 0:

$$ \frac{x^4 - 2x^3 - 2x^2 + 6x - 3}{x^2 - 2x + 1} = x^2 - 3 $$

Since the remainder is $0$, $x = 1$ is indeed a zero of multiplicity $2$.

b) Find all zeros:
Using the division above, $p(x)$ may now be written in factored form:

$$ p(x) = (x - 1)^2(x^2 - 3) $$

Factor the quadratic expression $x^2 - 3$ as a difference of squares:

$$ p(x) = (x - 1)^2 (x - \sqrt{3}) (x + \sqrt{3}) $$

Final Zeros:
$$ x = 1 \text{ (multiplicity 2)}, \quad x = \sqrt{3}, \quad x = -\sqrt{3} $$

c) Sketching the Graph:
With the factored form and zeros, we create a table of signs. Note that at $x=1$ (even multiplicity), the graph will touch the axis and turn around, whereas at $x=\pm\sqrt{3}$ (odd multiplicity), it will cross the axis.

We use the x-intercepts, the table of signs, the y-intercept $(0 , -3)$, and the end behavior (both ends go up since it's an even degree with a positive leading coefficient) to complete the graph:

Rational Zeros Theorem: If $p(x)$ is a polynomial with integer coefficients, any rational zero in the form $\frac{m}{n}$ (in lowest terms) must have $m$ as a factor of the constant term and $n$ as a factor of the leading coefficient.

Constant term = $2$. Factors of 2 are: $\pm 1, \pm 2$.
Leading coefficient = $6$. Factors of 6 are: $\pm 1, \pm 2, \pm 3, \pm 6$.

Possible rational zeros ($\frac{\text{factors of } 2}{\text{factors of } 6}$):

$$ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 2, \pm \frac{2}{3} $$

Because of the large list of possible zeros, we graph the polynomial and guess the zeros from the location of the $x$-intercepts. From the graph above, the zeros appear to be close to $-\frac{1}{3}$, $\frac{1}{2}$, and $2$.

We calculate $p\left(-\frac{1}{3}\right)$, $p\left(\frac{1}{2}\right)$, and $p(2)$ to verify:

$$ p\left(-\frac{1}{3}\right) = 6\left(-\frac{1}{3}\right)^3 - 13\left(-\frac{1}{3}\right)^2 + \left(-\frac{1}{3}\right) + 2 = 0 $$ $$ p\left(\frac{1}{2}\right) = 6\left(\frac{1}{2}\right)^3 - 13\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right) + 2 = 0 $$ $$ p(2) = 6(2)^3 - 13(2)^2 + (2) + 2 = 0 $$

Final Zeros: All three guesses are verified. The zeros are exactly:
$$ x = -\frac{1}{3}, \quad x = \frac{1}{2}, \quad x = 2 $$

1. Substitute to simplify:
Notice this is a polynomial in "quadratic form". Let $u = x^2$. The equation becomes:

$$ u^2 - 5u - 36 = 0 $$

2. Factor the quadratic:
We need two numbers that multiply to $-36$ and add to $-5$. These are $-9$ and $4$.

$$ (u - 9)(u + 4) = 0 $$

3. Substitute back to $x$:
Replace $u$ with $x^2$ to find the roots for $x$:

$$ x^2 - 9 = 0 \implies x^2 = 9 \implies x = \pm 3 $$ $$ x^2 + 4 = 0 \implies x^2 = -4 \implies x = \pm 2i $$

Final Zeros:
The polynomial has two real zeros and two complex zeros: $\mathbf{x = 3, -3, 2i, -2i}$.

1. Conjugate Root Theorem:
Since the polynomial has real coefficients, complex roots must occur in conjugate pairs. Therefore, if $x = 1 - i$ is a zero, $x = 1 + i$ must also be a zero.

2. Build the factored form:
The three zeros are $2, 1-i, 1+i$. The factors are $(x - 2)$, $(x - (1-i))$, and $(x - (1+i))$. The general equation is:

$$ P(x) = a(x - 2)(x - (1-i))(x - (1+i)) $$

3. Expand the complex factors:
Group the complex parts to use the difference of squares:

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

So the polynomial is $P(x) = a(x - 2)(x^2 - 2x + 2)$.

4. Solve for 'a' using the y-intercept:
We are given $P(0) = -4$. Plug in $x=0$:

$$ -4 = a(0 - 2)(0^2 - 2(0) + 2) $$ $$ -4 = a(-2)(2) \implies -4 = -4a \implies a = 1 $$

5. Expand to standard form:
$$ P(x) = 1(x - 2)(x^2 - 2x + 2) $$ $$ = x(x^2 - 2x + 2) - 2(x^2 - 2x + 2) $$ $$ = x^3 - 2x^2 + 2x - 2x^2 + 4x - 4 $$ $$ \mathbf{P(x) = x^3 - 4x^2 + 6x - 4} $$

1. Identify Possible Rational Roots:
Factors of constant (6): $\pm 1, \pm 2, \pm 3, \pm 6$.
Factors of leading coeff (2): $\pm 1, \pm 2$.
Possible roots: $\pm 1, \pm \frac{1}{2}, \pm 2, \pm 3, \pm \frac{3}{2}, \pm 6$.

2. Test roots using the Factor Theorem:
Let $p(x) = 2x^3 - x^2 - 7x + 6$. Test $x=1$:

$$ p(1) = 2(1)^3 - (1)^2 - 7(1) + 6 = 2 - 1 - 7 + 6 = 0 $$

Since $p(1) = 0$, $x = 1$ is a root, and $(x - 1)$ is a factor.

3. Use Synthetic Division to divide by $(x - 1)$:

  1 |   2   -1   -7    6
    |        2    1   -6
    ---------------------
        2    1   -6    0
                        

The quotient is the quadratic $2x^2 + x - 6$.

4. Factor the remaining quadratic:
We need to factor $2x^2 + x - 6 = 0$. We look for two numbers that multiply to $(2)(-6) = -12$ and add to $1$. Those numbers are $4$ and $-3$.

$$ 2x^2 + 4x - 3x - 6 = 0 $$ $$ 2x(x + 2) - 3(x + 2) = 0 $$ $$ (2x - 3)(x + 2) = 0 $$

This gives roots $x = \frac{3}{2}$ and $x = -2$.

Final Zeros:
$$ \mathbf{x = 1, \quad x = \frac{3}{2}, \quad x = -2} $$