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Highest Common Factor

Highest Common Factor or -12 and 6 ? is it 6 or -6
Original post by Sonile
Highest Common Factor or -12 and 6 ? is it 6 or -6


Well which ones a bigger number? 6 or -6?
Original post by Sonile
Highest Common Factor or -12 and 6 ? is it 6 or -6


It would be the largest of the two numbers.


Posted from TSR Mobile
Reply 3
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Sonile
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Original post by RDKGames
Well which ones a bigger number? 6 or -6?


well i was factorising a very long equation, and they had -6 on the outside instead of 6.
Check this out cp4.PNG
Reply 4
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Sonile
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My final answer was:
Capture5.PNG
Original post by Sonile
well i was factorising a very long equation, and they had -6 on the outside instead of 6.
Check this out cp4.PNG


The notation is ambiguous. Do you mean (4x+1)3[2(3x+2)⋅3]−(3x+2)2[3(4x+1)2⋅4](4x+1)6\displaystyle \frac{(4x+1)^3[2(3x+2)\cdot 3]-(3x+2)^2[3(4x+1)^2\cdot 4]}{(4x+1)^6}???
Reply 6
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Sonile
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Book Answer:

-6(3x+2)(2x+3)/ (4x+1)^4
Reply 7
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Sonile
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Original post by RDKGames
The notation is ambiguous. Do you mean (4x+1)3[2(3x+2)⋅3]−(3x+2)2[3(4x+1)2⋅4](4x+1)6\displaystyle \frac{(4x+1)^3[2(3x+2)\cdot 3]-(3x+2)^2[3(4x+1)^2\cdot 4]}{(4x+1)^6}???


Yes how did you write out the equation on TSR?
Original post by Sonile
Yes how did you write out the equation on TSR?


OK.

Well I'm not sure why you're focusing on the HCF specifically.

You get (3x+2)(−12x−18)(4x+1)4\displaystyle \frac{(3x+2)(-12x-18)}{(4x+1)^4} from simplifying then you'd want to get rid of the minuses in (−12x−18)(-12x-18) so you'd factor out −1-1. And then you can factor out 66 out of (12x+18)(12x+18) so you'd have −6-6 on the outside

Thus −6(3x+2)(2x+3)(4x+1)4\displaystyle \frac{-6(3x+2)(2x+3)}{(4x+1)^4}


To write like that on TSR, use http://www.thestudentroom.co.uk/wiki/LaTex
Reply 9
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Sonile
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Original post by RDKGames
OK.

Well I'm not sure why you're focusing on the HCF specifically.

You get (3x+2)(−12x−18)(4x+1)4\displaystyle \frac{(3x+2)(-12x-18)}{(4x+1)^4} from simplifying then you'd want to get rid of the minuses in (−12x−18)(-12x-18) so you'd factor out −1-1. And then you can factor out 66 out of (12x+18)(12x+18) so you'd have −6-6 on the outside

Thus −6(3x+2)(2x+3)(4x+1)4\displaystyle \frac{-6(3x+2)(2x+3)}{(4x+1)^4}

x3(x−3)4x^3\frac(x-3)^4



To write like that on TSR, use http://www.thestudentroom.co.uk/wiki/LaTex


Thnks,

Testing
x3(x−3)4x^3\frac (x-3)^4
(edited 6 years ago)
Reply 10
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alow
19
Original post by Sonile
Thnks,

Testing
x3(x−3)4x^3\frac (x-3)^4


To use a \frac you put the numerator and denominator in subsequent arguments, surrounded by braces:

7x−45y−3\frac{7x-4}{5y-3} = \frac{7x-4}{5y-3}

Also I'd recommend using \dfrac for a taller fraction when not writing equations inline:

7x−45y−3\dfrac{7x-4}{5y-3}
Reply 11
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Sonile
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Original post by alow
To use a \frac you put the numerator and denominator in subsequent arguments, surrounded by braces:

7x−45y−3\frac{7x-4}{5y-3} = \frac{7x-4}{5y-3}

Also I'd recommend using \dfrac for a taller fraction when not writing equations inline:

7x−45y−3\dfrac{7x-4}{5y-3}


x3(x−8)2\dfrac{x^3}{(x-8)^2}
(edited 6 years ago)

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