1. General Factoring

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.

2. Originally Posted by jay123

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 $\displaystyle x^2$ among the first two terms, and a common $\displaystyle 5$ among the last two, so pull those out.

$\displaystyle x^3 - 4x^2 + 5x - 20 = x^2 (x - 4) + 5(x - 4)$

now what can we do with that?

by the way, write powers using "^"

so you should write x^3 - 4x^2 + 5x - 20

3. I believe we can do this:

(x-4)^2 (x^2+5) right? wait a second can I make (x-4)^2?

4. 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 $\displaystyle (x - 4) \left( x^2 + 5 \right)$

5. 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?

6. 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:

$\displaystyle a^2 - b^2 = (a + b)(a - b)$

now, how do you think this question should be done

7. 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

8. 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: $\displaystyle x^4 - 81 = \left( x^2 \right)^2 - 81$

now what?

9. can you help me with this problem

(2y+3)^2-5(2y+3)+6

10. If we call $\displaystyle u=2y+3$, it remains to factorise $\displaystyle u^2-5u+6$

Factorise that, then back substitute.

11. 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:

$\displaystyle x^2 - 5x + 6$

now continue...

12. yes, i did.
I still don't understand...
x-5x+6 how do i solve this?

what rule should i follow?

13. 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

14. 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?

15. 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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