Knowing the zeroes makes it relatively easy to factor the polynomial.

The expression \(\displaystyle (x+6)(x+4)(x-2)\) fits the description of the zeroes.

Now we need to check the answer.

\(\displaystyle (x+4)(x-2)=x^2-2x+4x-8=x^2+2x-8\)

\(\displaystyle \small (x^2+2x-8)(x+6)=x^3+6x^2+2x^2+12x-8x-48=x^3+8x^2+4x-48\)

We are able to get back to the original expression, meaning that the answer is \(\displaystyle (x+6)(x+4)(x-2)\).

Since 6 is a triple root, and the degree of the polynomial is 3, the polynomial is \(\displaystyle (x-6)^{3}\), which we can expland using the cube of a binomial pattern.

\(\displaystyle (x-6)^{3} =x ^{3} - 3\cdot x^{2} \cdot 6 + 3\cdot x\cdot 6 ^{2} -6^{3}\)

\(\displaystyle x ^{3} - 18 x^{2}+ 108x -216\)

Since 5 is a single root and 7 is a double root, and the degree of the polynomial is 3, the polynomial is \(\displaystyle (x-5)(x-7)^{2}\). To put this in expanded form:

\(\displaystyle (x-7)^{2} = x^{2} - 2 \cdotx\cdot 7 + 7^{2} = x^{2} - 14x + 49\)

\(\displaystyle (x-5)(x-7)^{2} =(x-5) \left ( x^{2} - 14x + 49 \right )\)

\(\displaystyle =x \left ( x^{2} - 14x + 49 \right ) -5\left( x^{2} - 14x + 4 9 \right )\)

\(\displaystyle = x^{3} - 14x^{2} + 49x - \left( 5x^{2} - 70x + 245 \right )\)

\(\displaystyle = x^{3} - 14x^{2} + 49x - 5x^{2} + 70x -245\)

\(\displaystyle = x^{3} - 19x^{2} + 119x -245\)

When we are looking for the solutions of a quadratic, or the zeroes, we are looking for the values of \(\displaystyle x\) such that the output will be zero. Thus, we first factor the equation. 

\(\displaystyle 0 = x^{2} + x - 6 = (x+3)(x-2)\)

Then, we are looking for the values where each of these factors are equal to zero. 

\(\displaystyle x + 3 = 0\) implies \(\displaystyle x = -3\)

and \(\displaystyle x - 2 = 0\) implies \(\displaystyle x = 2\)

Thus, these are our solutions. 

First, we need to find all the possible rational roots of the polynomial using the Rational Roots Theorem:

\(\displaystyle \frac{\textup{factors of the constant}}{\textup{factors of the leading coefficient}}\)

Since the leading coefficient is just 1, we have the following possible (rational) roots to try:

±1, ±2, ±3, ±4, ±6, ±12, ±24

When we substitute one of these numbers for \(\displaystyle x\), we're hoping that the equation ends up equaling zero. Let's see if \(\displaystyle -1\) is a zero:

\(\displaystyle f(-1)=(-1)^{4}-4(-1)^{3}-7(-1)^{2}+22(-1)+24\)

\(\displaystyle f(-1)=1+4-7-22+24\)

\(\displaystyle f(-1)=0\)

Since the function equals zero when \(\displaystyle x\) is \(\displaystyle -1\), one of the factors of the polynomial is \(\displaystyle (x+1)\). This doesn't help us find the other factors, however. We can use synthetic substitution as a shorter way than long division to factor the equation.

\(\displaystyle \textup{-1 } | \textup{ 1 -4 -7 22 24}\)

             \(\displaystyle \small \textup{ -1 }\textup{ 5 }\textup{ 2 }\textup{ -24}\)

       \(\displaystyle \frac{ }{\textup{ 1 -5 -2 24 }\textup{ 0}}\)

Now we can factor the function this way:

\(\displaystyle f(x)=(x+1)(x^{3}-5x^{2}-2x+24)\)

We repeat this process, using the Rational Roots Theorem with the second term to find a possible zero. Let's try \(\displaystyle -2\):

\(\displaystyle f(-2)=[(-2)+1][(-2)^{3}-5(-2)^{2}-2(-2)+24]\)

\(\displaystyle f(-2)=(-1)(-8-20+4+24)=0\)

When we factor using synthetic substitution for \(\displaystyle x=-2\), we get the following result:

\(\displaystyle f(x)=(x+1)(x+2)(x^{2}-7x+12)\)

Using our quadratic factoring rules, we can factor completely:

\(\displaystyle f(x)=(x+1)(x+2)(x-4)(x-3)\)

Thus, the zeroes of \(\displaystyle f(x)\) are \(\displaystyle x=-2, -1, 3, 4.\)

To simplify the polynomial, begin by combining like terms:

\(\displaystyle (\frac{-a^{-2}b^{3}c^{-1}}{3a^{-4}b^{-1}c^{3}})^{-2}\)

\(\displaystyle (\frac{-a^{2}b^{4}}{3c^{4}})^{-2}\)

First, multiply the outside term with each term within the parentheses:

\(\displaystyle ab^{-2}(a^{2}b + a^{-1}b^{3}-a^{3}b^{-1})\)

\(\displaystyle a^{3}b^{-1} + b-a^{4}b^{-3}\)

Rearranging the polynomial into fractional form, we get:

\(\displaystyle \frac{a^{3}}{b}+b-\frac{a^{4}}{b^{3}}\)