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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 !

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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.

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

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  • $\begingroup$ As you supposed, i'm not good with complex numbers :) ! $\endgroup$ – ganzo db Dec 18 '18 at 16:54
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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 $$2(x-3)\Bigl(x-\frac12\Bigr)=(x-3)(2x-1).$$

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

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

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