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I know its:

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

But how do you come to that conclusion?

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    $\begingroup$ a general method is solve the equation 2x^2 - x - 3 = 0. There's a theory that says if x1, x2, ... x(n) which makes the polynomial P = 0 then P = a(x - x1)(x - x2)...(x - x(n)) $\endgroup$ Aug 20, 2013 at 7:52

6 Answers 6

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Here's a trick: multiply the first and last coefficients and see which factor pairs add to the middle coefficient.

In this case, $2\times (-3) = -6$. The factor pairs of $-6$ are $1,-6$, $2,-3$, $-1,6$, and $-2,3$. Since $2+(-3) = -1$, we can break up the middle coefficient and factor by grouping: \begin{align*} 2x^2 - x - 3 &= 2x^2 + (2-3)x - 3 \\ &= 2x^2 + 2x - 3x - 3 \\ &= 2x(x + 1) - 3(x + 1) \\ &= (2x - 3)(x+1). \end{align*}


The rule comes from the distributive rule: $(ax + b)(cx + d) = (ac)x^2 + (ad + bc)x + bd$. The first coefficient is $ac$ and the last is $bd$, they multiply together to $abcd$, and the middle coefficient is $ad + bc$, the sum of a factor pair of $abcd$.


Another way to do it is to use the quadratic formula to find the roots of the polynomial: $a = 2$, $b=-1$, and $c = -3$, so $$\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm\sqrt{1 - -24}}{4} = \frac{1\pm 5}{4} = \frac{3}{2}, -1,$$ which gives a factorization into the monic polynomials $x+1$ and $2x - 3$.

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Note that $2-3=-1,$ so that $$\begin{align}2x^2-x-3 &= 2x^2+(2-3)x-3\\ &= 2x^2+2x-3x-3\\ &= 2x(x+1)-3(x+1)\\ &= (x+1)(2x-3).\end{align}$$

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    $\begingroup$ Sweet, simple, and to the point. $\endgroup$
    – Edwin Buck
    Aug 20, 2013 at 5:30
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You can use several methods to solve the equation. Firstly you can use the quadratic formula

$$\frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$$

or you can use decomposition.

Let's try decomposition first. Look for two numbers that multiple to $-6$ and add to $-1$. These numbers would be $+2$ and $-3$. So, we would place those for $-x$ and have the following:

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

Now here you can factor out $x$ from the first two terms.

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

Notice the coefficient on $2x-3$ is $1$. So this means that our equation is $x+1$ and $2x-3$. Therefore, giving us the solution

$$ (2x-3) = 0 \, \text{or} \, (x+1) = 0$$

Now, if you want to use the quadratic formula then you can do this. $a = 2$, $b = -1$ and $c = -3$. So we will have:

$$\frac{6 \pm\sqrt{(-1)^2 - 4(2)(-3)}}{2(2)}$$

When you apply this formula twice, you will get the roots which are $x = \frac{3}{2}$ and $x = -1$. If you have any questions, let me know. I will try to clarify as much as possible.

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find the roots(if any). this is the most general answer to these problems.

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I have some more helpful and straight forward reasoning (Here is the step-by-step reasoning):

first we rewrite the term: Suppose that A=$2x^2-x-3$ then $$A=\frac{2(2x^2-x-3)}{2}$$ then simplify the last term to conclude a most-known identity: $$\frac{2(2x^2-x-3)}{2}=\frac{(2x)^2-(2x)-6}{2}$$ let $y=2x;$ then we can rewrite again: $$A=\frac{(y)^2-(y)-6}{2}$$ You can see that the numerator is a form of a most-known identity; then we have $y^2-y-6=(y+2)(y-3)$ and by rewriting again: $$A=\frac{((2x)+2)((2x)-3)}{2}=\frac{2((x)+1)((2x)-3)}{2}=\frac{2(x+1)(2x-3)}{2}$$ finally we conclude the desired result: $$A=(x+1)(2x-3)$$

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    $\begingroup$ how come the part $$\frac{2(2x-x-3)}{2}=\frac{(2x)^2-(2x)-6}{2}$$. isnt it should be (4x-2x-6)/2. I think you had right concept but some misprinting is there $\endgroup$
    – monalisa
    Aug 20, 2013 at 3:34
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    $\begingroup$ Thank you for your suggestions. I edited my answer. Writing math formulae is somewhat hard to me. $\endgroup$
    – RSh
    Aug 20, 2013 at 3:58
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    $\begingroup$ @monalisa: Simply, you could edit my typos(also you: Cameron!) :) $\endgroup$
    – RSh
    Aug 20, 2013 at 4:02
  • $\begingroup$ @I'mtoo.... Oh yeah...you are right... we could edit too :) sorry $\endgroup$
    – monalisa
    Aug 20, 2013 at 5:16
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Solution 1:

We usually do some tests by replacing x=0; x=1 and x=-1 in the equation (usual roots, you can take it as habit) here we can notice that -1 is a solution for the equation (root) so :

$$2x^2 - x - 3 =(x+1)\times 2 \times Q$$ and $$Q=ax+b;$$

Variables a and b can be extracted easily for this example (sometimes we can even do an euclidean division)

Result is :

$$ 2x^2 - x - 3 =(x+1)\times2\times(x-3)=2(x+1)(x-3)$$

Solution 2:

The equation is 2nd degree one so either it has 1 double root, 2 roots or no solution in R

(1) $$2x^2 - x - 3= ax^2 +bx+c=0$$ so a=2, b=-1 and c=-3

We calculate $$Delta=b^2-(4ac)=1-(4\times2\times(-3))=25 >0$$

We have two roots for equation 1 so we can write : $$2x^2 - x - 3 = a(x-x1)(x-x2)$$

$$x1=\frac{-b - \sqrt{Delta}}{2a} = \frac{(1-5)}{4}=-1;$$ $$x2=\frac{-b + \sqrt{Delta}}{2a} = \frac{(1+5)}{4}=3;$$

So

$$2x^2 - x - 3 = a(x-x1)(x-x2) = 2(x+1)(x-3)$$

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