# Algebraic Fractions

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#1
Simplify

Why does multiplying the fraction by 6 to get rid of the fractions in the denominator give the answer 6(x-1) and not (x-1).

Am I right to say I'm still correct by obtaining (x-1) as the answer since I multiplied each term in the numerator by 6 and then factorised the quadratic?
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7 years ago
#2
(Original post by Noah~)
Simplify

Why does multiplying the fraction by 6 to get rid of the fractions in the denominator give the answer 6(x-1) and not (x-1).

Am I right to say I'm still correct by obtaining (x-1) as the answer since I multiplied each term in the numerator by 6 and then factorised the quadratic?
right?

How did you factorize?

Edit: No, you're wrong. You forgot a 6 in your second bracket. (3x+2)(x-1) clearly does not give 18x^2.

Posted from TSR Mobile
0
7 years ago
#3
(Original post by Noah~)
Simplify

Why does multiplying the fraction by 6 to get rid of the fractions in the denominator give the answer 6(x-1) and not (x-1).

Am I right to say I'm still correct by obtaining (x-1) as the answer since I multiplied each term in the numerator by 6 and then factorised the quadratic?
18x^2 - 6x - 12 if you factorise it gives you 6(3x+2)(x-1) because if you expand (3x+2)(x-1) you get 3x^2 -x -2 so you just factorised it wrong
0
7 years ago
#4
(Original post by Noah~)

Am I right to say I'm still correct by obtaining (x-1) as the answer since I multiplied each term in the numerator by 6 and then factorised the quadratic?
As the others have said

You have not factorised the new quadratic
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#5
(Original post by Iowan)
18x^2 - 6x - 12 if you factorise it gives you 6(3x+2)(x-1) because if you expand (3x+2)(x-1) you get 3x^2 -x -2 so you just factorised it wrong
(Original post by TenOfThem)
As the others have said

You have not factorised the new quadratic
(Original post by majmuh24)
right?

How did you factorize?

Edit: No, you're wrong. You forgot a 6 in your second bracket. (3x+2)(x-1) clearly does not give 18x^2.

Posted from TSR Mobile
Factorising the new quadratic using the quadratic formula gives (when expanded this does not give the quadratic term factorised back). Since the quadratic formula assumes , whereas, in our case the term is not equal to 0 is the reason for me being incorrect (as I used the quadratic formula to factorise).

Posting this so you can confirm if I have the correct intuition behind this.
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7 years ago
#6
(Original post by Noah~)
Factorising the new quadratic using the quadratic formula gives (when expanded this does not give the quadratic term factorised back). Does the quadratic formula assume ? Whereas, in our case the term is not equal to 0.
No it doesn't. You can't just divide through unless it's an equation and equal to something, you have to take a constant outside of the brackets.

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0
7 years ago
#7
(Original post by Noah~)
Factorising the new quadratic using the quadratic formula gives (when expanded this does not give the quadratic term factorised back). Does the quadratic formula assume ? Whereas, in our case the term is not equal to 0.
You seem very confused

The quadratic formula gives solutions - it does not factorise

I am sure that you know that

And that, in fact

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#8
(Original post by TenOfThem)
You seem very confused

The quadratic formula gives solutions - it does not factorise

I am sure that you know that

And that, in fact

I see where I went wrong and that was using the quadratic formula to factorise (like you said it does not factorise but give solutions).

Hence

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