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: \(\displaystyle \,[3(x+2y+z)\, -\,4(x-2y+z)]\,\cdot\,[3(x+2y+x)\,+\,4(x-2y+z)]\)
\(\displaystyle \;\;\;= \;(3x\,+\,6y\,+\,3z\,-\,4x\,+\,8y\,-\,4z)\cdot(3x\,+\,6y\,+\,3z\,+\,4z\,-\,8y\,+\,4z)\)
\(\displaystyle \;\;\;= \;(-x\,+\,14y\,-\,z)\cdot(7x\,-\,2y\,+\,7z)\)
