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How would it be possible to factorize quadratics mentally, for example the following one? $$2x^2+7x+3$$ Maybe even something like $$3x^2+22x+24$$

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    $\begingroup$ Some methods here, I am not sure if anyone of these satisfies your criterion "mental" though. $\endgroup$ – user99914 Mar 6 '16 at 8:28
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    $\begingroup$ well for the first, the usual methods is easy to compute mentally. $\endgroup$ – Fabrice NEYRET Mar 6 '16 at 8:37
  • $\begingroup$ By mentally, I don't expect that you mean without using pencil and paper. Because it's not possible for every expression or every person. $\endgroup$ – Sufyan Naeem Mar 6 '16 at 13:02
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I use the "cross" method I was first taught when I was 13 or so, and I can do these quite easily in my head.

Consider the quadratic expression $Ax^2 + Bx + C$.

Visualise a big "X" with numbers at each corner (pole) - omit all the "$x$" terms. Left top pole = $a$, right top pole = $b$, left bottom pole = $c$, right bottom pole = $d$.

The rules are that:

1) on the left, the top and bottom poles will multiply to give you the coefficient of $x^2$ (i.e. $ac = A$) and on the right, the top and bottom will multiply to give you the constant term (i.e. $bd = C$).

2) when you "cross multiply" across the "X", i.e. top left with bottom right and top right with bottom left, the sum of those products will give you the coefficient of the $x$ term (i.e. $ad + bc = B$).

If you can achieve that, then the factorisation will be given by $Ax^2 + Bx + C = (ax + b)(cx + d)$.

It sounds much harder than it looks when you draw it out. Let's try out your examples.

For the first, the only possible values for the left poles are $\pm 2$ and $\pm 1$ and for the right are $\pm 3$ and $\pm 1$. It should be obvious that you only have to consider the positive values, and that you need $(2)(3) + (1)(1)$ to give $7$, so the factorisation is $(2x+1)(x+3)$.

For the second there are a few more possibilities to consider, but it should be quite easy to see that you need $(3)(6) + (4)(1) = 22$, so the factorisation is $(3x + 4)(x + 6)$.

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As $2\cdot3=6\cdot1$ and $6+1=7,$

$$2x^2+7x+3=2x^2+x+6x+3=2x(\underbrace{x+3})+1\cdot(\underbrace{x+3})=?$$

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  • $\begingroup$ I think the OP is asking for mental process behind, not the particular quadratics. $\endgroup$ – user99914 Mar 6 '16 at 8:23
  • $\begingroup$ @JohnMa that's right, the answer is good but it's a bit involved, whereas I was looking for something that's possible mentally $\endgroup$ – Skeleton Bow Mar 6 '16 at 8:25
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I would do it this way -

at $x=0$, the value is $3$

For positive $x$, The equation is positive.

So, try $x=-1$ , $x=-2$ etc... If you are far away from $0$, jump larger.

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  • $\begingroup$ It seems that your method only work when the factorization are of the form $(x - a)(bx+c)$ for $a, b, c\in \mathbb Z$. $\endgroup$ – user99914 Mar 6 '16 at 9:08
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Notice that the two roots of the quadratic equation $(ax^2+bx+c)$ are let's say roots $\alpha$ and $\beta$. The roots add together to give $\frac{-b}{a}$ and the roots multiply to give $\frac{c}{a}$

Yu would still have to use the facters of $a$ and $c$ in your quadratic equations to simplify.

Using example: $2x^2-7x+3$

Now addition of roots equal $\frac{7}{2}$.

Now you can fill in the blanks in the form $(2x-d)(x-e)$ with $1$ and $3$ since they are factors of 3 in this case since $c=3$. Note that 1 and 2 are factrs of 2 in this case $a=2$.

Now $\frac{d}{2}+\frac{e}{1}=\frac{7}{2}$ (Using the rule of roots addition)

$\frac{d+2e}{2}=\frac{7}{2}$

$d+2e=7$, using numbers $1$ and $3$

Now you can see clearly that $e=3$ and $d=1$ and now you have the simplified form $(2x-1)(x-3)$

Hope this helped.

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  • $\begingroup$ I'm not sure about encouraging this. Students in my experience struggle recognising the difference between expressions and equations and so there is potential for confusion. $\endgroup$ – Karl Mar 6 '16 at 11:32
  • $\begingroup$ Its just one of my methods. $\endgroup$ – Itakura Mar 8 '16 at 4:47
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What about taking out the factor multiplying the term $x^2$, then you need to consider something of the form $x^2 + Bx+ C$, then write this as $(x+b)^2 + c$.

When written like this you can almost read off the values for $b, c$, namely $2bx = Bx$ giving $b = B/2$ and $b^2 + c = C$, giving $c = C -B^2/4$.

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  • $\begingroup$ This is a bit hard to do mentally for my examples, but it's a nice alternative way to do it on paper, so thanks! $\endgroup$ – Skeleton Bow Mar 6 '16 at 10:07

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