How would I factor these:
1) $\displaystyle x^7+x$
2) x^12 - y^12
And I'm also having trouble with this problem:
1) If $\displaystyle x+y=1$ and $\displaystyle x^3+y^3=19$. What is $\displaystyle x^2+y^2$
1) divide by x to get: $\displaystyle x(1+6x)$
2) this is like a really complicated difference of squares. $\displaystyle x^2-y^2 = (x+y)(x-y)$, etc.
3) use the first equation to get $\displaystyle x = 1 - y$, then plug that into the second equation to figure out $\displaystyle y$. Once you have $\displaystyle y$, plug it into the first equation to find $\displaystyle x$, then plug them both into $\displaystyle x^2 + y^2$ to get your answer.
Question 1
$\displaystyle x^7+x=x(x^6+1)=x((x^2)^3+(1)^3)=x(x^2+1)(x^4-x^2+1)$
Question 2
$\displaystyle x^{12}-y^{12}=(x^6-y^6)(x^6+y^6)=((x^2)^3+(y^2)^3)(x^3-y^3)(x^3+y^3)$
$\displaystyle =(x^2+y^2)(x^4-x^2y^2+y^4)(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)$
Last Question
$\displaystyle (x+y)^3=x^3+y^3+3xy(x+y)$
Subbing in values:
$\displaystyle 1^3=19+3xy(1) \Leftrightarrow 3xy=-18 \Leftrightarrow xy=-6$
Also, $\displaystyle (x+y)^2=x^2+y^2+2xy$
Subbing in again,
$\displaystyle (1)^2=x^2+y^2+2(-6)$
$\displaystyle \Rightarrow x^2+y^2=13$