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In class my teacher showed us an alternative method for factorising quadratics which are more awkward (i.e. the $a$ in $ax^2+bx+c$ is greater than 1).

The method is:
1. Take your quadratic (e.g. $3x^2+12x-15=0$)
2.Taking the product of the last and first term, and the middle term (here $-45$ and $12$)
3. Find the numbers which multiply to $-45$ and add to $12$ (here $15$ and $-3$)
4. Use these numbers to form the equation $15x^2-3x+15x-3$
5. Factorise this to: $3x(5x-1)+3(5x-1)=0$
6. Finally rearrange this to: $(5x-1)(3x+3)=0$
7. This gives a result of $x$ as $\frac15$ and $-1$

How ever this is not write as if you plug the numbers into the original equation you don't get $0$. Where have I gone wrong (appologies if the answer is fairly obvious). I have done this method with all the questions and only the equation $3x^2-6x=0$ worked using this method.

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In your step 4, it should be $3x^2-3x+15x-15$. –  André Nicolas Dec 5 '13 at 17:28
    
There's something wrong in the 4th step. It should be: $3x^2+15x-3x-15$ –  Shubham Dec 5 '13 at 17:28
    
Maybe, I'm not sure - but the worked examples in class did it like that? Is that what I'm doing wrong? –  mathstudent Dec 5 '13 at 17:29

1 Answer 1

up vote 2 down vote accepted

In your $4$th step, you changed the coefficient of the $x^2$ term, and you changed the constant term of the original equation, thus changing the equation altogether. What you should have written is: $$3x^2\underbrace{\color{blue}{\bf -3}x+\color{blue}{\bf 15}x}_{\large 12x}-15 = 0$$

Note that doing it this way, since $-3x + 15x = 12x$, we are still working with the original equation, but have written it in a way that will facilitate factoring.

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Thank you - it solves it perfectly! –  mathstudent Dec 5 '13 at 17:39
    
You're welcome, mathstudent! –  amWhy Dec 5 '13 at 17:40
    
@amWhy: Needs a TU! Who thought this new UI looked good? It looks so hideous with the black banner at the top. –  Amzoti Dec 6 '13 at 2:22

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