Factorising 4x^2 - 7x +5. (1 Viewer)

[S1M0]

Seen the title? Of course you did. In case you didn't know, i need help. Again. Seems stupid of me to ask you of this, but i can't seem to factorise this. Help would be immensely appreciated. I forgot how to factorise when you have a co-efficent of 4 to the x^2 part of the equation.

So could you please, if you would be so kind factorise this please:

4x^2 - 7x + 5 = 0

Thankyou.

 

[pLuvia]

Have you tried using the quadratic formula?
You won't be able to find integer roots for this equation, nor will it be in the non-complex number system

 

[watatank]

The simplest I can factorise this is

4x^2 - 7x +5

= 4x^2 - 2x -5x +5

= 2x(2x-1) - 5 (x-1)

And that's as far as I can get. This one doesnt factorise any simpler than that. But of course that is useless. and wont help you.

 

[Kujah]

The simplest I can factorise this is

4x^2 - 7x +5

= 4x^2 - 2x -5x +5

= 2x(2x-1) - 5 (x-1)

And that's as far as I can get. This one doesnt factorise any simpler than that. But of course that is useless. and wont help you.

yeah, thats what I got. is there any other way to further factorise that? or do we leave it as it is?

 

[S1M0]

Have you tried using the quadratic formula?
You won't be able to find integer roots for this equation, nor will it be in the non-complex number system

tried.

Comes up with the weirdest numbers.

Okay forgot that.

The question that i'm working on, right now, is this:

1
--- > 4
x+1

I've multiplied both sides by (x+1)^2 to come up with this:

1(x+1) > 4 (x+1)^2

(x+1) > 4 (x^2 + 2x + 1)

(x+1) > 4x^2 + 8x +4

0 > 4x^2 + 7x + 3

Now i need to factorise 4x^2 + 7x + 3, find the two points, test them and make a statement which will follow the following format.

__ < x > __

 

[watatank]

No you don't. You've gone about the question wrong. This is how it shouldve been done...

TIP: Always keep things in its simplest form...expand if and only if you have no choice...factorise when you can....

Correct up to here so I'll carry on from here:

(x+1) > 4 (x+1)^2

4 (x+1)^2 - (x+1) < 0

Factorise!
(x-1)[4(x-1) -1] < 0

(x-1)[4x-4) -1] < 0

(x-1)[4x-5] < 0

Graph that and wherever the graph is below zero is your answer.

 

[S1M0]

thankyou thankyou thankyou!

You're the best guys!

 

[S1M0]

(x+1) > 4 (x+1)^2

4 (x+1)^2 - (x+1) < 0

Factorise!
(x-1)[4(x-1) -1] < 0

(x-1)[4x-4) -1] < 0

(x-1)[4x-5] < 0

Thanks mate, you're the best.

But i don't get how you managed to get from:

4 (x+1)^2 - (x+1) < 0

to here:

(x-1)[4(x-1) -1] < 0

And my apologies for annoying you, but this is frustrating me to breaking point. And my ext. class isn't for another week.

 

[SoulSearcher]

I'd reckon that's a typo, the - is supposed to be +.

 

[S1M0]

no its right when done, the answers correct.

Damn, dunno how he factorised it though......

 

[watatank]

Thanks mate, you're the best.

But i don't get how you managed to get from:

4 (x+1)^2 - (x+1) < 0

to here:

(x-1)[4(x-1) -1] < 0

And my apologies for annoying you, but this is frustrating me to breaking point. And my ext. class isn't for another week.

No worries!

Good thing you pointed out my small mistake...Sorry about that there's a typo there. So it should be....

4 (x+1)^2 - (x+1) < 0

(x+1)[4(x+1) -1] < 0

Okay disregard the other solution....

The new one is

(x+1)[4x+ 4 -1] < 0

(x+1)[4x+ 3] < 0

Then graph that...

 

[milton]

Seen the title? Of course you did. In case you didn't know, i need help. Again. Seems stupid of me to ask you of this, but i can't seem to factorise this. Help would be immensely appreciated. I forgot how to factorise when you have a co-efficent of 4 to the x^2 part of the equation.

So could you please, if you would be so kind factorise this please:

4x^2 - 7x + 5 = 0

Thankyou.

4 [ x - (7+i(31)^0.5)/8 ] [ x - (7-i(31)^0.5)/8 ]

 

[toadstooltown]

Yeah, I get the same as miltons and if you (are game enough to) sub it in it all cancels. This is definately not 2u though, it might be a typo or something in the question.

If it's 2u just write "descriminant < 0, hence no solutions over R, no real factors over R". If you've been taught complex then just use quad formula straight out.

 

[stephenchow]

This isnt 4unit and besides you dont even have to get into all the complex numbers. S1MO, you've done the whole question correct in the first place (in post #6) however you didnt quite finish it off.

--------------------------------------------------------------------------------------------------------------------------------

1/(x+1) > 4

x+1 > 4(x+1)^2

x+1 > 4(x^2 + 2x + 1)

x+1 > 4x^2 + 8x + 4

0 > 4x^2 + 7x + 3

4x^2 + 7x + 3 > 0

THIS IS WHERE YOU FACTORISE and it factorises very nicely, go back to your year eight work.

(4x + 3)(x+1) > 0

Draw a parabola so you know where the solutions lie.
Answer is : -1 < x < -3/4

 

[S1M0]

Alright guys, thanks for your help!

Seriously guys, helping guys like me out gets appreciated a lot, so i really owe all you guys one.

 

[S1M0]

This isnt 4unit and besides you dont even have to get into all the complex numbers. S1MO, you've done the whole question correct in the first place (in post #6) however you didnt quite finish it off.

--------------------------------------------------------------------------------------------------------------------------------

1/(x+1) > 4

x+1 > 4(x+1)^2

x+1 > 4(x^2 + 2x + 1)

x+1 > 4x^2 + 8x + 4

0 > 4x^2 + 7x + 3

4x^2 + 7x + 3 > 0

THIS IS WHERE YOU FACTORISE and it factorises very nicely, go back to your year eight work.

(4x + 3)(x+1) > 0

Draw a parabola so you know where the solutions lie.
Answer is : -1 < x < -3/4

Testament to my silly mistakes.

I forgot how to do all of this, the holidays have made me very rusty you see. If i knew that it was THAT easy i wouldn't have needed to ask you in the first place.

 

[jyu]

tried.

Comes up with the weirdest numbers.

Okay forgot that.

The question that i'm working on, right now, is this:

1
--- > 4
x+1

I've multiplied both sides by (x+1)^2 to come up with this:

1(x+1) > 4 (x+1)^2

(x+1) > 4 (x^2 + 2x + 1)

(x+1) > 4x^2 + 8x +4

0 > 4x^2 + 7x + 3

Now i need to factorise 4x^2 + 7x + 3, find the two points, test them and make a statement which will follow the following format.

__ < x > __

1
--- > 4
x+1

So x + 1 > 0 and x + 1 < 1/4

So x > -1 and x < -3/4

i.e. -1 < x < -3/4

:wave: