can anyone please help me with this question: Xwith a exponent of 3 -4x with exponent 2+5x-20..I dont know how to make the exponents small maybe someone can help me with that to.
Last edited by jay123; September 22nd 2007 at 06:01 PM. Reason: forgot exponent
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Originally Posted by jay123 can anyone please help me with this question: Xwith a exponent of 3 -4x with exponent 2+5x-20..I dont know how to make the exponents small maybe someone can help me with that to. Here we can factor by groups. note that there is a common among the first two terms, and a common among the last two, so pull those out. now what can we do with that? by the way, write powers using "^" so you should write x^3 - 4x^2 + 5x - 20
I believe we can do this: (x-4)^2 (x^2+5) right? wait a second can I make (x-4)^2?
Originally Posted by jay123 I believe we can do this: (x-4)^2 (x^2+5) right? wait a second can I make (x-4)^2? no, you factor out the common (x - 4). how did you make it squared? it is simply
i c. ok but what about when there is X^4-81? I know that since it is ^4 that 3 goes into 81 4 times. How would i write this ouT?
Originally Posted by jay123 i c. ok but what about when there is X^4-81? I know that since it is ^4 that 3 goes into 81 4 times. How would i write this ouT? no. this is the difference of two squares (do you see it?) we factor the difference of two squares as follows: now, how do you think this question should be done
sorry i dont understand....difference of squares yes...can i separate x^4-81 (x-3) (x+3) x^2+3x-3x-9= X^2-9
Originally Posted by jay123 sorry i dont understand....difference of squares yes...can i separate x^4-81 (x-3) (x+3) x^2+3x-3x-9= X^2-9 note that: now what?
can you help me with this problem (2y+3)^2-5(2y+3)+6
If we call , it remains to factorise Factorise that, then back substitute.
Originally Posted by jay123 can you help me with this problem (2y+3)^2-5(2y+3)+6 did you get the last one? the difference of two squares? note that for this question, we have a quadratic in 2y + 3. if you can't see that right away, try replacing 2y + 3 with a variable, say x, then you get: now continue...
yes, i did. I still don't understand... x-5x+6 how do i solve this? what rule should i follow?
lets see if i X^2-5x+6 = (x-2)(x-3)...then what????do i multiply (X-2)(X-3)(2y+3) but i still have the (2y+3)^2
Originally Posted by jay123 lets see if i X^2-5x+6 = (x-2)(x-3)...then what????do i multiply (X-2)(X-3)(2y+3) but i still have the (2y+3)^2 no, just replace the x with 2y + 3, that's what you used x to represent, remember?
lol...im still lost. do i X^2-5(2y+3)+6 X^2-10y-15+6 X^2-10y+1
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