1. ## Factorising

How would I factor these:

1) $x^7+x$

2) x^12 - y^12

And I'm also having trouble with this problem:

1) If $x+y=1$ and $x^3+y^3=19$. What is $x^2+y^2$

2. 1) divide by x to get: $x(1+6x)$

2) this is like a really complicated difference of squares. $x^2-y^2 = (x+y)(x-y)$, etc.

3) use the first equation to get $x = 1 - y$, then plug that into the second equation to figure out $y$. Once you have $y$, plug it into the first equation to find $x$, then plug them both into $x^2 + y^2$ to get your answer.

3. Originally Posted by acevipa
How would I factor these:

1) $x^7+x$

2) x^12 - y^12

And I'm also having trouble with this problem:

1) If $x+y=1$ and $x^3+y^3=19$. What is $x^2+y^2$
Question 1

$x^7+x=x(x^6+1)=x((x^2)^3+(1)^3)=x(x^2+1)(x^4-x^2+1)$

Question 2

$x^{12}-y^{12}=(x^6-y^6)(x^6+y^6)=((x^2)^3+(y^2)^3)(x^3-y^3)(x^3+y^3)$

$=(x^2+y^2)(x^4-x^2y^2+y^4)(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)$

Last Question

$(x+y)^3=x^3+y^3+3xy(x+y)$

Subbing in values:

$1^3=19+3xy(1) \Leftrightarrow 3xy=-18 \Leftrightarrow xy=-6$

Also, $(x+y)^2=x^2+y^2+2xy$

Subbing in again,

$(1)^2=x^2+y^2+2(-6)$

$\Rightarrow x^2+y^2=13$