A. (x-6)(2x+3)
B. (2x-9)(7x+4)
C. (x-1)(x+1)(x+3)
Factor:
A.) 2x^2-9x-18
B.) 14x^2-55x-36
C.) x^3+3x^2-x-3
2 answers
You must have learned a method which can be used for quadratics
One method which seems to be taught in many locations is the method of decomposition.
1. Multiply the first and last numbers:
2(-18) = -36
2. look for two factor of -36 which have a sum of -9 , (the middle term coefficient)
3.
-36
= -12(3)
3. now split the middle term of -9x into -12x + 3x
2x^2 - 12x + 3x - 18
partially factor it, that is, find a common facto for the first two terms and a common factor for the last two terms.
=2x(x-6) + 3(x-6)
You now must have a common factor, or else you made and error.
= (x-6)(2x+3)
A lot of students appear to like it, even though I have my own personal way which is much faster but rather hard to explain in this kind of forum.
Try the second one the same way, you should get the answer that Jai has given you.
The third one is similar to step #2 from above
One method which seems to be taught in many locations is the method of decomposition.
1. Multiply the first and last numbers:
2(-18) = -36
2. look for two factor of -36 which have a sum of -9 , (the middle term coefficient)
3.
-36
= -12(3)
3. now split the middle term of -9x into -12x + 3x
2x^2 - 12x + 3x - 18
partially factor it, that is, find a common facto for the first two terms and a common factor for the last two terms.
=2x(x-6) + 3(x-6)
You now must have a common factor, or else you made and error.
= (x-6)(2x+3)
A lot of students appear to like it, even though I have my own personal way which is much faster but rather hard to explain in this kind of forum.
Try the second one the same way, you should get the answer that Jai has given you.
The third one is similar to step #2 from above