Extraction and Grouping Common Factors help!

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December 4th 2010, 08:26 AM
zizou

Extraction and Grouping Common Factors help!

Hey guys, I am really confused with this work, I missed some time from college as I was ill and missed the work. Can anyone help me?

Question 1 says, use appropriate methods to completely factorise the following expression by extraction and grouping of common factors.

(A) 12a2b + 15ab2

(B) 10ab - 8ax + 15b - 12x

Question 2 says, solve a quadratic equation by factorising and one by the formula method.

a mojor leak is discovered on the surface of oil storage tank in the workshop. the supervisor decides that the best solution to this is to weld a reinforcement sheet of metal on the leaking area.
after taking measurements, the following conditions are decided.

- the are of rectangular sheet of metal should be 24000mm2
- the length of the sheet should be 10mm greater than the width

Calculate the dimensions of the reinforcement sheet.

What would be the dimensions if the area of the sheet was reduced wo 21987mm2?

Could someone please explain how to to this for me? Thank you!
December 4th 2010, 08:28 AM
zizou

its supposed to be 12a squared b + 15ab squared but the little 2 dont work for some reason
December 4th 2010, 09:18 AM
rtblue

For part A (i assume that is is what you intended to write):

$(12a)^2b+(15ab)^2$

$144a^2b+225a^2b^2$

$ba^2(144+225b)$ now we can take out a nine:

$9ba^2(16+25b)$ and I believe it cannot be simplified any further.

Part B:

$10ab-8ax+15b-12x$ now we group like terms together:

$10ab+15b-8ax-12x$ now we factor:

$b(10a+15)-x(8a+12)$ now more simplification:

$5b(2a+3)-4x(2a+3)$ now we can factor out a (2a+3):

$(5b-4x)(2a+3)$ and this is as simplified as the expression can get.
December 4th 2010, 09:25 AM
e^(i*pi)

Quote:

Originally Posted by zizou

Question 2 says, solve a quadratic equation by factorising and one by the formula method.

a mojor leak is discovered on the surface of oil storage tank in the workshop. the supervisor decides that the best solution to this is to weld a reinforcement sheet of metal on the leaking area.
after taking measurements, the following conditions are decided.
[/SIZE]
- the are of rectangular sheet of metal should be 24000mm2
- the length of the sheet should be 10mm greater than the width

Calculate the dimensions of the reinforcement sheet.

What would be the dimensions if the area of the sheet was reduced wo 21987mm2?

Could someone please explain how to to this for me? Thank you!

Use the caret (^) symbol to denote powers in plain text: a^2 = a squared. Better yet use LaTex

For Q2 use the fact that $A = lw$ , that $l = w+10$ and that $A = 2.4 \cdot 10^4$
December 4th 2010, 09:51 AM
zizou

im still not sure about question 2 :s
December 4th 2010, 10:11 AM
e^(i*pi)

Quote:

Originally Posted by e^(i*pi)

For Q2 use the fact that $A = lw$ , that $l = w+10$ and that $A = 2.4 \cdot 10^{4}$

Quote:

Originally Posted by zizou

im still not sure about question 2 :s

Substitute the second equation in for $l$ in the first equation:

$A = (w+10)w = 2.4 \cdot 10^4$

If you use the distributive property on $(w+10)w$ you get $w^2+10w$

Hence the equation can be written as $w^2+10w-(2.4 \cdot 10^4) = 0$

Solve using the quadratic formula and discard any negative solution since we don't deal with negative length in this problem
December 4th 2010, 05:36 PM
rtblue

I wanted to point out that the reuslt e^(i*pi) got can be factored as such:

$w^2+10w-24000=0$

$(w+160)(w-150)=0$

$w-150=0$

$w=150$

If the area were to change from 24000mm to 21987mm, we would use the fact that $Area=length*width$ and in this case, length=width+10:

$(w)(w+10)=21987$

$w^2+10w-21987=0$

I'm afraid factoring is not possible in this case, so using the quadratic formula we have:

$w=\frac{-10\pm\sqrt{88048}}{2}$

$w=\frac{-10\pm4\sqrt{5503}}{2}$

$w=-5+2\sqrt{5503}$

Note that I got rid of the negative solution, because width cannot have a negative value.
December 5th 2010, 07:03 AM
zizou

kinda get it. thanks ppl appreciate it