Factorizing quadratics mentally

Question

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

Answer 1

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 $(3x + 4)(x + 6)$.

Answer 2

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

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

Answer 3

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.

Answer 4

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.

Answer 5

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 $c = C -B^2/4$.