Hey guys, expanding factored equations and simplifying them has my head in a knot.

Any extremely helpful advice to get past it? (no need to resolve except for the last couple if you're generous.)

The questions format is varied, ie.

(2-x)(1-x) (struggling)

(3x-9y) -> got this type easily.

(1/2x + 1/4y) -> i got 1/4(1/4x+y) but think it's wrong.

(3x+1)(2x+1) -> which posed the question, how do I know if a number attributed to x or just a solo number was factored? (struggling)

1/3x + 1/6xy -> have not really attempted this in the last few days. But I can do it.

2/3pi^{3}+1/3pi ^{2}h. -> do solve please. I get that pi can be same as x in the equation but that isn't my issue trying to solve it.

also

2x^{2}(x-5)-x(x-2)-3x(x-5)

5x^{2}-4x-1 (can solve with lots of thoughts.

16x^{2}-1

Would love advice and ideas to solve.

Any extremely helpful advice to get past it? (no need to resolve except for the last couple if you're generous.)

The questions format is varied, ie.

(2-x)(1-x) (struggling)

(3x-9y) -> got this type easily.

(1/2x + 1/4y) -> i got 1/4(1/4x+y) but think it's wrong.

(3x+1)(2x+1) -> which posed the question, how do I know if a number attributed to x or just a solo number was factored? (struggling)

1/3x + 1/6xy -> have not really attempted this in the last few days. But I can do it.

2/3pi

also

2x

5x

16x

Would love advice and ideas to solve.

Stu Feb 8, 2014

#1**0 **

it is. You can't afford to not be able to do simple arithmetic like this if you're going into EE.

(1/2x + 1/4y) = 1/4(2x + y)

this is already factored. I'm not sure what you are asking here.

I don't know what this is. do you mean

(2/3) pi^{3} + (1/3) pi ^{2}h ?

if so it's (1/3)pi^{2}(2pi + h)

this is just arithmetic. just multiply everything out.

(5x+1)(x-1)

difference of 2 squares

(4x-1)(4x+1)

my advice to you is to practice these types of problems until you can do them as easily as you can add 2 and 2.

If you can't whip through this sort of stuff pre-calculus is going to be h**l.

Quote:Hey guys, expanding factored equations and simplifying them has my head in a knot.

Any extremely helpful advice to get past it? (no need to resolve except for the last couple if you're generous.)

The questions format is varied, ie.

(2-x)(1-x) (struggling)

(3x-9y) -> got this type easily.

(1/2x + 1/4y) -> i got 1/4(1/4x+y) but think it's wrong.

it is. You can't afford to not be able to do simple arithmetic like this if you're going into EE.

(1/2x + 1/4y) = 1/4(2x + y)

Quote:

(3x+1)(2x+1) -> which posed the question, how do I know if a number attributed to x or just a solo number was factored? (struggling)

this is already factored. I'm not sure what you are asking here.

Quote:

1/3x + 1/6xy -> have not really attempted this in the last few days. But I can do it.

2/3pi3+1/3pi2h. -> do solve please. I get that pi can be same as x in the equation but that isn't my issue trying to solve it.

I don't know what this is. do you mean

(2/3) pi

if so it's (1/3)pi

Quote:

also

2x2(x-5)-x(x-2)-3x(x-5)

this is just arithmetic. just multiply everything out.

Quote:

5x2-4x-1 (can solve with lots of thoughts.

(5x+1)(x-1)

Quote:

16x2-1

difference of 2 squares

(4x-1)(4x+1)

my advice to you is to practice these types of problems until you can do them as easily as you can add 2 and 2.

If you can't whip through this sort of stuff pre-calculus is going to be h**l.

Rom Feb 8, 2014

#2**0 **

(3x+1)(2x+1) = 6x ^{2}+5x+1 -> which posed the question, how do I know if a number attributed to x was factored or it was the whole number with no x that was factored.

Thanks for advice and yes, I will know it. Just getting on top of it so seeking some thoughts to more easily solve than I can. Just getting mixed up atm and only brushed over it in my little bit of study help. Still a couple weeks to practice left.

Thanks for advice and yes, I will know it. Just getting on top of it so seeking some thoughts to more easily solve than I can. Just getting mixed up atm and only brushed over it in my little bit of study help. Still a couple weeks to practice left.

Stu Feb 8, 2014

#3**0 **

I don't understand what you mean Stu. Maybe if I read all the other posts I might.

This has been simplified not factorised.

When you factorise you put the brackets in

when you simplify you take the brackets out.

Stu:(3x+1)(2x+1) = 6x

^{2}+5x+1 -> which posed the question, how do I know if a number attributed to x was factored or it was the whole number with no x that was factored.

Thanks for advice and yes, I will know it. Just getting on top of it so seeking some thoughts to more easily solve than I can. Just getting mixed up atm and only brushed over it in my little bit of study help. Still a couple weeks to practice left.

I don't understand what you mean Stu. Maybe if I read all the other posts I might.

This has been simplified not factorised.

When you factorise you put the brackets in

when you simplify you take the brackets out.

Melody Feb 8, 2014