# Urgent Maths help!!

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#1
Hi I really really need help, I've got this maths homework due tomorrow and I'm not sure what to do on a few and if I've even done a few right could someone please please have a look! Thank you!!
I can't attach images so I'll type it out:

a)5x^2 + 5x - 10
My answer: I divided by 5, x^2 + x - 2 answer = (x+2)(x-1)
b)15x^2 - 72x - 15
My answer: Divided by 3, 5x^2 - 24 - 5 = (-5+x)(5x-1)

Solve: 3 * 4^(2k+8) = 24
IDK WHAT TO DO!

Factorise fully (w+4)^3 - (w+4)^2(w+1)
HELP! This one I expanded until I eventually got 3w^2 +16w +48 but it doesn't facatorise and I'm gonna cry because someone I know got: 3(w+4)^2 and I don't understand!

Last question:
y = 16 * 10^(8k) where k is an integer
find an expression, in terms of k, for y^(5/4)
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1 month ago
#2
(Original post by FloryK)
Hi I really really need help, I've got this maths homework due tomorrow and I'm not sure what to do on a few and if I've even done a few right could someone please please have a look! Thank you!!
I can't attach images so I'll type it out:

a)5x^2 + 5x - 10
My answer: I divided by 5, x^2 + x - 2 answer = (x+2)(x-1)
b)15x^2 - 72x - 15
My answer: Divided by 3, 5x^2 - 24 - 5 = (-5+x)(5x-1)

Solve: 3 * 4^(2k+8) = 24
IDK WHAT TO DO!

Factorise fully (w+4)^3 - (w+4)^2(w+1)
HELP! This one I expanded until I eventually got 3w^2 +16w +48 but it doesn't facatorise and I'm gonna cry because someone I know got: 3(w+4)^2 and I don't understand!

Last question:
y = 16 * 10^(8k) where k is an integer
find an expression, in terms of k, for y^(5/4)
You can't divide if it's an expression. Factorize. So the first one is
5(x+2)(x-1)

For 3l) write 24 as a product of factors 3 and 4. What is the 4 exponent?
Last edited by mqb2766; 1 month ago
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1 month ago
#3
(Original post by FloryK)
Hi I really really need help, I've got this maths homework due tomorrow and I'm not sure what to do on a few and if I've even done a few right could someone please please have a look! Thank you!!
I can't attach images so I'll type it out:

a)5x^2 + 5x - 10
My answer: I divided by 5, x^2 + x - 2 answer = (x+2)(x-1)
b)15x^2 - 72x - 15
My answer: Divided by 3, 5x^2 - 24 - 5 = (-5+x)(5x-1)

Solve: 3 * 4^(2k+8) = 24
IDK WHAT TO DO!

Factorise fully (w+4)^3 - (w+4)^2(w+1)
HELP! This one I expanded until I eventually got 3w^2 +16w +48 but it doesn't facatorise and I'm gonna cry because someone I know got: 3(w+4)^2 and I don't understand!

Last question:
y = 16 * 10^(8k) where k is an integer
find an expression, in terms of k, for y^(5/4)
q3 divide by 3, then take log base 4
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#4
(Original post by mqb2766)
You can't divide if it's an expression. Factorize. So the first one is
5(x+2)(x-1)
Okay thank you I thought that was the case but for some reason my maths teacher wrote a hint saying "you need to divide through by a common factor first) so I was really confused as I wasn't really sure what she meant. What about the rest?
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1 month ago
#5
(Original post by FloryK)
Okay thank you I thought that was the case but for some reason my maths teacher wrote a hint saying "you need to divide through by a common factor first) so I was really confused as I wasn't really sure what she meant. What about the rest?
You can't divide an expression (and it means the same). You can an equation. Factorize the common part out, don't divide.
edited previous post for next one.
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#6
(Original post by mqb2766)
You can't divide if it's an expression. Factorize. So the first one is
5(x+2)(x-1)

For 3l) write 24 as a product of factors 3 and 4. What is the 4 exponent?
I don't understand what you mean how do I write it as a product of factors 3 and 4? 4 * 3 * 2??
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#7
(Original post by asianism)
q3 divide by 3, then take log base 4
How do you take log base 4?
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1 month ago
#8
(Original post by FloryK)
I don't understand what you mean how do I write it as a product of factors 3 and 4? 4 * 3 * 2??
How are 2 and 4 related
2 = 4^?
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#9
(Original post by mqb2766)
How are 2 and 4 related
2 = 4^?
Oh uhh is it 4^1/2 ??
But then what do I do?
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1 month ago
#10
(Original post by FloryK)
Oh uhh is it 4^1/2 ??
Yes so 24 = 3*4^(3/2)
So what must
2k+8 = ...?
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#11
(Original post by mqb2766)
Yes so 24 = 3*4^(3/2)
So what must
2k+8 = ...?
oh um so would 2k+8 = 3/2 ?
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1 month ago
#12
(Original post by FloryK)
oh um so would 2k+8 = 3/2 ?
Yes, so solve...
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#13
(Original post by mqb2766)
Yes, so solve...
Thank you so much!
So would k = -3.25 ?
How would I do the third and fourth ones?
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1 month ago
#14
(Original post by FloryK)
Thank you so much!
So would k = -3.25 ?
How would I do the third and fourth ones?
For 3) there is common expression in each of the two terms? What is it, and factorize it out (don't divide).
Generally, never expand our then attempt to factorize from scratch.
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#15
(Original post by mqb2766)
For 3) there is common expression in each of the two terms? What is it, and factorize it out (don't divide).
I can see that (w+4)^2 is the common factor but how do you mean factorise it out?
0
1 month ago
#16
(Original post by FloryK)
I can see that (w+4)^2 is the common factor but how do you mean factorise it out?
Write the two terms as
(w+4)^2 * ( a + b)
Where a and b are what remains from the original two terms.
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#17
(Original post by mqb2766)
Write the two terms as
(w+4)^2 * ( a + b)
Where a and b are what remains from the original two terms.
ummm so what remains from the original two terms after I'd expanded and subtracted them? so 3w^2 + 16w +48?? I don't understand
0
1 month ago
#18
(Original post by FloryK)
ummm so what remains from the original two terms after I'd expanded and subtracted them? so 3w^2 + 16w +48?? I don't understand
For the first term
a(w+4)^2 = (w+4)^3
So a = ...

For the second
b(w+4)^2 = -(w+4)^2(w+1)
So b = ...
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#19
(Original post by mqb2766)
For the first term
a(w+4)^2 = (w+4)^3
So a = ...

For the second
b(w+4)^2 = -(w+4)^2(w+1)
So b = ...
um would a = (w+4) and b = (w+1) ???????
0
1 month ago
#20
(Original post by FloryK)
um would a = (w+4) and b = (w+1) ???????
Almost b = -(w+1).
That's what it means to factorize. Take out the common parts and combine what's left.
So what is the final answer?
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