The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Factorising quadractics 'harder problems'

I was refreshing my knowledge of GCSE Maths by factorising quadractics. I did 6 problems and then checked if my answers are right. I only got the last 5 problems right. I can't work my head around to getting the answer of the first problem (sad I know :frown:).

The problem is: 3x^2+5x+2.

My 2 embarrassing attempts:

Spoiler

Reply 1
Avatar for MartyO
MartyO
7
When factoring ax2+bx+cax^2+bx+c, you need to find two numbers whose sum equals bxbx and whose product equals acx2acx^2.
In your attempts, you found the numbers 6x,−x6x,-x, but their product does not equal 6x26x^2.
(edited 7 years ago)
It's a pain thinking of numbers to put in the brackets; I find it much easier to just complete the square or even use the quadratic formula.

You'll get x=−5±16x = \frac{-5 \pm 1}{6}



Giving you (x+1)(x+23)(x+1)(x+\frac{2}{3}) or (3x+2)(3x+2)
(edited 7 years ago)
The answer to that problem is: (3x+2) (x+1)

I don't get how I got the other 5 right, when I followed the same method (used on MathsWatch, since my teacher wasn't helpful). I'll show you how I did one other problem and got it right.

IMG_0365.JPG
Reply 4
Avatar for Zacken
Zacken
22
Original post by 34908seikj
It's a pain thinking of numbers to put in the brackets; I find it much easier to just complete the square or even use the quadratic formula.

Giving you (x+1)(x+23)(x+1)(x+\frac{2}{3})


And this is why simply using the formula without thought goes wrong. This is incorrect.
I just worked it backwards with the answer and see where I went wrong. Apparently, I should have used 2 and 3, instead of 6 and -1.

Since 6 times -1 equals -6, when multiplying the coefficient of x^2 which is 3 times the constant of 2 equals 6...

Rookie mistake by me. :banghead:
Original post by 34908seikj
It's a pain thinking of numbers to put in the brackets; I find it much easier to just complete the square or even use the quadratic formula.

You'll get x=−5±16x = \frac{-5 \pm 1}{6}



Giving you (x+1)(x+23)(x+1)(x+\frac{2}{3}) or (3x+2)(3x+2)

Erm multiply those 2 expressions together and see what you get...
Reply 7
Avatar for Helg
Helg
8
Original post by ColossalAtom
The answer to that problem is: (3x+2) (x+1)

I don't get how I got the other 5 right, when I followed the same method (used on MathsWatch, since my teacher wasn't helpful). I'll show you how I did one other problem and got it right.

IMG_0365.JPG


Awesome handwriting
Original post by Helg
Awesome handwriting


I have to take high dosage of immunosuppressants, so having tremors is to be expected. :frown:
Reply 9
Avatar for Helg
Helg
8
Original post by ColossalAtom
I have to take high dosage of immunosuppressants, so having tremors is to be expected. :frown:


Oh didn't know that, my cousin has the same. I personally have dyslexia which is a big obstacle for math [especially with algorithms] so I was allowed extra time in exams.
Original post by ColossalAtom
I was refreshing my knowledge of GCSE Maths by factorising quadractics. I did 6 problems and then checked if my answers are right. I only got the last 5 problems right. I can't work my head around to getting the answer of the first problem (sad I know :frown:).

The problem is: 3x^2+5x+2.

My 2 embarrassing attempts:


I just use my calculator's quadratic solver and then turn them into the factorised form because I'm too lazy to solve them by hand :^)
Original post by 34908seikj
It's a pain thinking of numbers to put in the brackets; I find it much easier to just complete the square or even use the quadratic formula.

You'll get x=−5±16x = \frac{-5 \pm 1}{6}



Giving you (x+1)(x+23)(x+1)(x+\frac{2}{3}) or (3x+2)(3x+2)


nah.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you