We must begin by factoring out $\displaystyle x^2$ from each term.

$\displaystyle x^2(x^2-8x+12)$

Next, we must find two numbers that sum to $\displaystyle -8$ and multiply to $\displaystyle 12$.

$\displaystyle (-6)+(-2)=-8$

$\displaystyle -6*-2=12$

Thus, our final answer is:

$\displaystyle x^2(x-6)(x-2)$

$\displaystyle x^2-x-12$

To trinomial is in $\displaystyle ax+bx+c$ form

In order to factor, find two numbers whose procuct is $\displaystyle c$, in this case $\displaystyle -12$, and whose sum is $\displaystyle b$, in this case $\displaystyle -1$

Factors of $\displaystyle -12$:

$\displaystyle -1, 12; 1, -12; -3,4; 3,-4;-2,6; 2,-6$

Which of these pairs has a sum of $\displaystyle -1$?

$\displaystyle 3$ and $\displaystyle -4$

Therefore the factored form of $\displaystyle x^2-x-12$ is:

$\displaystyle (x+3)(x-4)$

Our factors will need to have a product of $\displaystyle (-14)$, and a sum of $\displaystyle (-5)$, so our factors must be $\displaystyle (-7)$ and $\displaystyle 2$.

$\displaystyle (x^2+y^2+z^2)(x^2+y^2+z^2)$

Use the distributive property:

$\displaystyle x^2(x^2+y^2+z^2)+y^2(x^2+y^2+z^2)+z^2(x^2+y^2+z^2)$

$\displaystyle x^4+x^2y^2+x^2z^2 + x^2y^2+y^4+y^2z^2+x^2z^2+y^2z^2+z^4$

Combine like terms:

$\displaystyle x^4+y^4+z^4+2x^2y^2+2y^2z^2+2x^2z^2$

Find the product:

$\displaystyle (x^2+7x+4)(x^2-3x-2)$

Step 1: Use the Distributive Property

$\displaystyle x^2(x^2-3x-2)+7x(x^2-3x-2)+4(x^2-3x-2)$

$\displaystyle x^4-3x^3-2x^2+7x^3-21x^2-14x+4x^2-12x-8$

Step 2: Combine like terms

$\displaystyle x^4+4x^3-19x^2-26x-8$

Find the product:

$\displaystyle (2x^2+3y+6)(7x^2+4y+1)$

Use the distributive property:

$\displaystyle 2x^2(7x^2+4y+1)+3y(7x^2+4y+1)+6(7x^2+4y+1)$

$\displaystyle 14x^4+8x^2y+2x^2+21x^2y+12y^2+3y+42x^2+24y+6$

$\displaystyle 14x^4+29x^2y+44x^2+12y^2+27y+6$

All operations are addition, so we can first remove the parentheses:

$\displaystyle (4x^2+6x+5)+(3x^3+10x+7)$

$\displaystyle =4x^2+6x+5+3x^3+10x+7$

Now rearrange the terms so that like terms are next to each other:

$\displaystyle =3x^3+4x^2+(6x+10x)+(5+7)$

Combine like terms:

$\displaystyle =3x^3+4x^2+16x+12$

$\displaystyle \left ( 4x^{2}+ 9y^{2} \right ) + \left ( 2x^{2}- 4xy + 7y\right )$

$\displaystyle = 4x^{2}+ 2x^{2} + 9y^{2} - 4xy + 7y$

$\displaystyle = 6x^{2} + 9y^{2} - 4xy + 7y$

$\displaystyle \left ( 4x^{2}+ 2xy + 7y^{2} \right ) + \left ( 2x^{2}+7y\right )$

$\displaystyle = 4x^{2}+ 2x^{2} + 2xy + 7y^{2} + 7y$

$\displaystyle = 6x^{2} + 2xy + 7y^{2} + 7y$

$\displaystyle \left ( \frac{1}{2}x^{2} + \frac{2}{3}x - \frac{1}{5}\right ) + \left ( \frac{1}{4}x^{2} - \frac{1}{4}x - \frac{1}{4}\right )$

$\displaystyle = \frac{1}{2}x^{2} + \frac{2}{3}x - \frac{1}{5} + \frac{1}{4}x^{2} - \frac{1}{4}x - \frac{1}{4}$

$\displaystyle = \frac{1}{2}x^{2} + \frac{1}{4}x^{2} + \frac{2}{3}x - \frac{1}{4}x- \frac{1}{5} - \frac{1}{4}$

$\displaystyle =\left ( \frac{1}{2} + \frac{1}{4} \right )x^{2} +\left ( \frac{2}{3} - \frac{1}{4} \right )x- \left (\frac{1}{5}+\frac{1}{4} \right )$

$\displaystyle =\left ( \frac{2}{4} + \frac{1}{4} \right )x^{2} +\left ( \frac{8}{12} - \frac{3}{12} \right )x- \left (\frac{4}{20}+\frac{5}{20} \right )$

$\displaystyle = \frac{3}{4} x^{2} + \frac{5}{12}x - \frac{9}{20}$