# Method to factor an expression

As the title says, i want to factorize an expression, but i don't have any clue how to proceed.

Here is the expression :

$$2x² -7x +3$$

And here is the factorized form :

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

My question is, which method or rule to use to go from first to second ?

please note that I am a beginner, and the only question that i found which is closer to mine is this post.

Thank you for your help !

Rewrite the expression into the form:

$$2x^2-6x-x+3$$ ,

then group the first two terms together and the last two terms together:

$$(2x^2-6x)-(x-3)=2x(x-3)-(x-3)$$ ,

next extract the common factor:

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

and you are done.

• I like the simplicity of your method, thank's :) – ganzo db Dec 18 '18 at 16:41
• @ganzodb You are welcome – Matko Dec 18 '18 at 18:24

If it concerns quadratic polynomial $$ax^2+bx+c$$ then start with calculating discriminant: $$D:=b^2-4ac$$

If $$D$$ is negative then give up (unless you are familiar with complex numbers already).

If $$D$$ is nonnegative then: $$ax^2+bx+c=a(x-x_1)(x-x_2)$$ where $$x_1=\frac{-b+\sqrt D}{2a}$$ and $$x_2=\frac{-b-\sqrt D}{2a}$$.

Especially if $$D$$ is a perfect square (as in your case, where $$D=25$$) then there is reason to cheer.

• As you supposed, i'm not good with complex numbers :) ! – ganzo db Dec 18 '18 at 16:54

It relies on the following property of quadratic polynomials:

If the quadratic polynomial $$\;ax^2+bx+c\;(a\ne 0)$$ has roots $$\xi_0$$ and $$\xi_1$$ (real or complex, distinct ot not), it can factored as $$ax^2+bx+c=a(x-\xi_0)(x-\xi_1).$$

This property is a consequence of a more general property of polynomials (of any degree) and the ring of polynomials over a field being a euclidean domain:

If a polynomial $$p(x)$$ has root $$\xi$$, it is divisible by $$x-\xi$$.

In the present case, the discriminat of $$2x² -7x +3$$ is $$\;\Delta=49-4\cdot 2\cdot 3=25$$, so its roots are $$\;\frac{7\pm 5}4==\bigl\{3,\frac 12\bigr\}$$, and the factorisation is $$2(x-3)\Bigl(x-\frac12\Bigr)=(x-3)(2x-1).$$

• i wish i could understand the last one ( \frac{7\pm 5}4 ). – ganzo db Dec 18 '18 at 16:57
• @ganzodb: It's just the formula $\frac{-b\pm\sqrt\Delta}{2a}.$. – Bernard Dec 18 '18 at 18:22

In general ,if $$x_1$$ and $$x_2$$ are roots of $$\underbrace{a}_{\neq 0}x^2+bx+c=0$$ then $$ax^2+bx+c=k(x-x_1)(x-x_2)=k[x^2-(x_1+x_2)x+x_1x_2]$$ comparing the coefficients, $$a=k,b=-k(x_1+x_2),c=kx_1x_2$$ Consequently $$\text{sum of the roots}=-\frac{b}{a}\;\;\&\;\;\text{product of the roots}=\frac{c}{a}$$

So your case, $$x_1+x_2= \frac{7}{2}$$ and $$x_1x_2=\frac{3}{2}$$

Now solve these to get $$x_1$$ and $$x_2$$ to finish your conclusion

To find $$x_1$$ and $$x_2$$, $$2x^2-7x+3=0$$ implies $$x^2-\frac{7}{2}x+\frac{3}{2}=0$$ which means $$x^2-2\left(\frac{7}{4}\right)x=-\frac{3}{2}$$ which is same as $$x^2-2\left(\frac{7}{4}\right)x+\frac{49}{16}=-\frac{3}{2}+\frac{49}{16}=\frac{25}{16}$$ so $$\left(x-\frac{7}{4}\right)^2=\frac{25}{16}$$ and so $$x-\frac{7}{4}=\pm \sqrt{\frac{25}{16}}=\pm \frac{5}{4}$$ so $$x=\frac{7}{4} \pm \frac{5}{4}=\frac{7\pm 5}{4}$$

• Thank you for the time spent in the explanation, i'm still unfamiliar with some mathematical concepts. – ganzo db Dec 18 '18 at 16:53