# Quadratic form to regular form

Let's say that I have an equation in quadratic form

$$x^2-3x+2$$

Is there a method in turning that into:

$$(x-1)(x-2)$$

And what is the name of the form above? I realize that this has probably been asked before, but I just don't know the name of the second form.

• The process is called factorisation, and in this particular case you could use the so-called 'cross method'. See the link below for a good explanation/visualisation. math.stackexchange.com/questions/685748/… – Jack Jun 2 '17 at 6:12

Note that $(x+a)(x+b) = x^2+(a+b)x+ab$. So, the coefficient of $x$ is the sum of the two numbers we want to find, and the constant term is their product. So when we see a quadratic polynomial like $x^2-3x+2$, we have to try and find two numbers which add up to $-3$ and multiply out to $2$. The numbers we get are $-1$ and $-2$, and so the factorization of $x^2-3x+2$ (that's what it's called) is $(x-1)(x-2)$.

As another example, try factorizing the following: $x^2-x-6$. That is, find two numbers which multiply to $-6$ and add up to $-1$. The answer is below.

$(x-2)(x+3)$

Factorization is harder when the leading coefficient (that is, the coefficient of $x^2$) is not $1$, as the above method doesn't work. Consider, for example, the polynomial $3x^2-7x+2$. Here's a fun method for factoring these. First, make sure you factor out any constants that can be factored out (e.g. if given the polynomial $4x^2-10x+4$, you would want to factor out the $2$ to obtain $2(2x^2-5x+2)$, and then perform the following process on the non-constant factor).

Using the example of $3x^2-7x+2$, take the leading coefficient out and multiply it into the constant term to give us $x^2-7x+6$. Now use the above method to factor this polynomial as $(x-1)(x-6)$. Remember that $3$ we took out of the leading term? Put it as the denominator of the constant terms to give us $\left( x-\frac 1 3\right)\left(x - \frac 6 3\right)$. Now, reduce the fractions to get $\left( x-\frac 1 3\right)\left(x-\frac 2 1\right)$. Finally, put whatever is in the denominators next to the $x$ term in each factor, giving us the final answer of $(3x-1)(x-2)$.

The procedure is called factorization.

We need to expand a product with unknown roots $r_1,r_2$ to match our polynomial

$$(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_{1}r_{2}=x^2-3x+2$$

and identify coefficients of the same order

$$r_1+r_2=3$$ $$r_{1}r_{2}=2$$

Now consider integer factors of $2$, which are $-2,-1,1$ and $2$. We need to choose two of them such that the product is $2$ and the sum is $3$. Since the product has to be positive the roots $r_1$ and $r_2$ must have the same sign. Since the sum has to be positive, at least the one with larger absolute value has to be positive too. As a conclusion, both are positive. We may write $r_1=1,r_2=2$ or $r_1=2,r_2=1$.

Finally, let us check the result. $$(x-1)(x-2)=x^2-1·x-2·x+(-1)(-2)=x^2-3x+2$$