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###### Factorising quadratics when a isn't 1

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2 weeks ago

I got taught then when factorising a quadratic when a isn't 1 to just basically do trial and error and I wanted to know if there was a more efficient method than just trying a bunch of combinations.

Reply 1

2 weeks ago

Original post by RandomPancake

I got taught then when factorising a quadratic when a isn't 1 to just basically do trial and error and I wanted to know if there was a more efficient method than just trying a bunch of combinations.

What do you mean by trial and error?

Assuming roots of a quadratic are rational (meaning your leading coefficient is tyically not 1) then you can factorise it.

Let's say the roots are $x = -\dfrac{p}{q}$ and $x = - \dfrac{r}{s}$ and these correspond to linear factors $(qx+p)$ and $(sx+r)$ being zero.

Therefore, the two roots satisfy the quadratic equation $(qx+p)(sx+r)=0$, i.e. $qs x^2 + (qr+ps) x + pr = 0$

The leading coefficient is $qs$.

Here $qr,ps$ are two magic numbers which add to give the middle coefficient, but they also multiply to give $qrps$ which is exactly the same as the leading coefficient $qs$ multiplying the constant term $pr$.

All this is to say that there some logic involved here in factorising. When trying to factorise $ax^2 + bx + c = 0$ you should be splitting your $b$ into two magic numbers which add to $b$ yet multiply to $ac$.

As they multiply to $ac$, you just need to look at the factor pairs of this product and see what adds to $b$.

Example: Factorise $2x^2 + 5x + 2$.

Look for two numbers which add to $5$ and multiply to $2\times 2 = 4$.

These are $1,4$. Then rewrite your quadratic as $(2x^2 + x) + (4x+2)$ [we have split $5x=x+4x$] and factorise fully each bracket to get

$x(2x+1) + 2(2x+1)$

Finally note that $(2x+1)$ is a bracket occuring twice in this expression therefore it can be factored out overall to give

$(2x+1)(x+2)$

This is the factorisation. No trial and error needed. Just find the two magic numbers by looking at factor pairs of $ac$ which add to $b$.

(edited 2 weeks ago)

Reply 2

2 weeks ago

Original post by RDKGames

What do you mean by trial and error?

Assuming roots of a quadratic are rational (meaning your leading coefficient is tyically not 1) then you can factorise it.

Let's say the roots are $x = -\dfrac{p}{q}$ and $x = - \dfrac{r}{s}$ and these correspond to linear factors $(qx+p)$ and $(sx+r)$ being zero.

Therefore, the two roots satisfy the quadratic equation $(qx+p)(sx+r)=0$, i.e. $qs x^2 + (qr+ps) x + pr = 0$

The leading coefficient is $qs$.

Here $qr,ps$ are two magic numbers which add to give the middle coefficient, but they also multiply to give $qrps$ which is exactly the same as the leading coefficient $qs$ multiplying the constant term $pr$.

All this is to say that there some logic involved here in factorising. When trying to factorise $ax^2 + bx + c = 0$ you should be splitting your $b$ into two magic numbers which add to $b$ yet multiply to $ac$.

As they multiply to $ac$, you just need to look at the factor pairs of this product and see what adds to $b$.

Example: Factorise $2x^2 + 5x + 2$.

Look for two numbers which add to $5$ and multiply to $2\times 2 = 4$.

These are $1,4$. Then rewrite your quadratic as $(2x^2 + x) + (4x+2)$ [we have split $5x=x+4x$] and factorise fully each bracket to get

$x(2x+1) + 2(2x+1)$

Finally note that $(2x+1)$ is a bracket occuring twice in this expression therefore it can be factored out overall to give

$(2x+1)(x+2)$

This is the factorisation. No trial and error needed. Just find the two magic numbers by looking at factor pairs of $ac$ which add to $b$.

Assuming roots of a quadratic are rational (meaning your leading coefficient is tyically not 1) then you can factorise it.

Let's say the roots are $x = -\dfrac{p}{q}$ and $x = - \dfrac{r}{s}$ and these correspond to linear factors $(qx+p)$ and $(sx+r)$ being zero.

Therefore, the two roots satisfy the quadratic equation $(qx+p)(sx+r)=0$, i.e. $qs x^2 + (qr+ps) x + pr = 0$

The leading coefficient is $qs$.

Here $qr,ps$ are two magic numbers which add to give the middle coefficient, but they also multiply to give $qrps$ which is exactly the same as the leading coefficient $qs$ multiplying the constant term $pr$.

All this is to say that there some logic involved here in factorising. When trying to factorise $ax^2 + bx + c = 0$ you should be splitting your $b$ into two magic numbers which add to $b$ yet multiply to $ac$.

As they multiply to $ac$, you just need to look at the factor pairs of this product and see what adds to $b$.

Example: Factorise $2x^2 + 5x + 2$.

Look for two numbers which add to $5$ and multiply to $2\times 2 = 4$.

These are $1,4$. Then rewrite your quadratic as $(2x^2 + x) + (4x+2)$ [we have split $5x=x+4x$] and factorise fully each bracket to get

$x(2x+1) + 2(2x+1)$

Finally note that $(2x+1)$ is a bracket occuring twice in this expression therefore it can be factored out overall to give

$(2x+1)(x+2)$

This is the factorisation. No trial and error needed. Just find the two magic numbers by looking at factor pairs of $ac$ which add to $b$.

Thanks for the explanation!

We got taught to find factors of a and put them into the brackets first and then fill in the rest of the bracket with factors of c and do trial and error until the x values add up to b.

Reply 3

2 weeks ago

Original post by RandomPancake

Thanks for the explanation!

We got taught to find factors of a and put them into the brackets first and then fill in the rest of the bracket with factors of c and do trial and error until the x values add up to b.

We got taught to find factors of a and put them into the brackets first and then fill in the rest of the bracket with factors of c and do trial and error until the x values add up to b.

OK, I see.

That works too, but this above method you might prefer it. Significantly fewer options to check when looking at factors of ac in one go rather than choosing a factor pair of a then looking at factors of c. This is great because it always works, although you might find the trial and error to be quicker if there are not many factors to consider.

Try both methods on this one example:

$12x^2 + 7x + 12$

to get $(3x+4)(4x-3)$.

See this video to help you with this method (called method 2 in the video):

https://www.youtube.com/watch?v=cbbEFYpeWV8

(edited 2 weeks ago)

Reply 4

2 weeks ago

Original post by RDKGames

OK, I see.

That works too, but this the above method you might prefer it. Significantly fewer options to check when looking at factors of ac in one go rather than choosing a factor pair of a then looking at factors of c.

Try both methods on this one example:

$12x^2 + 7x + 12$

to get $(3x+4)(4x-3)$.

That works too, but this the above method you might prefer it. Significantly fewer options to check when looking at factors of ac in one go rather than choosing a factor pair of a then looking at factors of c.

Try both methods on this one example:

$12x^2 + 7x + 12$

to get $(3x+4)(4x-3)$.

The method you suggested definitely makes it a lot less time consuming and also makes cases where a and c have a ridiculous amount of factors (eg 12) a lot easier to work with. Using the method I got taught to factorise it took a lot longer than the method you showed and also took a lot more working out. Thanks once again!

You can actually reduce any factorisation problem (quadratic) to factorising a quadratic with leading term 1 - meaning the coefficient of $x^2$ is 1.

Say you have $ax^2+bx+c$. Then multiply by $a$ which gives $a^2x^2+abx+ac$ which is also $(ax)^2+b(ax)+ac$.

Now substitute $y=ax$ which gives us $y^2+by+ac$. This has coefficient 1 and so just factorise it as you know how. For the final step just divide by an and this will cancel with the factors in some way.

EXAMPLE

$2x^2-3x-2$.

The value of $a$ is 2 so we multiply by 2 giving $4x^2-6x-4$. Now we set $y=2x$ and we have $y^2-3y-4$.

You factorise this easily as $(y-4)(y+1)=(2x-4)(2x+1)$.

Therefore we have $4x^2-6x-4 =(2x-4)(2x+1)$ but we can’t to factorise half of this, namely $2x^2-3x-2$ and so we just divide by 2 giving

$2x^2-3x-2=(x-2)(2x+1)$.

This method is not necessarily the most efficient, but all I’m showing is that actually it always can be boiled down to something you can easily factorise.

Say you have $ax^2+bx+c$. Then multiply by $a$ which gives $a^2x^2+abx+ac$ which is also $(ax)^2+b(ax)+ac$.

Now substitute $y=ax$ which gives us $y^2+by+ac$. This has coefficient 1 and so just factorise it as you know how. For the final step just divide by an and this will cancel with the factors in some way.

EXAMPLE

$2x^2-3x-2$.

The value of $a$ is 2 so we multiply by 2 giving $4x^2-6x-4$. Now we set $y=2x$ and we have $y^2-3y-4$.

You factorise this easily as $(y-4)(y+1)=(2x-4)(2x+1)$.

Therefore we have $4x^2-6x-4 =(2x-4)(2x+1)$ but we can’t to factorise half of this, namely $2x^2-3x-2$ and so we just divide by 2 giving

$2x^2-3x-2=(x-2)(2x+1)$.

This method is not necessarily the most efficient, but all I’m showing is that actually it always can be boiled down to something you can easily factorise.

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