really simple question but no matter what i do i can't get it! please help

a) Factorise 2n^2 + 5n +3
okay this part i was fine with and i got (n+1) (2n+3)

BUT then it said:
b) hence, or otherwise (im assuming this means to use your answer from part a) write 253 as the product of two prime factors

i am completely stumped.
thanks

Reply 1

username3154254

16

I assume you just use a prime factor tree, but yh from the wording of the question it says hence or otherwise so it has to be related to the first question. I have no idea to be honest, hoping someone else replies cause i wanna know the answer :P

Reply 2

S.H.Rahman

15

Good start to part (a) and well done on acknowledging that you had to use part a to get your answer to part b!

A method you could have used is trial and error through putting in different 'n' values until you get an answer that is 253. It works out that when you put 'n' in as 10 it produces 253 as 11 and 23 are its two prime factors.

Alternatively, the way I knew straight away that 11 was a factor of 253 is because of the 2 outer numbers, 2 and 3, add to make the inner number, 5. Numbers that have this are always divisible by 11.

Hence I looked out how one of the brackets could be equivalent to 11 and saw when n=10 it would produce 253 hence, the answer would be 11 x 23.

The trial and error method would probably be the best method I can think for the majority for this question as a safe bet.

Hope this helped.

(edited 6 years ago)

Reply 3

Shaanv

17

Original post by razzy02

a) Factorise 2n^2 + 5n +3
okay this part i was fine with and i got (n+1) (2n+3)

BUT then it said:
b) hence, or otherwise (im assuming this means to use your answer from part a) write 253 as the product of two prime factors

i am completely stumped.
thanks

Via inspection i can tell u that n=10.

As 2n^2+5n=250.

U can sub this back into ur factorised version of the quadratic and get ur answer.

Have u not tried solving the quadratic equal to 253, i haven’t tried it because i don’t have time but i think it should give u n=10

Reply 4

S.H.Rahman

15

Original post by Shaanv

Via inspection i can tell u that n=10.

As 2n^2+5n=250.

U can sub this back into ur factorised version of the quadratic and get ur answer.

Have u not tried solving the quadratic equal to 253, i haven’t tried it because i don’t have time but i think it should give u n=10

Exactly what I thought initially but when I tried factorizing it wasn't working out using the method I was using. So after just went for what I said in my post above.

Edit: Guess it would have worked out if I used the quadratic equation but don't have a calculator around D:

(edited 6 years ago)

Reply 5

Kalo7854

15

I would have just used a prime factor tree, and see what it divides by, as I'm a little confused why people are saying you would have to use part (a), as surely when it says 'hence, or otherwise' the 'or otherwise' implies you could do it a different way? (Honest question)
The way I would loop it back in to part (a) is just by checking it by subbing it in or something of the sort.

Reply 6

username3555092

14

Notice that

253=200+50+3=2 \times 10^2 + 5 \times 10 + 3

.

Now look back at your expression:

2n^2+5n+3

. Notice a similarity? Once you find the right value of

n

, then you can sub that into your factorisation to get your needed prime factors.

Reply 7

username3555092

14

Original post by Kalo7854

x

You are correct, using part a is optional. However in this case, unless you're familiar with a trick for spotting certain multiples, it might be easier to use the result from part a.

Reply 8

JustAPhilosopher

3

I would do it in a factor tree but you there s a way to do it where you sub back into the equation however I think that the simplest is just a factor tree.

Reply 9

S.H.Rahman

15

Original post by Kalo7854

I would have just used a prime factor tree, and see what it divides by, as I'm a little confused why people are saying you would have to use part (a), as surely when it says 'hence, or otherwise' the 'or otherwise' implies you could do it a different way? (Honest question)
The way I would loop it back in to part (a) is just by checking it by subbing it in or something of the sort.

Whenever it says 'hence or otherwise' you always take it as hence and not otherwise. It's just a known kinda rule to follow in exams that exam boards are weird about. 100% of questions using this phrase that I've seen or done always require you to use 'hence'. My teacher likes to say the 'otherwise' is only there for fools :laugh:

Reply 10

Shaanv

17

Original post by S.H.Rahman

Whenever it says 'hence or otherwise' you always take it as hence and not otherwise. It's just a known kinda rule to follow in exams that exam boards are weird about. 100% of questions using this phrase that I've seen or done always require you to use 'hence'. My teacher likes to say the 'otherwise' is only there for fools :laugh:

Really, u wouldnt lose marks if you did otherwise, surely.

Especially if the question says hence or otherwise, i understand that u want to use ur previous answer to answer question but surely u don’t get penalised for using otherwise.

Happy new year btw

Reply 11

Notnek

21

Original post by Shaanv

Really, u wouldnt lose marks if you did otherwise, surely.

Especially if the question says hence or otherwise, i understand that u want to use ur previous answer to answer question but surely u don’t get penalised for using otherwise.

Happy new year btw

No you don't lose marks as long as your method is correct. It's usually best to ignore the "otherwise" at least initially because the quickest method will be to use the previous part of the question.

Although there have been one or two occasions where "otherwise" is an easier (but slower) method for most students.

really simple question but no matter what i do i can't get it! please help

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