Use the $\displaystyle ac$-method to split the middle term into the sum of two terms whose coefficients have sum $\displaystyle -31$ and product $\displaystyle 6*35 = 210$. These two numbers can be found, using trial and error, to be $\displaystyle -21$ and $\displaystyle -10$.

$\displaystyle -21*-10=210$ and $\displaystyle -21+(-10)=-31$

Now we know that $\displaystyle 6x^{2}-31x+35$ is equal to $\displaystyle 6x^{2}-10x-21x+35$.

Factor by grouping.

$\displaystyle (6x^{2}-10x)-(21x-35)$

$\displaystyle 2x(3x-5)-7(3x-5)$

$\displaystyle (2x-7)(3x-5)$

Since both terms are perfect cubes $\displaystyle \left (343 = 7^{3} \right )$, the factoring pattern we are looking to take advantage of is the sum of cubes pattern. This pattern is

$\displaystyle A^{3}+ B^{3} = (A + B) (A^{2}-AB +B^{2})$

We substitute $\displaystyle t$ for $\displaystyle A$ and 7 for $\displaystyle B$:

$\displaystyle t^{3} + 343 = t^{3} + 7^{3}$

$\displaystyle = (t+ 7) (t^{2}-t \cdot 7+7^{2})$

$\displaystyle = (t+ 7) (t^{2}-7t+49)$

The latter factor cannot be factored further, since we would need to find two integers whose product is 49 and whose sum is $\displaystyle -7$; they do not exist. This is as far as we can go with the factoring.

$\displaystyle t^{6}$ is the cube of $\displaystyle t^{2}$. Therefore, if $\displaystyle N$ is a perfect cube, the expression $\displaystyle t^{6}+N$ is factorable as the sum of two cubes. All four of the choices are perfect cubes - 8, 27, 64, and 125 are the cubes of 2, 3, 4 and 5, respectively. The correct response is that none of the choices are correct.

$\displaystyle 9t^{2}$ is a perfect square term - it is equal to $\displaystyle \left (3t \right )^{2}$. All of the values of $\displaystyle N$ given in the choices are perfect squares - 25, 36, 49, and 64 are the squares of 5, 6, 7, and 8, respectively. 

Therefore, for each given value of $\displaystyle N$, the polynomial is the sum of squares, which is normally a prime polynomial. However, if $\displaystyle N = 36$ - and only in this case - the polynomial can be factored as follows:

$\displaystyle 9t^{2}+36 = 9 \cdot t^{2} + 9 \cdot 4 = 9 (t^{2}+ 4)$.

Call $\displaystyle P(x) = x^{4} - 11x^{2} + 23x - 18$.

By the Rational Zeroes Theorem, since $\displaystyle P (x)$ has only integer coefficients, any rational solution of $\displaystyle P (x) = 0$ must be a factor of 18 divided by a factor of 1 - positive or negative. 18 has as its factors 1, 2, 3, 6, 9, and 18; 1 has only itself as a factor. Therefore, the rational solutions of $\displaystyle P (x) = 0$ must be chosen from this set:

$\displaystyle \left \{ 1, 2, 3, 6, 9, 18 -1,- 2,- 3,- 6, -9, -18 \right \}$.

By the Factor Theorem, a polynomial $\displaystyle P (x)$ is divisible by $\displaystyle x- c$ if and only if $\displaystyle P(c) = 0$ - that is, if $\displaystyle c$ is a zero. By the preceding result, we can immediately eliminate $\displaystyle x-4$ and $\displaystyle x- 5$ as factors, since 4 and 5 have been eliminated as possible zeroes.

Of the three remaining choices, we can demonstrate that $\displaystyle x- 2$ is the factor by evaluating $\displaystyle P(2)$:

$\displaystyle P(x) = x^{4} - 11x^{2} + 23x - 18$

$\displaystyle P(2) = 2^{4} - 11 \cdot 2 ^{2} + 23 \cdot 2 - 18$

$\displaystyle = 16 - 11 \cdot 4 + 23 \cdot 2 - 18$

$\displaystyle = 16 - 44 + 46 - 18$

$\displaystyle = 0$

By the Factor Theorem, it follows that $\displaystyle x- 2$ is a factor.

As for the other two, we can confirm that neither is a factor by evaluating $\displaystyle P(1)$ and $\displaystyle P(3)$:

$\displaystyle P(1) = 1^{4} - 11 \cdot 1 ^{2} + 23 \cdot 1 - 18$

$\displaystyle = 1- 11 \cdot 1 + 23 \cdot 1 - 18$

$\displaystyle = 1- 11 + 23 - 18$

$\displaystyle = -5$

$\displaystyle P(3) = 3^{4} - 11 \cdot 3 ^{2} + 23 \cdot 3 - 18$

$\displaystyle = 81 -11\cdot 9 + 23 \cdot 3 - 18$

$\displaystyle = 81 -99 + 69 - 18$

$\displaystyle =33$

Set $\displaystyle y = x^{2}$. Then $\displaystyle x^{4 } = \left (x^{2} \right )^{2} = y ^{2}$. 

$\displaystyle 2x^{4} - 13x^{2} +21 = 0$ can be rewritten as

$\displaystyle 2 \left (x^{2} \right )^{2} - 13x^{2} +21 = 0$

Substituting $\displaystyle y^{2}$ for $\displaystyle x^{4 }$ and $\displaystyle y$ for $\displaystyle x^{2}$, the equation becomes 

$\displaystyle 2 y^{2} - 13y +21 = 0$,

a quadratic equation in $\displaystyle y$.

This can be solved using the $\displaystyle ac$ method. We are looking for two integers whose sum is $\displaystyle -13$ and whose product is $\displaystyle 2 (21)= 42$. Through some trial and error, the integers are found to be $\displaystyle -7$ and $\displaystyle -6$, so the above equation can be rewritten, and solved using grouping, as

$\displaystyle 2 y^{2} - 7y - 6y +21 = 0$

$\displaystyle (2 y^{2} - 7y )- (6y -21 )= 0$

$\displaystyle y(2 y - 7 )- 3(2 y - 7 ) = 0$

$\displaystyle (y - 3 )(2 y - 7 ) = 0$

By the Zero Product Principle, one of these factors is equal to zero:

Either:

$\displaystyle y- 3 = 0$

$\displaystyle y - 3 + 3 = 0 + 3$

$\displaystyle y = 3$

Substituting $\displaystyle x^{2}$ back for $\displaystyle y$:

$\displaystyle x^{2} = 3$

Taking the positive and negative square roots of both sides:

$\displaystyle x = \pm \sqrt{3}$.

Or:

$\displaystyle 2y- 7 = 0$

$\displaystyle 2y- 7+ 7 = 0 + 7$

$\displaystyle 2y = 7$

$\displaystyle \frac{2y}{2} =\frac{ 7}{2}$

$\displaystyle y =\frac{ 7}{2}$

Substituting back:

$\displaystyle x^{2} =\frac{ 7}{2}$

Taking the positive and negative square roots of both sides, and applying the Quotient of Radicals property, then simplifying by rationalizing the denominator:

$\displaystyle x =\pm \sqrt{\frac{ 7}{2}} = \pm \frac{\sqrt{7}}{\sqrt{2}} = \pm \frac{\sqrt{7} \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2} } =\pm \frac{\sqrt{14}}{2}$

The solution set is $\displaystyle \left \{ - \frac{\sqrt{14}}{2}, - \sqrt{3}, \sqrt{3} ,\frac{ \sqrt{14}}{2} \right \}$.

Define $\displaystyle g(x)= f(x) - 4 = x^{2} + \cos 3x - 4$. Then, if $\displaystyle g(c)= f(c) - 4 = 0$, it follows that $\displaystyle f(c) = 4$.

By the Intermediate Value Theorem (IVT), if $\displaystyle g(x)$ is a continuous function, and $\displaystyle g(a)$ and $\displaystyle g(b)$ are of unlike sign, then $\displaystyle g(c) = 0$ for some $\displaystyle c \in (a, b)$.  $\displaystyle g(x)$ is a continuous function, so the IVT applies here.

Evaluate $\displaystyle g(x)$ for each of the following values: $\displaystyle \left \{ 1.5, 1.6, 1.7,1.8, 1.9, 2.0 \right \}$

$\displaystyle g(x)= x^{2} + \cos 3x - 4$

$\displaystyle g(1.5)= 1.5^{2} + \cos 3 (1.5) - 4$

$\displaystyle = 2.25 + \cos 4.5 - 4$

$\displaystyle \approx 2.25 + (-0.21 )- 4$

$\displaystyle \approx - 1.96$

$\displaystyle g(1.6)= 1.6^{2} + \cos 3 (1.6) - 4$

$\displaystyle = 2.56 + \cos 4.8 - 4$

$\displaystyle \approx 2.56 +0.09 - 4$

$\displaystyle \approx - 1.35$

$\displaystyle g(1.7)= 1.7^{2} + \cos 3 (1.7) - 4$

$\displaystyle =2.89 + \cos 5.1- 4$

$\displaystyle \approx 2.89 + 0.38- 4$

$\displaystyle \approx - 0.73$

$\displaystyle g(1.8)= 1.8^{2} + \cos 3 (1.8) - 4$

$\displaystyle = 3.24 + \cos 5.4 - 4$

$\displaystyle \approx 3.24 + 0.63 - 4$

$\displaystyle \approx - 0.13$

$\displaystyle g(1.9)= 1.9^{2} + \cos 3 (1.9) - 4$

$\displaystyle =3.61 + \cos 5.7 - 4$

$\displaystyle \approx 3.61 + 0.83 - 4$

$\displaystyle \approx 0.44$

$\displaystyle g(2.0)= 2.0^{2} + \cos 3 (2.0) - 4$

$\displaystyle = 4 + \cos 6 - 4$

$\displaystyle \approx 4 +0.96 - 4$

$\displaystyle \approx 0.96$

Only in the case of $\displaystyle (1.8, 1.9)$ does it hold that $\displaystyle g (x)$ assumes a different sign at each endpoint - $\displaystyle g(1.8) < 0 < g(1.9)$. By the IVT, $\displaystyle g(c) = 0$, and $\displaystyle f(c) = 4$, for some $\displaystyle c \in (1.8, 1.9)$.

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Two imaginary zeroes are given that are each other's complex conjugate -  $\displaystyle 2 + 3i$ and $\displaystyle 2 -3i$. Since the polynomial is cubic - of degree 3 - it has one other zero, which must be real. However, no information is given about that zero. Therefore, the polynomial cannot be determined.

Define 

$\displaystyle g(x) = p(x) - q(x)$

$\displaystyle = 7x- 5^{x}$

Then if $\displaystyle g(c) = 0$,

it follows that

$\displaystyle p(c) - q(c) = 0$,

or, equivalently,

$\displaystyle p(c) = q(c)$.

By the Intermediate Value Theorem (IVT), if $\displaystyle g(x)$ is a continuous function, and $\displaystyle g(a)$ and $\displaystyle g(b)$ are of unlike sign, then $\displaystyle g(c) = 0$ for some $\displaystyle c \in (a, b)$. 

Since polynomial $\displaystyle p(x)$ and exponential function $\displaystyle g(x)$ are continuous everywhere, so is $\displaystyle g(x)$, so the IVT applies here.

Evaluate $\displaystyle g(x)$ for each of the following values: $\displaystyle \left \{ 1. 0 , 1.1, 1.2, 1.3, 1.4, 1.5 \right \}$:

$\displaystyle g(x) = 7x- 5^{x}$

$\displaystyle g(1) = 7 (1)- 5^{1} = 7 - 5 = 2$

$\displaystyle g(1.1) = 7 (1.1)- 5^{1.1} \approx 7.7 - 5.9 \approx 1.8$

$\displaystyle g(1.2) = 7 (1.2)- 5^{1.2} \approx 8.4 - 6.9 \approx 1.5$

$\displaystyle g(1.3) = 7 (1.3)- 5^{1.3} \approx 9.1 - 8.1 \approx 1$

$\displaystyle g(1.4) = 7 (1.4)- 5^{1.4} \approx 9.8 - 9.5 \approx 0.3$

$\displaystyle g(1.5) = 7 (1.5)- 5^{1.5} \approx 10.5- 11.2 \approx - 0.7$

Only in the case of $\displaystyle (1.4, 1.5)$ does it hold that $\displaystyle g (x)$ assumes a different sign at both endpoints - $\displaystyle g(1.4)> 0 > g(1.5)$. By the IVT, $\displaystyle g(c) = 0$, and $\displaystyle p(c) = q(c)$, for some $\displaystyle c \in (1.4, 1.5)$.

A cubic polynomial has three zeroes, if a zero of multiplicity $\displaystyle n$ is counted $\displaystyle n$ times. Since its lead term is $\displaystyle x^{3}$, we know that, in factored form,

$\displaystyle p(x) = (x-b_{1}) (x-b_{2}) (x-b_{3})$, 

where $\displaystyle b_{1}$, $\displaystyle b_{2}$, and $\displaystyle b_{3}$ are its zeroes.

Since 2 is a zero of multiplicity 1, its only other zero, 3, must be a zero of multiplicity 2.

Therefore, we can set $\displaystyle b_{1} = 2$, $\displaystyle b_{2} = b_{3}= 3$,  in the factored form of $\displaystyle p(x)$, and 

$\displaystyle p(x) = (x-2) (x-3) (x-3)$,

or

$\displaystyle p(x) = (x-2) (x-3) ^{2}$

To rewrite this, firs square $\displaystyle x -3$ by way of the square of a binomial pattern:

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

$\displaystyle (x-3)^{2} = x^{2} - 6x + 9$

Thus,

$\displaystyle p(x) = (x-2) ( x^{2} - 6x + 9)$

Multiplying:

                 $\displaystyle x^{2} - 6x + 9$ 

                             $\displaystyle \underline{x - 2}$

      $\displaystyle -2x^{2} +12x - 18$

$\displaystyle \underline{x^{3} - 6x^{2} + 9x}$________

$\displaystyle x^{3} - 8x^{2} + 21x - 18$,

the correct polynomial.