factoring y^5 - y^4 + y^3 - y^2 + y - 1

[Guest]

How do you factor this?

. . .y^5 - y^4 + y^3 - y^2 + y - 1

I did:

. . .y^4(y - 1) + y^2(y - 1) + y - 1
. . .(y - 1)(y^4 + y^2 + 1)

But the answer is supposed to be:

. . .(y^3 - 1)(y^2 - y + 1)

What did I do wrong?

And how do you factor this:

. . .9(x + 2y + z)^2 - 16(x - 2y + z)^2

 

[galactus]

\(\displaystyle \L\\(y-1)(y^{4}+y^{2}+1)\) is correct.

Now, factor again:

\(\displaystyle \L\\y^{4}+y^{2}+1=(y^{2}+y+1)(y^{2}-y+1)\)

So, you have: \(\displaystyle \L\\(y-1)(y^{2}+y+1)(y^{2}-y+1)\)

Notice: \(\displaystyle \L\\(y-1)(y^{2}+y+1)=y^{3}-1\)

So, you do have:\(\displaystyle \L\\(y^{3}-1)(y^{2}-y+1)\)

 

[soroban]

Re: factoring

Hello, bittersweet!

How do you factor this? $\displaystyle \,y^5\,-\,y^4\,+\,y^3\,-\,y^2\,+\,y\,-\,1$

I did: $\displaystyle \,y^4(y\,-\,1)\,+\,y^2(y\,-\,1)\,+\,(y\,-\,1)$

$\displaystyle \;\;= \;(y\,-\,1)(y^4\,+\,y^2\,+\,1)$ . . . correct!

but the answer is suppose to be: $\displaystyle \,(y^3\,-\,1)(y^2\,-\,y\,+\,1)$ ??
What did I do wrong? $\displaystyle \;$ . . . nothing!

I don't like their answer at all!
$\displaystyle \;\;$I can't believe they didn't factor that difference-of-cubes.

If "they" aren't required to factor completely, neither are we
$\displaystyle \;\;$and your answer is correct.

In fact, you may be "more correct" since further factoring is harder to see,
$\displaystyle \;\;$while "they" were very sloppy in leaving $\displaystyle y^3\,-\,1$.


It takes "experience" to know that $\displaystyle y^4\,+\,y^2\,+\,1$ can be factored.
Here's how it is done . . .

Add and subtract $\displaystyle y^2:\;\;y^4\,+\,y^2\,$+ y²$\displaystyle \,+\,1\,$- y² $\displaystyle \,=\;y^4\,+\,2y^2\,+\,1\,-\,y^2$

and we have: $\displaystyle \,(y^2\,+\,1)^2\,-\,y^2\;$ . . . a difference of squares

which factors: \(\displaystyle \,\left([y^2\,+\,1]\,-\,y\right)\cdot\left([y^2\,+\,1]\,+\,y) \;= \;(y^2\,-\,y\,+\,1)\cdot(y^2\,+\,y\,+\,1)\)

and how do you factor this: $\displaystyle \,9(x+2y+z)^2\,-\,16(x-2y+z)^2$

Look again . . . we have: $\displaystyle \,9a^2\,-\,16b^2$ . . . a difference of squares.
$\displaystyle \;\;$The factoring would be: $\displaystyle (3a\,-\,4b)(3a\,+\,4b)$

So we have: $\displaystyle \,[3(x+2y+z)\, -\,4(x-2y+z)]\,\cdot\,[3(x+2y+x)\,+\,4(x-2y+z)]$

$\displaystyle \;\;\;= \;(3x\,+\,6y\,+\,3z\,-\,4x\,+\,8y\,-\,4z)\cdot(3x\,+\,6y\,+\,3z\,+\,4z\,-\,8y\,+\,4z)$

$\displaystyle \;\;\;= \;(-x\,+\,14y\,-\,z)\cdot(7x\,-\,2y\,+\,7z)$

 

[galactus]

Re: factoring

I don't like their answer at all!
$\displaystyle \;\;$I can't believe they didn't factor that difference-of-cubes.

I agree Soroban. I was thinkin' the same thing.

The $\displaystyle y^{4}+y^{2}+1$ is harder to see(possibly that was their intent).