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.