Factor Polynomial
- Thread starter Stine
- Start date
\(\displaystyle \H\
75x^2 - 90xy + 27y^2 =0\)
First simplify (divide by 3):
\(\displaystyle \H\
25x^2 - 30xy + 9y^2=0\)
Notice that the last term (9y^2) is positive. This means that either to positive numbers were multiplied, or two negative numbers were multiplied.
We also see that the middle term (-30xy) is negative. This probably means that two negative numbers were added together, and that the two numbers multiplied together to get the last term are negatives.
With these two factors in mind:
\(\displaystyle \H\
(5x - 3y)(5x - 3y)=0\)
75x^2 - 90xy + 27y^2 =0\)
First simplify (divide by 3):
\(\displaystyle \H\
25x^2 - 30xy + 9y^2=0\)
Notice that the last term (9y^2) is positive. This means that either to positive numbers were multiplied, or two negative numbers were multiplied.
We also see that the middle term (-30xy) is negative. This probably means that two negative numbers were added together, and that the two numbers multiplied together to get the last term are negatives.
With these two factors in mind:
\(\displaystyle \H\
(5x - 3y)(5x - 3y)=0\)
Hello, Stine!
calchere did a good job, but he thought you had an equation . . .
Always look for a common factor first.
Each term has a factor of 3.
Factor it out: \(\displaystyle \:3\,\cdot\,(25x^2\,-\,30xy\,+\,9y^2)\)
Now we work on the trinomial . . .
Do the two letters bother you?
Just "split" them: \(\displaystyle \
\;x\;\;\;y)(\;x\;\;\;y)\)
Here's a hint . . .
The first term is a square: \(\displaystyle 25x^2\)
The last term is a sqiare: \(\displaystyle 9y^2\)
Quite often (not always), this can mean that the trinomial is a square,
. . a factor times itself.
Let's try it: \(\displaystyle \;(5x\;\;3y)(5x\;\;3y)\)
Multiply and we have: \(\displaystyle \;25x^2\;\;15xy\;\;15xy\;\;9y^2\)
Hey, this will work! . . . We need: \(\displaystyle \;25x^2\) - \(\displaystyle 15xy\) - \(\displaystyle 15xy \,+\,9y^2\)
So the factors are: \(\displaystyle \5x\,-\,3y)\) and \(\displaystyle (5x\,-\,3y)\)
Answer: \(\displaystyle \:3(5x\,-3y)^2\)
calchere did a good job, but he thought you had an equation . . .
Always look for a common factor first.
Each term has a factor of 3.
Factor it out: \(\displaystyle \:3\,\cdot\,(25x^2\,-\,30xy\,+\,9y^2)\)
Now we work on the trinomial . . .
Do the two letters bother you?
Just "split" them: \(\displaystyle \
\;x\;\;\;y)(\;x\;\;\;y)\)Here's a hint . . .
The first term is a square: \(\displaystyle 25x^2\)
The last term is a sqiare: \(\displaystyle 9y^2\)
Quite often (not always), this can mean that the trinomial is a square,
. . a factor times itself.
Let's try it: \(\displaystyle \;(5x\;\;3y)(5x\;\;3y)\)
Multiply and we have: \(\displaystyle \;25x^2\;\;15xy\;\;15xy\;\;9y^2\)
Hey, this will work! . . . We need: \(\displaystyle \;25x^2\) - \(\displaystyle 15xy\) - \(\displaystyle 15xy \,+\,9y^2\)
So the factors are: \(\displaystyle \
5x\,-\,3y)\) and \(\displaystyle (5x\,-\,3y)\)Answer: \(\displaystyle \:3(5x\,-3y)^2\)