2u unit maths help,basic algebra (sort of tricky questions) (1 Viewer)

[Riot09]

Basically,im doing the excersices from the fist chapter of my 2u excell book for this year an theirs are few questions i'm stuck on.(working out and solutions are provided at the back but-these are tricky questions that i can't get.So please explain whats going on here step wise.Thnx in advance.

All the solutions are already here but i don't get some of the steps involved to them as these are the "difficult" questions toward the end.so please provide an explanation.Thnx in advance.

Note: ------ ^2= squared. e.g,8m^8=8m squared-----------

32.c simplify

8m^8n^6/12m^5n^9 =2m^3/3n^3

50.e factorise

a^3 b^7-a^4b^6=a^3b^6(b-a)

52.c factorise

m^2-3m+2mn-6n=m(m-3)+2n(m-3)
=(m-3)(m+2n)
56.c factorise

8-125k^3=2^3-125^3
=(2-5k)(4+10k+25k^2)

56.d factorise

64p^3+34q^3=(4p+7q)(16p^2-28pq+49q^2)

58.d factorise fully

m^4-16=(m^2-4)(m^2+4)
=(m-2)9m+2) (m^2+4)

58.e factorise fully

k^2-9+7km-21m=(k-3)(k+3)+7m(k-3)
=(k-3)(k+3+7m)


58.f factorise fully

pqr+3pr-5pq-15p=p(qr+3r-5q-15)
=p[r(q+3)-5(q+3)]
=p(q+3)(r-5)

 

[Arcorn]

Why are you using the power to symbol for multiplication?

 

[Aerath]

I think the power button was to symbolise power...like k^2 is k squared.

 

[lpodnano]

sorry i can't help, not using the power sign and stuff really confuses me.

 

[mioumiou]

Try the maths forum! They have these mathematics symbols that you can click on. It's much easier and way more readable. I can't help cuz the power signs are a bit confusing.

 

[Riot09]

Why are you using the power to symbol for multiplication?

what do you mean?

 

[Riot09]

I think the power button was to symbolise power...like k^2 is k squared.

yep,isn't thats how ur supposed 2 do it?

 

[Riot09]

I added the question verb,if that helps.e.g factorise and simplify.

 

[ninetypercent]

look mate, you should've done these in junior years, but its nice to see you studying

32.c simplify

8m^8n^6/12m^5n^9 =2m^3/3n^3

simplify it. 8/12 = 2/3
and then bring the powers down m^8 divided by m^5 is m^3
n^6 divided by n^3 is n^3

50.e factorise

a^3 b^7-a^4b^6=a^3b^6(b-a)

a^3 x b^6 are common factors. take them out. and then you are left with b -a

52.c factorise

m^2-3m+2mn-6n=m(m-3)+2n(m-3)
=(m-3)(m+2n)

group in pairs. and then factorise by taking out (m-3)

56.c factorise

8-125k^3=2^3-125^3
=(2-5k)(4+10k+25k^2)

(a^3 - b^3) = (a-b)(a^2 + ab + b^2)
use this, with a being 2 and b being 5k

56.d factorise

64p^3+34q^3=(4p+7q)(16p^2-28pq+49q^2)

same with question above

58.d factorise fully

m^4-16=(m^2-4)(m^2+4)
=(m-2)9m+2) (m^2+4)

difference of two squares in first line. difference of two squares again for m^2 -4
and the last line should say (m-2)(m+2)(m^2 + 4)

58.e factorise fully

k^2-9+7km-21m=(k-3)(k+3)+7m(k-3)
=(k-3)(k+3+7m)

group in two pairs, then take out (k-3) as common factor

58.f factorise fully

pqr+3pr-5pq-15p=p(qr+3r-5q-15)
=p[r(q+3)-5(q+3)]
=p(q+3)(r-5)

take out p as common factor. group in pairs for the inside, then factorise by taking out (q+3)

 

[Riot09]

thnx,ninety

 

[adriano_negr0]

yer ninety's right, completely

 

[x_cp3]

lol i can't do anything im sorry.. i dun get jack shit of the thing and confusing me lol

 

[xMaFF]

58.f factorise fully

pqr+3pr-5pq-15p=p(qr+3r-5q-15)
=p[r(q+3)-5(q+3)]
=p(q+3)(r-5)

This one caught my eye.

pqr+3pr-5pq-15p = p(qr+3r-5q-15)

pr(q+3)-5p(q+3) = p[r(q+3)-5(q+3)]

(pr-5p)(q+3) = p(r-5)(q+3)


Aaand, that's about as far as I can go, haha. T.T
I hope this helps in some way.

If in any procedure I am incorrect, (I was never known to be mathematically-inclined) please do correct me.

Is the answer just that, or..?

 

[ohexploitable]

Use LaTeX.

 

[ohexploitable]

But at least replace the ^ with *.

 

[AAEldar]

This one caught my eye.

pqr+3pr-5pq-15p = p(qr+3r-5q-15)

pr(q+3)-5p(q+3) = p[r(q+3)-5(q+3)]

(pr-5p)(q+3) = p(r-5)(q+3)


Aaand, that's about as far as I can go, haha. T.T
I hope this helps in some way.

If in any procedure I am incorrect, (I was never known to be mathematically-inclined) please do correct me.

Is the answer just that, or..?

Judging by the way that that question was set out, there isn't a LHS=RHS, it's just the one equation. It should be:

pqr+3pr-5pq-15p
=p(qr+3r-5q-15)
=p[r(q+3)-5(q+3)]
=p(r-5)(q+3)

As you can see from your working, you could still simplify the LHS and get the same answer as the RHS Unless you did know all this, in which case disregard my post.