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

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

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.