$\displaystyle \small -x(4x+2)=(-x)(4x)+(-x)(2)=-4x^2+(-2x)=-4x^2-2x$

$\displaystyle \small x^2+5x+4=(x+a)(x+b)$

where

$\displaystyle \small a+b=5\ and\ ab=4$

The numbers $\displaystyle \small 4$ and $\displaystyle \small 1$ fit those criteria. Therefore,

$\displaystyle \small x^2+5x+4=(x+1)(x+4)$

You can double check the answer using the FOIL method

Factor $\displaystyle y^{4}- 81$ all the way to its prime factorization.

$\displaystyle y^{4}- 81$ can be factored as the difference of two perfect square terms as follows:

$\displaystyle y^{4}- 81$

$\displaystyle =\left ( y^{2} \right )^{2} - 9^{2}$

$\displaystyle =\left ( y^{2} + 9 \right )\left ( y^{2} - 9 \right )$

$\displaystyle y^{2} + 9$ is a factor, and, as the sum of squares, it is a prime. $\displaystyle y^{2} - 9$ is also a factor, but it is not a prime factor - it can be factored as the difference of two perfect square terms. We continue:

$\displaystyle \left ( y^{2} + 9 \right )\left ( y^{2} - 9 \right )$

$\displaystyle =\left ( y^{2} + 9 \right )\left ( y^{2} - 3^{2} \right )$

$\displaystyle =\left ( y^{2} + 9 \right )\left (y+3 \right )\left ( y-3\right )$

Therefore, all of the given polynomials are factors of $\displaystyle y^{4}- 81$, but $\displaystyle y^{2} - 9$ is the correct choice, as it is not a prime factor.

$\displaystyle n^{4} + 30n^{2} +225$ can be seen to fit the pattern 

$\displaystyle A ^{2} + 2AB + B^{2}$:

$\displaystyle n^{4} + 30n^{2} +225 = \left (n^{2} \right ) ^{2}+ 2 \cdot n^{2} \cdot 15 + 15 ^{2}$

where $\displaystyle A = n^{2} , B = 15$

$\displaystyle A ^{2}+2AB + B^{2}$ can be factored as $\displaystyle (A+B ) ^{2}$, so

 $\displaystyle n^{4} + 30n^{2} +225 = \left (n^{2} \right ) ^{2}+ 2 \cdot n^{2} \cdot 15 + 15 ^{2} = (n^{2}+15)^{2}$.

$\displaystyle n^{2}+15$ does  not fit into any factorization pattern, so it is prime, and the above is the complete factorization of the polynomial. Therefore, $\displaystyle n^{2}+15$ is the correct choice.

Divide termwise:

$\displaystyle \left ( 12 x ^{5} + 9x^{3}+ 6x^{2}+36 x \right ) \div 6x$

$\displaystyle = \frac{ 12 x ^{5} + 9x^{3}+ 6x^{2}+36 x }{6x}$

$\displaystyle = \frac{ 12 x ^{5} }{6x} + \frac{ 9x^{3} }{6x} + \frac{ 6x^{2} }{6x}+ \frac{ 36 x }{6x}$

$\displaystyle = \frac{ 12 }{6} x ^{5-1} + \frac{ 9 }{6 }x ^{3-1} + \frac{ 6 }{6}x^{2-1}+ \frac{ 36 }{6 }$

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

$\displaystyle \left (100Y^{2}+ 110Y + 121 \right )(10Y-11)$

$\displaystyle =\left [\left (10Y \right) ^{2}+ 10Y \cdot 11 + 11^{2} \right] (10Y-11)$

This product fits the difference of cubes pattern, where $\displaystyle A = 10Y, B = 11$:

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

so

$\displaystyle \left [\left (10Y \right) ^{2}+ 10Y \cdot 11 + 11^{2} \right] (10Y-11)$

$\displaystyle = \left ( 10Y\right )^{3} -11^{3} = 1,000Y^{3}-1,331$

A quadratic trinomial is a perfect square if and only if takes the form

$\displaystyle A^{2} \pm 2AB + B^{2}$ for some values of $\displaystyle A$ and $\displaystyle B$.

$\displaystyle 121x^{2}- Nx + 49= (11x)^{2} - Nx + 7^{2}$, so 

$\displaystyle A = 11x$ and $\displaystyle B = 7$. 

For $\displaystyle 121x^{2}- Nx + 49$ to be a perfect square, it must hold that 

$\displaystyle Nx = 2AB = 2 \cdot 11x \cdot 7 = 154x$,

so $\displaystyle N = 154$. This is the correct choice.

Perhaps the easiest way to identify the factor is to take advantage of the factor theorem, which states that $\displaystyle x-a$ is a factor of polynomial $\displaystyle P(x)$ if and only if $\displaystyle P(a)= 0$. We substitute 1, 2, 4, and 9 for $\displaystyle x$ in the polynomial to identify the factor.

$\displaystyle x=1$:

$\displaystyle 1^{4}-13 \cdot 1^{2}+ 36$

$\displaystyle = 1 - 13 + 36$

$\displaystyle = 24$

$\displaystyle x=2$:

 $\displaystyle 2^{4}-13 \cdot 2^{2}+ 36$

$\displaystyle = 16- 13 \cdot 4 + 36$

$\displaystyle = 16 -52 + 36$

$\displaystyle = 0$

$\displaystyle x =4$:

$\displaystyle 4^{4}-13 \cdot 4^{2}+ 36$

$\displaystyle = 256- 13 \cdot 16 + 36$

$\displaystyle = 256 -208 +36$

$\displaystyle = 84$

$\displaystyle x=9$:

$\displaystyle 9^{4}-13 \cdot 9^{2}+ 36$

$\displaystyle = 6,561- 13 \cdot 81 + 36$

$\displaystyle = 6,561- 1,053+ 36$

$\displaystyle = 5,544$

Only $\displaystyle x = 2$ makes the polynomial equal to 0, so among the choices, only $\displaystyle x-2$ is a factor.

$\displaystyle x^{6} +1$ is the sum of two cubes:

$\displaystyle x^{6} +1 =\left ( x^{2} \right ) ^{3}+1^{3}$

As such, it can be factored using the pattern 

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

where $\displaystyle A = x^{2}, B = 1$;

$\displaystyle x^{6} +1 =\left ( x^{2} \right ) ^{3}+1^{3}$

$\displaystyle = (x^{2}+1)((x^{2})^{2}-x^{2} \cdot 1 + 1^{2})$

$\displaystyle = (x^{2}+1)(x^{4}-x^{2} + 1)$

The first factor,as the sum of squares, is a prime.

We try to factor the second by noting that it is "quadratic-style" based on $\displaystyle x^{2}$. and can be written as

$\displaystyle = (x^{2})^{2}-x^{2} + 1$;

we seek to factor it as $\displaystyle (x^{2}+\; \; \; )(x^{2}+\; \; \; )$

We want a pair of integers whose product is 1 and whose sum is $\displaystyle -1$. These integers do not exist, so $\displaystyle x^{4}-x^{2} + 1$ is a prime. 

$\displaystyle (x^{2}+1)(x^{4}-x^{2} + 1)$ is the prime factorization and the correct response is $\displaystyle x^{2}+1$.

Perhaps the easiest way to identify the factor is to take advantage of the factor theorem, which states that $\displaystyle x-a$ is a factor of polynomial $\displaystyle P(x)$ if and only if $\displaystyle P(a)= 0$. We substitute $\displaystyle \pm2$ and $\displaystyle \pm 4$ for $\displaystyle x$ in the polynomial to identify the factor.

$\displaystyle x = 2$:

$\displaystyle x^{6} - 66x^{3} + 128$

$\displaystyle 2^{6} - 66 \cdot 2^{3} + 128$

$\displaystyle = 64 - 66 \cdot 8 + 128$

$\displaystyle = 64 - 528 + 128$

$\displaystyle = -336$

$\displaystyle x = -2$:

$\displaystyle x^{6} - 66x^{3} + 128$

$\displaystyle \left (-2 \right )^{6} - 66 \cdot \left (-2 \right )^{3} + 128$

$\displaystyle = 64 - 66 \cdot\left ( -8 \right )+ 128$

$\displaystyle = 64 + 528 + 128$

$\displaystyle = 720$

$\displaystyle x = 4$:

$\displaystyle x^{6} - 66x^{3} + 128$

$\displaystyle 4^{6} - 66 \cdot 4^{3} + 128$

$\displaystyle = 4,096- 66 \cdot 64 + 128$

$\displaystyle = 4,096- 4,224 + 128$

$\displaystyle = 0$

$\displaystyle x = -4$:

$\displaystyle x^{6} - 66x^{3} + 128$

$\displaystyle \left (-4 \right )^{6} - 66 \cdot \left (-4 \right )^{3} + 128$

$\displaystyle = 4,096- 66 \cdot\left ( -64 \right )+ 128$

$\displaystyle = 4,096+4,224 + 128$

$\displaystyle = 8,448$

Only $\displaystyle x = 4$ makes the polynomial equal to 0, so of the four choices, only $\displaystyle x - 4$ is a factor of the polynomial.