|
Question 182397: Factor each polynomial. a^2-2a-35
Factor completely. -3t^3+ 3t^2-6t
Factor polynomial completely. 10a^2+ab-2b^2
Factor completely. 4m^2+20m+25
Answer by J2R2R(94)   (Show Source):
You can put this solution on YOUR website! Factor each polynomial. a^2-2a-35
Here, the two number inside the brackets must add up to -2 and multiply together to give -35
-7 and +5 satisfy the criteria
a^2-2a-35 = (a-7)(a+5)
Factor completely. -3t^3+ 3t^2-6t
Here we have a common factor -3t
-3t^3+ 3t^2-6t = -3t(t^2-t+2)
(t^2-t+2) cannot be factored any further (using real numbers).
Factor polynomial completely. 10a^2+ab-2b^2
Here we have the two coefficient numbers multiplied together to give 10 (from 10a^2) and the two other numbers multiplied together give -2 (from -2b^2)
Then we have a fiddling act where crossover multiplying and adding together give 1 (from ab = 1ab).
5a and 2a are useful to start with and 2 and 1 (with one of these being a minus which we will decide on the next line).
You should see that 5*1 and 2*2 when a minus is in front of the 2 will make 5*1-2*2=1, so the 2 has a minus. These are the crossovers (5a with b and -2b with 2a)
10a^2+ab-2b^2 = (5a-2b)(2a+b)
Factor completely. 4m^2+20m+25
Similarly to above but just observe that 4m^2 and 25 are both squares, so think of using 2m and 5.
2m times 2m gives 4m^2 and 5 times 5 gives 25. Now the crossovers 2m times 5 = 10m and the same again the other way 5 times 2m = 10m, combined makes 20m, so we have
4m^2+20m+25=(2m+5)(2m+5) = (2m+5)^2
Whenever there is a square factor, just double the crossover since it is the sum of the product of the two numbers in both directions e.g. 2 times 5 plus 5 times 2 or simply 2 times 2 times 5. With this in mind you should be able to spot square factors like this. 2^2=4, 5^2=25, 2x2x5=20.
|
|
|
| |
