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Algebraic Fractions

Simplify

\dfrac{3x^2 - x - 2}{x/2 + 1/3}

\dfrac{18x^2 - 6x - 12}{3x + 2}

\dfrac{(3x+2)(x-1)}{(3x + 2)}

{(x-1)}

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?

(edited 10 years ago)

Reply 1

interstitial

Original post by Noah~

Simplify

\dfrac{3x^2 - x - 2}{x/2 + 1/3}

\dfrac{18x^2-6x-2}{3x+2}

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.

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(edited 10 years ago)

Reply 2

Iowan

Original post by Noah~

Simplify

\dfrac{3x^2 - x - 2}{x/2 + 1/3}

\dfrac{18x^2 - 6x - 12}{3x + 2}

\dfrac{(3x+2)(x-1)}{(3x + 2)}

{(x-1)}

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

Reply 3

TenOfThem

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

Reply 4

Noah~

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

\dfrac{18x^2-6x-2}{3x+2}

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

18x^2 - 6x - 12

using the quadratic formula gives

(3x+2)(x-1)

(when expanded this does not give the quadratic term factorised back). Since the quadratic formula assumes

Ax^2 + Bx + C = 0

, 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.

(edited 10 years ago)

Reply 5

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Original post by Noah~

Factorising the new quadratic

18x^2 - 6x - 12

using the quadratic formula gives

(3x+2)(x-1)

(when expanded this does not give the quadratic term factorised back). Does the quadratic formula assume

Ax^2 + Bx + C = 0

? 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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Reply 6

TenOfThem

Original post by Noah~

Factorising the new quadratic

18x^2 - 6x - 12

using the quadratic formula gives

(3x+2)(x-1)

(when expanded this does not give the quadratic term factorised back). Does the quadratic formula assume

Ax^2 + Bx + C = 0

? 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

18x^2 - 6x - 12 \not= (3x+2)(x-1)

And that, in fact

18x^2 - 6x - 12 = 6(3x^2 - x - 2) = 6(3x+2)(x-1)

Reply 7

Noah~

Original post by TenOfThem

You seem very confused

The quadratic formula gives solutions - it does not factorise

I am sure that you know that

18x^2 - 6x - 12 \not= (3x+2)(x-1)

And that, in fact

18x^2 - 6x - 12 = 6(3x^2 - x - 2) = 6(3x+2)(x-1)

I already know that.

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).