\(\displaystyle \frac{3xy \cdot 7xy}{12y^{3}}\)

\(\displaystyle = \frac{3 \cdot 7 \cdot x \cdot x \cdot y \cdot y}{12y^{3}}\)

\(\displaystyle = \frac{21 x^{2} y^{2} }{12y^{3}}\)

\(\displaystyle = \frac{21 x^{2} }{12y^{3-2}}\)

\(\displaystyle = \frac{21 x^{2} }{12y }\)

\(\displaystyle = \frac{21 x^{2} \div 3 }{12y \div 3}\)

\(\displaystyle = \frac{7 x^{2} }{4y }\)

\(\displaystyle (8x ^{2} -7x + 4) -(x^{2} - 5)\)

\(\displaystyle = 8x ^{2} -7x + 4 -x^{2} + 5\)

\(\displaystyle = 8x ^{2} -1x^{2} -7x + 4 + 5\)

\(\displaystyle =\left ( 8-1 \right )x ^{2} -7x +\left ( 4 + 5 \right )\)

\(\displaystyle =7x ^{2} -7x +9\)

\(\displaystyle (8x ^{2} -7x + 4) +(x^{2} - 5)\)

\(\displaystyle =8x ^{2} + x^{2} -7x + 4 - 5\)

\(\displaystyle =\left ( 8+1 \right ) x ^{2} -7x +\left ( 4 - 5 \right )\)

\(\displaystyle =9x ^{2} -7x -1\)

First, factor out the greatest common factor of the terms, which is \(\displaystyle 5y\):

\(\displaystyle 5y^{3}- 20y^{2}- 105y\)

\(\displaystyle = 5y \cdot y^{2}- 5y \cdot 4y- 5y \cdot 21\)

\(\displaystyle = 5y \left (y^{2}- 4y- 21 \right )\)

The quadratic trinomial can be factored as \(\displaystyle (x+A)(x+ B)\) where \(\displaystyle A = -21\) and \(\displaystyle A + B = -4\); by trial and error we find that the numbers chosen are \(\displaystyle A = -7, B = 3\), so

\(\displaystyle 5y \left (y^{2}- 4y- 21 \right )\)

\(\displaystyle = 5y (y-7)(y + 3)\)

Apply the power of a quotient rule:

\(\displaystyle \left (\frac{3}{t} \right )^{3} = \frac{3^{3}}{t^{3}} = \frac{27}{t^{3}}\)

\(\displaystyle x\) is a common factor of both terms, so factor it out:

\(\displaystyle x^{3} + 121x\)

\(\displaystyle = x \cdot x ^{2} +x \cdot 121\)

\(\displaystyle = x \left ( x ^{2} + 121 \right )\)

\(\displaystyle = x ^{2} + 121\) cannot be factored, so this is the complete factorization.

First, we find two integers whose sum is 19 and whose product is \(\displaystyle 12 \times 4 = 48\). Through trial and error we find these integers are 3 and 16. We use these numbers to split the middle term, then we factor using the grouping method:

\(\displaystyle 12x^{2} + 19x + 4\)

\(\displaystyle = 12x^{2}+ 3x + 16x + 4\)

\(\displaystyle = \left (12x^{2}+ 3x \right )+ \left (16x + 4 \right )\)

\(\displaystyle = \left (3x \cdot 4x+ 3x \cdot 1 \right )+ \left (4 \cdot 4x + 4 \cdot 1 \right )\)

\(\displaystyle = 3x \left (4x+ 1 \right )+ 4\left (4x+ 1 \right )\)

\(\displaystyle = \left (3x + 4 \right )\left (4x+ 1 \right )\)

Factor by grouping as follows:

\(\displaystyle 5x ^{3}+ 25x^{2} - 7x - 35\)

\(\displaystyle =\left (5x ^{3}+ 25x^{2} \right )- \left (7x + 35 \right )\)

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

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

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

Factor by grouping as follows:

\(\displaystyle x ^{3} + 7x ^{2} - 9x - 63\)

\(\displaystyle = \left (x ^{3} + 7x ^{2} \right )- \left (9x + 63 \right )\)

\(\displaystyle = \left (x ^{2} \cdot x + x ^{2} \cdot 7 \right )- \left (9 \cdot x + 9 \cdot 7\right )\)

\(\displaystyle = x ^{2} \left ( x + 7 \right )- 9 \left ( x + 7\right )\)

\(\displaystyle =\left ( x ^{2}- 9 \right ) \left ( x + 7 \right )\)

The first factor is the difference of squares, so further factoring can be done:

\(\displaystyle =\left ( x ^{2}- 3 ^{2} \right ) \left ( x + 7 \right )\)

\(\displaystyle =\left ( x + 3 \right )\left ( x - 3 \right ) \left ( x + 7 \right )\)

The polynomial fits the perfect square pattern:

\(\displaystyle 49x^{2} - 70x+ 25\)

\(\displaystyle \left (7x \right )^{2} - 2 \cdot 7x \cdot 5 + 5^{2}\)

This can be factored using the pattern

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

with \(\displaystyle A = 7x, B = 5\):

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