1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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/… $\endgroup$ – Jack Jun 2 '17 at 6:12
0
$\begingroup$

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 $(3x-1)(x-2)$.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.